One of my favorite things I studied were the bizzaro function. Such as those that are continuous everywhere but differential nowhere. The real fun isn’t the functions themselves. But to go back in history and read how angry they made (some) mathematicians is fascinating.
I would argue that the Weierstrass function is not as bizarre as it sounds. When you look at what continuity really is, you know that f : U -> R is continuous at p iff for all real ε > 0, there exists some δ > 0 (which may depend on p and ε), such that for all x in U, if |x - p| < δ, then |f(x) - f(p)| < ε. Meanwhile, f is differentiable at p if lim (f(x) - f(p))/(x - p) (x -> p) exists, which translates to saying there exists some real k such that for all real ε > 0, there exists some δ > 0 (which may depend on p and ε), such that for all x in U, if |x - p| < δ, then |(f(x) - f(p))/(x - p) - k| < ε, but the latter is equivalent to |f(x) - f(p) - k·(x - p)| < ε·|x - p|, which is to say (|k| + ε)·|x - p| < |f(x) - f(p)| < (|k| + ε)·|x - p|. This clearly looks like it should be a stronger condition than the condition for continuity, which is simply that |f(x) - f(p)| < ε. Now, of course, that does not prove it is stronger, but it should not be unexpected at all if it were for some functions at some p. And it being the case for some function at all p is not such a stretch from there. So, I would argue that the existence of such functions conceptually is actually sort of intuitive. The only reason people call it bizarre is because people think of continuity and differentiability in terms of graphs, rather than in terms of what they actually are.
It's bad enough that there is always a real number between any two rationals, but it's far worse that there's always a rational number between any two reals.
@@egberthuizinga4408 Agreed. It's easy to have a mental model of the rationals and irrationals zipping together alternately - but that's so misleading. The uncountability of the irrationals is just so big we have the logical tools for it but not the mental models or analogies.
@@egberthuizinga4408 I was thinking of the techniques that Cantor used in defining a hierachy of infinities or the logic within this video. Careful reasoning from axioms, construction of examples or counter examples that sort of thing. So logically we prove the reals are uncountable (a bigger infinity) but at the same time any 2 rationals have an irrational between them and vica versa - but at this point I have conclusions and no mental model of where those "extra" irrationals are hiding.
@@andrewharrison8436 The extra irrational is hiding in the infinities. There is a countably infinite number of rationals between 0 and 1. There is also countably infinite number of rational numbers between any two rational interval. So in conclusion there is infinity in between any pair of rational number. The number line is a bad model of understanding this thing because in reality real numbers are define as an infinite set of rationals. By definition it follows that for every real number there is a rational below it and a rational above it otherwise two real numbers without a unique rational between them is define as equivalent.
Lol this is bizarre! If you asked me whether linearity implied a continuous function, I would have said yes and had been 100% confident and 100% wrong.
Right?! You wouldn't have batted an eye! The tricky bit is your intuition is correct for "R-linear" functions, it is a standard proof in any intro real analysis class. But not for "Q-linear".
@@DrTrefor This mind-blowing! I remember proving “linearity implies continuity” using the epsilon-delta definition, could you shed some light on this? Or is there any implicit assumptions I missed?
@@mingmiao364 Yes! You would have used that the coefficients are real numbers. That whole epsilon-delta proof falls apart when you restrict to only rationals.
@@DrTrefor Thanks for pointing it out! Is is true that such example motivates the definition for topological vector spaces? As linearity and continuity really are seperate concepts
@@mingmiao364 The important hypothesis to say linearity implies continuity is finite dimension. Even with R Vectorial spaces, if you have an infinite basis, some function can be linear but not continous.
Some follow-ups to common questions! 1) The set S is certainly infinite, in fact it is uncountable. Nevertheless, the definition of a basis being used here (and guaranteed from axiom of choice) is we can write any x as a FINITE linear combination of things from S. So each choice is finite even if the set S is infinite. This means that f(x) always does give a rational number not an infinite series. 2) The big thing we relaxed from the "usual" notion of linearity is that we are only using rational coefficients, what I called "Q-Linear". For an "R-linear" function it is a standard theorem in any analysis course to prove that this does imply continuity. You might have initially guessed that since Q is dense in R that only Q-linear was needed, but no! 3) A number of people have suggested alternatives along the lines of f(x)=x for x irrational and f(x)=0 for x rational. It's tempting, but this actually fails the additivity property of linearity. Consider for instance x=1-pi and y=pi. So f(x+y)=0 which is not equal to f(x)+f(y=)=1-pi +pi=1.
W.r.t (1) can you link to a video or proof that explains why every x in reals is a *finite* linear combination of q in rationals? My intuition would have expected each x in irrationals to be only the limit of a series, i.e. an infinite linear combination.
@@drewtmacha34 Your intuition is correct, but that is not what the bit at 10:40 is saying. It is claiming that every real number x is a finite Q-linear combination of stuff from a set S. The (uncountable) set S may consist of rationals, irrationals, etc. . The finiteness of the linear combinations comes from the definition of a Hamel basis (as opposed to Schauder basis, which talks about infinite linear combis)
You got were my mind was thinking with Number 1, and then I was thinking is there any x in R, such that there was an infinite number of basis used to create x, and thus have an infinite summation that may be divergent. I quickly got that S was uncountable, if S was countable, then Q x S is countable, but we defined Q x S to cover R, which is uncountable, therefore either Q or S is uncountable, and we already know Q is countable.
@@strikeemblem2886 To be fair, the basis exists only if we accept the Axiom of Choice. We cannot ever fully construct a complete set, but any finite independent set will extend if AC is accepted. But once we choose a starting sequence of basis elements and locked them in, we can't t know what else the AC picked for us. The end result is not very concrete, but the AC itself sounds believable.
When I saw the thumbnail, It did seem to me you would need to restrict c to the rationals for a discontinuous function to be possible, and that you would use a basis for R as a vector space over Q. Interestingly, when you restrict c to be rational, the condition f(cx)=cf(x) becomes redundant; i.e. f(x+y)=f(x)+f(y)⇒f(cx)=cf(x) for c∈Q. For c=0: Setting x=y=0, we get f(0)=f(0)+f(0), so f(0)=0, and f(0x)=f(0)=0=0f(x). For c∈Z⁺, a simple inductive argument does the trick. The base case c=1 is trivial. For c=-1: Seting y=-x, we get f(0)=f(x)+f(-x). Since f(0)=0, we get f(x)=-f(-x). For c∈Z⁻, c≤-2, we have f(cx)=f(-|c|x)=-f(|c|x) (using the result for c=-1) =-|c|f(x) (using the result for c∈Z⁺) =cf(x). So the result is true for c∈Z. If c∈Q, we can write c=p/q where p∈Z and q∈Z⁺. So pf(x)=f(px)=f(qcx)=qf(cx) and f(cx)=p/q f(x) (as q≠0) =cf(x). So the result is true for c∈Q. Done.
Oh wow so the axiom of choice is the math equivalent of the "tree falls in the woods" question. If there's no way for us to explicitly construct this 'linear' but discontinuous function, does it still exist? Of course it does!
I work on discrete dynamical models in biology (esp. gene regulatory networks), and Boolean linear functions have been getting some attention lately. The classic example is the "exclusive or" function xor(x,y)=x+y mod 2. This function is linear (mod 2), but is not monotonic! Note that because ordered fields can't be finite, the definition of monotonic is somewhat different* in this context than in, e.g., the reals, but basically the observation is that "increasing" x from 0 to 1 can either "increase" xor(x,y) form 0 to 1 (if y=0) or "decrease" it from 1 to 0 (if y=1). Boolean linear functions are, according to multiple formal definitions, the most "sensitive" Boolean functions. Their non-monotonic nature, however, makes them unlikely candidates for describing how the presence or absence of transcription factors impacts gene expression. This has some interesting implications for a bunch of concepts related to the robustness of genetic circuitry, such as canalization, damage spread, and criticality. * For completeness, a Boolean function f is monotonic if for all input variables x, across all input configurations, changing x from 0 to 1 never changes f from 1 to 0 OR changing x never changes f from 0 to 1. (Some authors, e.g., on Wikipedia, use a stricter definition that leaves off the part after the "OR", but I don't like that because it's more analogous to the definition of non-decreasing functions.)
2:44 , I'm currently taking real analysis now and it still baffles me that numbers behave so nicely in this way. So much that one of the definitions of a real number is similar to this (ie, letting S be all the non-positive powers of 10, or 2, or ...whatever base you want).
@@zachpence8283 Thank you for sharing! Number theory and mathematics intrigue me as theoretical ideas, but I am struggling to choose a major as a high school student applying to university (looking for majors more like applied math/engineering fields). If I may, what do you plan to do after you get your degree? Thank you for your time.
Are you missing a condition on S? By your definition why can't you just choose S=R? Then there's only one coefficient (1) and your function f is just 1 everywhere Edit: nevermind, did not catch that the decomposition must be *unique*
First I thought: uh? Then I remembered: linear functions over finite dimensional (normed) R or C vector spaces are continuous, so one has to go into infinite dimension or change the underlying number field. So your answer is correct. The thing is: you play with the ambiguity of not naming the underlying number field.
@@DrTreforYou can also take a more general definition of continuity and then say that your function returns to being continuous - just choose the right topology and everything will work out fine...
@@eofirdavid Does it? If you take the definition based on neighbourhoods it's not clear to me why the function would be continuous. Image you construct your set such that both 1 and pi are in the basis, now all rational numbers just map to themselves, but 10pi maps to 10, so all neighbourhoods around 10pi will always map to a set containing at least 10pi and 10, so you can never find one that maps to a (9;11) despite that being a neighbourhood of f(10pi)
@@chalkchalkson5639 You can define a metric on R (in fact on any set), with d(x,y)=1 if x!=y and d(x,x)=0 (this induces the discrete topology). Using that metric every function from R to any topologic space is trivially continuous.
@@chalkchalkson5639 Whenever you have a function f:A->B (and in this case both A and B were the set of real numbers), you can always make it continuous by either giving B the trivial topology (the only open sets are the empty sets and B) or giving A the discrete topology (all the sets are open. This topology also comes from the discrete metric). You can then try to create some more interesting topologies by looking for stronger topologies on B or weaker topologies on A. Not sure how many of them will be "interesting" but there sure are many of them.
2 missleading aspects: 1: Not saying that S has to be infinite. 2: Drawing f as a line, like we are used from linear functions. Let 2 basevectors be e1=1 and e2=pi. Then f(3)=3, f(pi)=1, f(4)=4. It is mathematically linear, but looks not like a line.
The second property of linearity you left up to us can be found through factoring by grouping , no? Awesome video, I’m taking college classes but still a high school junior and find your videos quite impressive and intuitive. Thank you!
I sort of imagine constructing S as follows: Start with S_0 = Ø, repeat ad infinitum the following procedure: pick any nonzero real number and check if it can be written as a finite linear combination (with rational coefficients) of elements that are already in S_n thus far. If not, then add it to S_n, to make S_n+1. This way we construct our basis S (in principle). Although f is linear as proved, it may be a bit misleading to depict it as a line. My intuition tells me it comes arbitrarily close to any point on the plane. Edit to backup my intuition a bit: Suppose a and b are different elements of S. Then, for any point (x,y) on the R2- plane, we can choose q1 in Q arbitrarily close to (x-by)/(a-b) and q2 in Q arbitrarily close to (-x+ay)/(a-b). Then q1 a + q2 b is arbitrarily close to x, whereas f(q1 a + q2 b) = q1+q2 is arbitrarily close to y.
And guess the missing part is: pick your favorite well-ordering of real numbers, and construct the basis by transfinite recursion in the order of well-ordering. The only part left is to well-order the reals, and while the existence of well-ordering is a consequence of the axiom of choice, no formula can be proven in ZFC to actually define one (in the sense that 1) there is exactly one set satisfying the formula, and 2) that set is a well-ordering relation on reals).
And if you don't accept the axiom of choice, it can be shown that the existence of such a function cannot be proved. This is even crazier to me that you can construct a setting in which you can prove that every Q-linear real function is continuous!
Did you claim that you did not accept the axiom of choice? I think you meant you accept that the axiom of choice is false. It’s different, because if you don’t accept the axiom of choice is true then you can’t imply the axiom of choice is false. To sum it up, there are 3 possibilities: we can claim axiom of choice is true, we can claim axiom of choice is false, we can neither claim it’s true neither claim it’s false.
@@becomepostal I do not claim anything about me. I personally think it's very interesting to consider theories with the axiom of choice as well as theories where it is false and to compare them. To answer your question, I chose not to get into details but you're absolutely right that working without the axiom of choice doesn't mean it's false. What I meant is that there exists a coherent theory containing ZF in which you can prove that every Q-linear real function is continuous (and therefore it is the multiplication by a real number). Such a theory must contradict the axiom of choice. This is due to Cohen and it uses the method of forcing. To clarify, I'm not saying that refusing the axiom of choice would imply anything. However even if you consider a theory in which the axiom of choice is false, you could very well have non continuous Q-linear real functions. It is not sufficient to say that the axiom of choice is false to get to the result I stated. In other words, what Cohen proved is that "every Q-linear real function is continuous" could itself be taken as an axiom instead of the axiom of choice and it would be exactly as coherent. In a way, both possibilities coexist (if either one actually exists, which is another question...)!
I think a much easier to understand example is for example the differential operator from and to the set of smooth functions on an interval [a,b] with the supremum norm, D_x : C_inf[a,b]-->C_inf[a,b]. It is obviously linear and well defined. But you can find a sequence like f_n(x)=sin(n²x)/n, with D(f_n)(x) = cos(n²x)*n, and ||f_n|| --> 0, but ||D(f_n)|| --> inf.
In fact, the "additive linearity" implies the rational multiplication linearity, because you can reproduce all the process, i.e: f(2x)=f(x+x)= f(x)+f(x)= 2f(x)... f(x+(-x))=f(x)+f(-x)=f(0)=0 and therefore f(-x)=(-1)f(x)...f(x)=f(x/3+x/3+x/3)=3f(x/3)... This extension of linearity from addition to multiplication is automatic for the rational field of the "big field".
Nice video. I'm glad you took the time out to describe the Axiom of Choice, and how this is one of the many non-constructive examples that falls out of assuming it (AoC) to be true. I might rank the Cantor function above this one, simply because it's constructive and is the first time I bumped into a function that was continuous but couldn't be drawn with an "unbroken stroke of pencil on paper" (the way continuity is usually described, pre-calculus). Its differentiablility almost everywhere with derivative zero, despite the ascent from 0 to 1 was equally baffling. That was not the first time my measure theory course broke my brain for a moment or three.
Maybe I misunderstood something but: if your function was R-linear, i.e. f(cx)=cf(x) for all real x and c, then f would be onto. This is a contradiction since f(R)=Q. What do I miss? If I'm not wrong what you say is that R is a Q-vector space, and f is a Q-linear map. The map you described is not R-linear. I wonder if there is a discontinuous additive real function f, such that f(cx)=cf(x) for all real x,c
0:30 - 0:38 There is something that needs to be cleared up. In general, when non-mathematicians talk about linear functions, they are imagining an equation of the form y = m·x + b (which partially stems from the misconception that a function is a kind of equation, which is not true at all). Those who are more careful will instead say that f is a linear function if it satisfies f(x) = a·x + b everywhere for some a, b, and the inspiration in calling this linear is that if you graph f on a Cartesian coordinate system, the set of points is a line. However, this is still incorrect. Such a function f is not linear unless b = 0. This is confusing, because the functions that we use to graph lines on a plane are not themselves called linear functions, they are called affine functions. If b = 0, then the affine function is also linear. But what is the difference? Well, as stated in the video, the actual definition of a function being linear is that it satisfies the property in the video. To be more precise: it only makes sense to talk about linear functions in the context of vector spaces (or if you want to get more general, modules). If you have two vector spaces V, W over the same field scalar field F, then f : V -> W is called linear iff f(v0 + v1) = f(v0) + f(v1) for all v0, v1 in V, and f(α·v) = α·f(v) for all α in F and v in V. In this context, though, since we are dealing with calculus, linearity means that the vector space in question is R, the field of real numbers. In particular, we are taking R to be a field over itself. R acts as both the vector space and the scalar field, and it is a 1-dimensional vector space. Applying the definition of linearity, then, looks like this: f : R -> R iff f(A·x + y) = A·f(x) + f(y) = x·f(A) + f(y) for all real A, x, y. Understanding that linearity is about functions of vector spaces specifically (and not just arbitrary sets) is key to understand the results in the video. 2:30 - 3:12 By the way, the proof of the existence of S usually uses Zorn's lemma (though you can get it from weaker assumptions), and Zorn's lemma is actually equivalent to the axiom of choice. Also, S is not unique, generally speaking. Also, I want everyone to pay close attention to what is being said here. S is some set of real numbers, an uncountably infinite set, and if you take some finite number of these real numbers, scale each of them *by a rational number,* and add them, you get a real number x, and this is possible for *all real numbers* x. What this is really saying is that if you take the real numbers R to be a vector space over the scalar field Q, the field of rational numbers, then S is a basis for this vector space. Notice that this is different from how we usually handle R as a 1-dimensional vector space over itself. Here, R is a Beth(1)-dimensional vector space over Q. This is very important. 6:14 - 6:31 In other words: it is very important that R is being considered a vector space over Q, rather than over R itself. In fact, you could not define f otherwise, to begin with, but even if you could, it would not be linear. Here is the point: usually, when we talk about linear functions, we are talking about one vector space, and one vector space only: R as a vector space over itself. We talk about functions f : R -> R being linear, and because this vector space is 1-dimensional, it works out to also be continuous. However, when we switch vector spaces, say R with R' (this is just me abusing notation: the set of real numbers is the same set, but now it is being considered as a vector space over Q, so I am denoting it as R'), and we talk about linear functions g : R' -> R', since R' is Beth(1)-dimensional, and yet the scalar field, Q, only has cardinality Beth(0) = Aleph(0), it is impossible for it to be continuous. The more general lesson here is that if you have a vector space V over a field F and you consider linear functions h : V -> V in general, whether these functions are continuous or not depends largely on the properties of V and F and how they interact together. Continuity is guaranteed if V is finite dimensional, but not so if V is infinite dimensional. Intuitively speaking: If F is "too small" compared to V, the topologies may be "incompatible," in a certain sense, resulting in these linear functions being not continuous. And if you are familiar with some basic category theory, you can think of it this way: we are talking about endormorphisms in the category of topological vector spaces over a field F. There is a relationship between these endomorphisms and the endomorphisms in the category of *topological* vector spaces over a field F: there is a functor between these categories. If an endomorphism in the category of vector spaces is also a monomorphism (generalizing the idea of an injective function), then it will be an isomorphism, and thus the functor maps this endomophism to a homeomorphism. However, the endomorphism f that was "constructed" in the video (I am using the word "constructed" very loosely here, you guys should know what I mean) is not a monomorphism, clearly, since its range is Q, of cardinality smaller than R. So it also cannot be a homeomorphism in the category of topological vector spaces over F. In other words, it cannot be continuous. 7:30 - 8:15 For those who are curious what an exact statement of the axiom of choice looks like, it is like this: consider a set S, and consider all the sets X that are elements of S. We are gonna to index the sets of S with some other set I, and to index them, you have a surjective (onto) function I -> S. This indexing allows me to talk about each set X in S as being labeled by some i, so we can denote as X(i). Why do I care about this labeling? Because now I can state the axiom of choice: the axiom of choice is simply saying that there is some set T such that for all X(i), there is some x(i) in X(i) such that x(i) in T, and this is true for all S, and regardless of the indexing chosen. This is more or less how you make formal the intuitive idea explained in the video. 8:50 - 9:15 Without the axiom of choice, you can still prove all *finite dimensional* vector spaces have a basis. But remember: the vector space defined in this video is Beth(1)-dimensional (the dimension is uncountably infinite). To prove that infinite dimensional vector spaces always have a basis, you need something like Zorn's lemma, which is equivalent to the axiom of choice. 11:30 - 11:48 Once again, this comes down to the fact that Q is countable, but R is not. Q is dense in R *even though* it is countable and not uncountable. As a result, the order topology of Q is the discrete topology, but the order topology of R is not the discrete topology. These are "incompatible" in some sense, which is why you get this discontinuity.
Although axiom of choice is not constructive, I would like to know if my understanding is correct: is that S = {1, 2^(1/2), 2^(1/3), ..., 3^(1/2), 3^(1/3), ..., pi, e, ...} that includes all the "irrationality" basis and then f(x) is kind of "counting" the "irrationality". For example f(pi+1/3*e) = 1+1/3=4/3, f(sqrt(2) - 1/2*sqrt(3)) = 1-1/2=1/2. f(c) = c if c is rational. If that is the case it is easy to see it is only Q-linear but not R-linear: f(sqrt(2)*sqrt(2))=f(2)=2 but sqrt(2)*f(sqrt(2))=sqrt(2)*1=sqrt(2)
At 5:00… does not satisfying the intermediate value theorem for every interval mean it’s discontinuous? I’m really confused because it seems like it could still satisfy the delta epsilon definition of continuity Answer: yeah not satisfying IVT does mean it’s discontinuous, which I think means the delta epsilon definition isn’t satisfied but the internet says the explanation is complicated
Exactly! IVT says "if it is continuous, then it must hit all intermediate values". As it fails to do so, it is not continuous. We don't need to open up delta epsilon definition for this.
I understand that the axiom of choice implies that such set S exists, because R is a vector space, and Q is the field of scalars over that vector space. However, You assumed that this set S is finite, and I don't quite understand why that should be the case. And S must be finite, or alternatively all the possible series sum_of{q_i} must be convergent for f to be well-defined, and I don't see the reason why either of those must be true
I encountered a similar reasoning when a friend challenged me to write the identity function f(x)=x as the sum of two periodic functions. Find a Q-base of R, define g(x) as the projection of x on b_1, the first element of the base and h(x) as the projection of x on span(all the others elements of the base). Then you have naturally that f(x)=g(x)+h(x) by definition of the projection. Moreover, g(x) is periodic of period b_1 while h(x) is periodic of any period b_i where i=/=1. That’s crazy.
I have one confusion, why is f not continuous if its range is cut by irrational numbers? By definition of continuous, (1) f is defined on R, (2) f(x) -> f(c) when x -> c for every c on R. I can choose rational numbers q = f(x) that are infinitely close to f(c) when x is close to c. So the second condition is also satisfied.
Continuity means more than just that it gets close in the limit, it actually has to have a value c so f(c) equals any number, and that won't work for irrationals.
If we’re just looking for a Q-linear function that isn’t continuous, can’t we make a much simpler function? Just say f(x)=x for rational numbers and f(x)=0 for irrationals. Then f is Q-linear since for all rational c, f(cx)=cx=cf(x) if x is rational and f(cx)=0=cf(x) of x is irrational. Furthermore f isn’t continuous for the same reason mentioned in the video: it can’t output irrational numbers.
I don’t understand the proof for discontinuity. Why does the fact that f only takes values in Q implies discontinuity? Counter example: suppose f is the constant function that is always equal to 1. This function is continuous and only takes value in Q.
I was using the intermediate value theorem which says for continuous functions it attains all values between any two given values. But your example only attains one value so I can’t reject continuity via contradicting IVT
the main propose of this video is to show you a change of paradigm... the linearity or continuity is only an abstract way from our language and how this lead in your maths... so its okay to make clear the idea .... at all not everyone gives importance to this because it seems not relevant in the diary life... but is thoutght involving itself in a infinite cycle jeje
Not only is there at least one irrational between any two rationals, there are infinite rationals and irrationals between any two rationals. Any interval of nonzero length on the real line contains "a whole real line" of its own, because there are bijections that can be defined between the open interval we get by excluding the endpoints, no matter how small it is, and the entire real line.
I don't know that much about core mathematics so maybe the 2 conditions are the definition of linearity but I would expect the second one (f(cx) = cf(x)) to use c real (even irrational) because for rational c's it easily follows from f(x+y) = f(x) + f(y).
That absolutely is the ‘normal’ expectation. In almost all cases, it is going to work out the same so we don’t talk about it much, but this example really shows we do have to be careful about whether the c is real or rational.
how can we know that f(x) converges ? since S is infinite we are adding up an infinite amount of terms, it is not clear at all if f(x) should converge or not.
While S is indeed uncountably infinite, the definition of basis used here (and guaranteed by AOC) is that any x an be written as a FINITE linear combination
@@DrTrefor It has cardinality of the continuum, right? (Assuming axiom of choice.) * Consider the set of all finite subsets of the set S. This set has the same cardinality as S. * For each set consider the set of all sequences of rational numbers with the same number of elements. Of these there are countably many. * Every such sequence defines a real number, and by assumption every real number can be obtained as a linear combination of finitely many elements of S, with rational coefficients. * It follows: The set of all linear combinations is a union of S-many countable sets, which has the same cardinality as S. * Therefore: S has cardinality of the continuum. (I don't think that the argument would work without axiom of choice, or at least some weaker consequence thereof - even assuming that the basis exists. Without choice, I can't even be sure that a union of countably many countable sets is countable.)
The big "cheat" is that the function is only linear over the rational field, which is only mentioned in the middle of the video. The existence of such a function is not surprising at all. There are very few rational numbers compared to reals. So linearity over the rationals is a quite weak condition on a real function.
In contrast, rationals are dense in R, so one might have guessed that this was sufficient. Indeed, we can’t construct such a function so not so easy to do, need non-constructive proofs via axiom of choice to make it work
@@DrTrefor Density is a topological notion. We did not make any assumptions that takes into consideration the topology or the metric of the reals (e.g., that the function be bounded, or upper semi-continuous, or open, or whatever). So we are simply looking for a linear operator on the continuum dimensional Q-vector space R. It is no surprise that there are a lot of them. Discontinuity of most of these functions is again not surprising for the same reason: the vector space structure is oblivious to the proximity of real numbers. Basically, in order to make this construction sound interesting, you have to bring up concepts that were never mentioned in the problem, and point out that those concepts are violated. Nobody expected them to hold to begin with. It's like saying "I'm, gonna mix water and salt, and believe it or not, the mixture is not gonna sing Beethoven's fifth symphony."
A common knowledge says that any linear functional on finite dimensional vector space is continuous. In infinite dimensions it is a very common thing to have discontinuous linear functional. This function you provided is not R linear (because if you multiply x by irrational number you may have a whole new set of coefficients q), but it is Q linear, and R is infinite dimensional space over Q, so there is nothing strange in your result:)
0:30 By that definition, f(x)=x+1 isn't linear, because c*f(x) =/= f(c*x). Example: c=2, x=0 f(x)=x+1 2*(0+1)=(2*0)+1 2=/=1 I suppose it's only linear additively, not linear multiplicitatively.
How do you know the basis (set S) isnt just the whole R. Such that the coeficients are always 0 except from 1, which equals 1. Hence the function would be just the constant function 1 which is continous?
the moment you mentioned a vector space over Q, I stopped the video, typed this. if you're allowed to use a nonstandard VP, I would also be allowed to make up some topology in the codomain for any function y=ax such that it's not linear. for example, discrete topology
The functions "f(x)= p(x)×n^kx, f(x)= p(x)×n^|kx| and f(x)= p(x)×n^-kx" with n as negative real number, k a non negative and nor zero real number and p(x) a polynomial of x fascinates me. The functions can be linear or curved and all real set are included but the numbers alternate between positive. But this occurs in Rational numbers. But I thought a little about irrational numbers in this case and those are perfectly possible to apply but we don't know if f(x) for a irrational x would be positive or negative. That's a in indetermination? The domain must be just the rationals? If irrational is really included the function is a non continuous but delimited in two curves well described I don't speak English so I'm sorry. Hahaha.
If 1_Q is the characteristic function over the rationals (1 if x in Q, 0 if not), then f(x) = x * 1_Q(x) is also Q-linear and not continuous over R. EDIT: it is continuous at 0.
@@estebanmarco8755 ah, my bad, I missed that part It still fails though because of the reasons person before me said. If you have 2 irrational numbers x and y that add up to a rational number q different than 0, you will have f(x)+f(y)=0+0=0 but f(x+y) = f(q) = q = x+y ≠ 0
This seems like a bit of slight of hand. You've made a linear function that maps real numbers to rational numbers. The rational numbers by definition are not continuous, so of course your function is not continuous. Also, the way you define x seems to me a type of field extension with regard to Q? I'm still in my Abstract Algebra course, but isn't each of the s_a going to either be an element within Q are an irrational element that acts as a field extension?
I think you can think of it as a field extension, but this is not the canonical structure to consider (i.e as a vector space over Q). The video is a bit slight of handy, but only a bit. The wierd property is that axiom of choice gives us this funky basis. The function itself is just repackaging that funky basis into a "clickbait" kind of way, but the existence of the funky basis is the real meat of the problem.
Choiciness feels like diet "False". In a similar way to how False implies anything in logic, choice implies choiciness. There's like a bajillion unintuitive things as a consequence of choice, I'm not surprised that there are other things that are unintuitive and choicy, if you assume choice. :P
OK. Would you prefer it that a set can be split into more subsets - disjoint and non-empty - than it has elements? Let's visualize this: imagine Cantor's Hotel (a competitor of Hilbert's Hotel), where rooms are indexed by real numbers. Of course, the hotel has infinitely many floors, and infinitely many rooms on each floor: on each floor there are numbers which differ from each other by a rational number. In absence of axiom of choice, it's consistent that the hotel has more floors than rooms. ("Say what? Of course it has no more floors than rooms." Okay, what does that mean? "It means that the set of floors can be mapped one-to-one with a subset of rooms." And can you do that? "Sure, just pick a single room from each floor..." Oops. That's precisely what axiom of choice says.) On the other hand, it can be proven even without axiom of choice that the hotel has no less floors than rooms. And that's not going to Feferman-Levy model of set theory, where real numbers (obviously an uncountable set) are a union of countably many countable sets. (Really, don't go there, because for obvious reasons in this model Lebesgue measure is not countably additive, with a number of other unfortunate consequences for calculus.)
Thank you, i was actually searching an example like this last year. I was wondering if f(x+y)=f(x)+f(y) could imply linearity and I easily proved that Q-linearity is indeed implied. Then I showed that if f is also continuos, than it's linear. I did guess that some kind of "loose continuity" property was indeed needed, but I failed to find an example where f failed to be linear. Your example works very nicely, even though it requires uncountable AC.
if I take (f(x+h)-f(x))/h, I get (f(x) + f(h) - f(x)) / h Which gives f(h)/h And, since f(cx) = cf(x), f(h)/h = h f(1)/h = f(1) So it seems a linear function is differentiable, so long as f(1) exists, and that implies continuity. Am I wrong?
why are you using y as in input variable? Why wouldn't you instead use x_1 and x_2? You can't use y as both an independent variable and dependent variable (as measured along the vertical axis) in 2 space.
Thinking of R as a vector space over Q, the existence of this functional actually shows that R is infinite dimensional over Q because if R were finite dimensional then all functional would be continuous, right?. Also, since R is a banach space then by the Baire Category theorem the hamel basis is uncountable. This is really cool. Is there any other way to show that R is infinite dimensional without constructing the discontinuous functional? And do you know any book where i can learn more about similar results? I mean books that collect results related to the axiom of choice. By the way. This is a great video, don't be discouraged by the negative comments. This piece of mathematics is really beautiful and it also confused me when i first learned it. Another similar beautiful theorem is that if you have an additive funtion over the reals that is also lebesgue measurable then it is continuous. This can be proven by luzin theorem or using mollifications. It's very cool because luzin theorem intuitively tells us that measurable functions are almost continuous and if you add additivity you obtain continuity! That could be great topic for a video.
Is the existance of this function completely dependent on the axiom of choice? Or is it actually imposible to construct the set? I'm always unconfortable with theorems that are undecidable without this axiom, as you can have objects that exist, but are "unfindable". Like, is pi on the set? can we prove it?
@@bautibunge737 Or more like: there are uncountably many such functions, so you can pick the values rather arbitrarily. For example, it may be the case that pi is an element of the basis, so pi f(pi)=1. Or 2*pi is an element of the basis, in which case f(pi)=1/2. Or no rational multiple of pi is an element of the basis, instead 1 and pi-1 is. (In that case f(pi) is 2.)
Linear operators in infinite dimensions are kinda discontinuous by default (continuous = bounded for those). This baffled me when I first learned about it. For instance, the derivative operator is discontinuous because functions of fixed norm can still have arbitrarily large derivatives. P.S. Note that the reals are an infinitely-dimensional vector space over the rationals.
First of all, the set S has to be infinite or we've just proven that real numbers are countable. And if S is infinite, then vectors over it have infinite number of coefficients as well. And there's even no guarantee their sum is a rational number. Pi is sum of countable number of rational numbers too.
It is true that S is infinite. Actually it is necessarily uncountably infinite! However, the fact that we have a basis means every x can be written as a FINITE linear combination from the the set S. So the sum is a finite sum.
@@DrTrefor That one got me as well, upon checking, yes "finite linear combination" is part of basically every standard definition, but I guess it's just a part used so rarely that it's easy to forget :D Same with the finite part of span.
@@chalkchalkson5639 ya you probably never explicitly state it if you are working in something like R^n which is already finite dimensional and has a finite basis.
You already implicitly restricted the domain to declare linearity, didn't you,@@DrTrefor? That it's somehow unsurprising in this light was kinda my point. You eventually clarify with the weaker statement that the original R->Q function is "Q-linear," though this just amounts to restricting the linearity condition rather than restricting the function itself--which _is_ different but _not really_ . If it's unsurprising that the function is discontinuous while its restriction to Q is linear, isn't it equally unsurprising that the function is discontinuous while merely Q-linear?
My first thought when I saw your thumbnail was the Dirac delta but then I remembered you're a math professor and you guys tend to insist that the Dirac delta function doesn't exist.
The main thing left out is the proof that Axiom of Choice implies that every vector space has a basis, a theorem that can be found in plenty of places. Everything else is more or less rigorous even if stated casually.
@@DrTrefor Thanks, but that actually wasn't the point I initially ran into confusion from. The part I wanted more rigor was in the assumption that there must be an irrational number between any two values of f(x). This proof that f(x) is not continuous relies on the fact that f(x) has at least two distinct elements in its image, how do we know that f(x) is not a constant function? Intuitively, it's obvious, but that isn't enough.
@@varmituofm By construction of the function, f(0) must be equal 0. Now let x be some number which belongs to the basis; then by construction f(x)=1 (and by definition of a basis, any real number can be produced as a linear combination of elements of the basis in only one way).
I'm fairly certain it would be continuous in that sense. I feel like the linearity condition treating R as the multidimensional Q^|S| space while continuity condition framed as treating R as the one-dimensional R space is the crux of the clash here. "R as a vector space over Q" is fundamentally different from R as we normally view it, and the metric we normally use for R doesn't map well onto the structure of R as a vector space over Q. Here's my attempt at a proof (which includes an explicit hand-wave) using the preimage definition of continuity. Let I = (a, b) be any open interval in R. We want to show that the preimage f^-1(I) is open in Q^|S|. This preimage is precisely {real x: sum of coefficients of S in presentation of x is strictly between a and b}. It seems clear to me that the boundary points of the preimage is precisely {real x: sum of coefficients of S in presentation of x is equal to a or equal to b}, and the preimage contains none of these, and thus must be open.
What are the indices alpha in s sub alpha sub 1 ... s sub alpha sub k at 10:27 ? I don't think these can be integers if S is uncountable but I'm not aware how you'd index an uncountable set. Unless this is just some abstract shorthand for "some member of the set S, exactly which is undefined"
There is a small gap in the proof that the function is not linear. If the function outputs two different values for two different inputs, then it cannot be continuous. But perhaps the function always outputs the same number for any input. It's not hard to show that this cannot be, but without that step the proof is incomplete.
Isnt the numberline f(x)=0 also linear? (funny question to ask if a line is linear lol). So if you have f(x)=0 and x ∈ Q, it is going to be discontinous, and the R numbers is indeed linear combinations of Q. I dont get how this is so exciting
A physicist, an engineer, and a mathematician are all staying in the same hotel. By sheer coincidence, a fire breaks out in each of their rooms. The physicist wakes up and sees the fire. He assumes the fire is perfectly ideal, calculates the amount of water needed to put it out, measures that much water, puts it out, and goes back to bed. The engineer wakes up and sees the fire. He grabs the ice bucket, fills it with water, and dumps it on the fire. He fills the bucket again, and gives it another dousing. Now it's definitely out. The mathematician wakes up, sees the fire and the ice bucket, exclaims, "there is a solution!" and goes back to bed.
I don't like the arguing for it not being continuous. I'd state any sequence which basically is an function from the natural numbers into some set, or let's say some metric space is continuous. However there are "holes" between 1 and 2 in the identity-function of the natural numbers...
OK, I have watched the whole video - the next week's insomnia is about rationalising this proof or is it irrationalising this proof? Every irrational number has a countably infinite set of friends formed by multiplying by a rational number - pick one. There must be an uncountably infinite number of such sets, however we have combinations of these. Suppose we pick pi and e as representatives of their sets then we now have all rational combinations of pi and e for example pi*2 + e*13/27 (Excel convention of * for multiplication) - but that's still only a countable number of irrational numbers. If we only pick a finite number of irrationals and take linear combinations then we still only have a countable number of irrationals. So we must use the axiom of choice an infinite number of times - is that a countably infinite number of times or an uncountably infinite number of times? If it's uncountable has it the same cardinality as the reals (surely it should be smaller)? I fully accept the overall logic but the consequences feel more like Lovecraft than mathematics. Thank you?
I'm not sure if I understand your question right, I think what you're asking is, is a countably infinite set of real numbers big enough for the set of all their linear combinations (over Q) to be uncountably infinite? I think the answer is yes. Take such a countably infinite set M={r1,r2,r3,...} of reals. Now lets look at one particular linear combination x=q1*r1+q2*r2+q3*r3+... of that set. We will now construct a subset P of M where r1 is in P if and only if q1 is equal to 1, r2 is in P if and only if q2 is equal to 1 and so on. (Suppose for example that M={e,pi,sqrt(2),...} and we take the linear combination x=1*e+2*pi-1.3*sqrt(2)+... Then P contains e, but not pi or sqrt(2).) You can get all subsets of M by setting the coefficients of all numbers from M that you want to be in the subset to 1 and the rest to something else, for example 0. So the cardinality of the set of linear combinations of the numbers from M is at least the cardinality of subsets of M. But because M is countably infinite, that cardinality is the same as the number of subsets of natural numbers and thus the cardinality of real numbers, i.e. uncountably infinite.
@@dorisch8038 Yes, I agree we have some countable and uncountable sets - and that the set of all subsets of a set of size Q is uncountable. What I was thinking about is the unresolved question: are there sets of size > the size of the integers and less than the size of the reals? The question can be phrased as: is Aleph1 the cardinality of the real numbers? Aleph1 is definitely uncountable as it is greater than Aleph0. The reals are uncountable but are they the same uncountable? Better minds than mine have shown that either would be consistent with set theory so don't take my question too seriously. However the sets involved in this video have such strange properties that it is difficult to think about sizes hence my reference to Lovecraft and his eldritch horrors.
The video shows that the function is not continuous, gaps at all the irrationals - that's a lot of gaps. It gets worse, the function takes every rational value in every interval. Outline of the proof: Take 2 irrationals in different sets that map to none zero rationals a & b For any rational c there are an infinite number of ways of combining rational multiples of a & b to give c Each combination goes back to a rational multiple of our original irrationals which gives an infinite number of irrationals that map to c Every interval will contain some of those irrationals. Thats hideous, plot it out and you are shading the whole plane but somehow with zero density.
Is S an infinite set? If so, what's stopping me from saying S = R - {pi + 1}, so S is a proper subset of the reals. Now the statement is that x = f(q) covers all reals, so for any rational number in S we can construct it by just picking a single q_i=1 corresponding to that number and then nothing else. Then add in pi + 1, the last real we need to cover R and define it with q_i=1 for pi and q_j = 1 for 1. I'm not seeing what's stopping us from just assigning q's and s's so we can do the trivial thing I just outlined. Perhaps there is a constraint I'm not seeing? Again it would help to know if this property is true when S is a finite set.
I must be missing something in my proof, some underling assumption that i am not taking into account If you use the definition for checking if a function is continuous using only the first two properties you end up showing that any linear function must be continuous. so let's show that: for all epsilon > 0 exist a delta > 0 st: || f(x+delta) - f(x) || < epsilon || f(x+delta) - f(x) || = (for property [a] ) = || f(x) + f(delta) - f(x) || = ||f(delta) || Now if we take delta = epsilon^b || f(x+delta) - f(x) || = epsilon^b ||f(1)|| and taking a appropriate b we can show that 0 < || f(x+delta) - f(x) || = epsilon^b ||f(1)|| < epsilon Creating a angry mathematician ( i am not a mathematician)
Oh, Vitalli sets how you haunt my darkest nightmares. You just couldn't leave me alone after you first traumatized me in measure theory, and now look what you have done: I like my vectorspaces to have bases, but I like my linear functions continuous, too. Ah, but boundedness shall save me.
When I clicked on the video I was expecting something about the differentiation operator and was thinking "okay this is gonna be harsh to explain in a YT video". Then, during the first half of the video I was like "wtf isn't he taking a finite dimensional vector space his linear function can't be discontinuous" and then you went "we only consider Q-linearity" and I was like "Aaaaaaah of course"... About that set S, if we remove the uniqueness assumption and only ask for existence, it's quite obvious that such a set exists, for example take the set of irrational numbers together with {1}; but this would only make S a spanning set and not a basis, so it's not suitable to define your function f. However, from this S we could think of removing elements to get a basis, couldn't we? Wouldn't some sort of quotient do the trick? Or is that so that we can't possibly construct any such S by removing elements from the irrational numbers?
is there any way to choose this magic set such that this function always converges? and also, it wasn't said explicitly, but in order to have discontinuity you must also have that the function takes on multiple values, which isn't obvious...
Assuming axiom of choice, there is a basis of real numbers over rational numbers; that is, there is a set of real numbers with the following properties: 1) Every real number can be expressed as a linear combination of finitely many elements of the set: x = a1*x1 + a2*x2 + x3*x3 + ... + an*xn, where a1, a2, ..., an are rational numbers and x1, x2, ..., xn are elements of the basis; and 2) the set is linearly independent - no element of the set can be expressed as a linear combination of finitely many of the other elements (as above; otherwise, you could have trivially picked the set of all real numbers as a basis). So every real number x can be expressed as a sum of finitely many terms, and therefore the value of f(x) is a sum of finitely many rational numbers. That the way to express a real number as a linear combination of the elements of the basis is unique follows from the linear independence. And what do you mean by "in order to have discontinuity you must also have that the function takes on multiple values"? This is a function in one variable, and for a given x the function has only one value f(x) (that's by definition of a function). Really, a function is a set of ordered pairs: if [a,b] is an element of the set, then we say that f(a)=b.
@@MikeRosoftJH firstly: is it per definition of a basis that every real is a linear combination of finitely many elements of the set? Secondly, by takes on multiple values, I mean that the set of outputs of the function (the values that the function takes on) is not a singleton
@@blizzard_inc Yes, the usual notion of a basis of a vector space is that every element of the space can be expressed as a linear combination of finitely many elements of the basis. For precision, this notion of a basis can be called 'Hamel basis', as opposed to 'Schauder basis', which allows a combination of infinitely many elements (see the corresponding Wikipedia articles). This has an interesting consequence: Take the set of all infinite sequences of all real numbers (R^N). If we pick a countably infinite basis [1], [0,1], [0,0,1], ... (by convention, I omit a trailing infinite sequence of numbers 0), the resulting vector space (over reals) is the set of all sequences of real numbers which have finitely many non-zero elements. The space R^N as a whole doesn't have an easily defined basis (it's uncountably infinite, and its existence is a consequence of the axiom of choice - similar to the space of real numbers over rational numbers). That the range of the function has more than one element is fairly obvious: by construction (and by linear independence of the basis), f(0)=0; and if x belongs to the basis, then f(x)=1. (Similarly, if y=a*x, where x is an element of the basis, and a is a rational number, then f(y)=a.)
I wish we received a bit more information about this set S. It's not actually surprising at all that there is such a set, and one can be constructed easily right away, say, the real numbers. Or, if you want it to be a proper subset, just remove tau. Of course, what would make this interesting is just how much you can cut out of S. What seems obvious to me is that you could just take the real numbers modulo multiplication by rationals, so just pick one element from each unique set that looks like: {y: y = q*x, q \in Q}. Well, here's my problem, suddenly you are writing a finite sum! Now *that's* a real shocker! There is such a set S that x can be written as a *finite* superposition (over the rational numbers) of the elements of S and also *S is finite* ??? Hoyly moly
But is such set really finite? That would imply, that there is not infinitely many "independent" real numbers. But just take sqrt(n) for every integer n. There's infinitely many of these that aren't rational multiples of any combination of the previous ones, just take n to be a prime number. How would you accomplish such decomposition with a finite basis? If it is finite, what is the smallest such basis? If it is infinite, then your function may contain infinitely many terms as the value, does it converge? Can you sketch the function for some choice of the basis?
The set obviously is not finite; it's uncountably infinite. But by definition of a basis, every real number can be expressed as a linear combination of finitely many elements of the basis. And the basis can't really be constructed in the usual sense; its existence follows from the axiom of choice. (Well, you can do the following: pick your favorite well-ordering of real numbers, and then construct the basis by transfinite recursion, with the following recursion step: if the current element is linearly independent with the set constructed so far, then add it to the set, otherwise, do nothing. However, that real numbers can be well-ordered is a consequence of the axiom of choice.)
Well the basis is of infinite length so there is no quarantee that the sum of the q factors doesn't have to converge => the function isn't necessarily defined. It is also clearly not linear with irrational factors as that would imply an irrational output ov the function.
The basis is uncountably infinite; but by definition of a (Hamel) basis every real number can be expressed as a linear combination, with rational coefficients, of finitely many elements of the basis. The other condition is that the basis must be linearly independent; from this follows that any real number can be expressed as a linear combination of finitely many elements of the basis in only one way. Therefore: the function is well-defined, and is finite and rational, for every real number.
Why does axiom of choice imply there exists a basis for any vector space? EDIT: I ask because it's not trivial to me that there would be a finite set of bases to satisfy the property you outline. But I want to understand!
Without going into details, AC is equivalent to Zorn’s Lemma: “a partially ordered set containing upper bounds for every chain (that is, every totally ordered subset) necessarily contains at least one maximal element.” From here you can take all linearly independent sets of vectors, and order them by inclusion. For example, in R3 you could take A= {(1,0,0)} B={(1,0,0), (2,3,0)} C={(2,3,0)} D={(0,0,9), (2,3,0)} … Then, A
AC implies that the real numbers are "well ordered" (actually they could be well ordered even without AC...) Start the basis with, say B = {1}. Then take S = R \ span B and use well ordering to pick a maximal element, and add it to B. Continue this process until R = span B. (the rigorous way to do this is 'transfinite induction', but it gets the idea across).
I read about the axiom of choice a long time ago but I didn't get the point of it, like yea of course you can choose any number you want and why did this only come up so late? After watching this it's quite obvious that it was invented around the time people started to play with infinity more formally and it makes so much more sense now. The name is so misleading
One mathematician has explained the axiom of choice as follows: axiom of choice is not required to pick one of each from infinitely many pairs of shoes (you can always pick the left shoe); but it is required to pick one of each from infinitely many pairs of socks (there is no obvious way to distinguish between socks in a pair). So for example consider the following relation on reals: x~y, if x-y is a rational number. This is an equivalence relation, which splits the real numbers into equivalence classes. Does there exist a set which contains exactly one element from each equivalence class? That it does is a consequence of the axiom of choice. Going back to the "shoes/socks" example: what if you have a collection of infinitely many pairs of real numbers? Obviously, a choice set exists; just pick a minimum from each set. But what if you have an arbitrary infinite collection of two-element sets (without telling what are these elements)? Then it's consistent without axiom of choice that there is such a collection for which no choice set exists. So: axiom of choice is important if you need to pick one of each from infinitely many sets (even infinitely many finite sets). Conversely, for a collection of finitely many sets (even finitely many infinite sets) it is trivial. In particular, for one set, it becomes a truism: a non-empty set has an element.
@@MikeRosoftJH Thank you for taking your time to explain it (to me or to anyone else reading). I say the naming of the axiom of choice is a little misleading because it is not just about the ability to make a choice, but about being able to make infinitely many arbitrary choices.
@@DrTrefor I didn't see the pinned comment right away, as phones tend to hide important information. (otherwise, I would've seen my question was already answered)
If I understand, the basis of the reals over the rations is (uncountably) infinite and so the function f is generally an infinite sum. Do we have to worry about this sum always converging? (ie is f always well defined)? Cool video! Thanks for taking the time to present this idea
Even when an infinite base is generating a vector space, the meaning is still that every vector in the vector space can be written as a linear combination of a finite number of vectors from that infinite base. The sum has a finite number of terms.
Interesting to notice that, in general linearity implies continuity in any finite-dimensional vector space. The real numbers are infinite-dimensional over Q, which is why this fails!
If you allow infinite sets, you can build R with coefficients from Z and elements of S from Q via the diagonal argument. The most obvious set has Sa=10^(-a), zero indexed. I'm not really qualified to know, but my intuition is that this means this is one of those cases where the axiom of choice breaks math.
Actually, if we choose a set S such that one of its elements is 0 OR where some linear combination of elements of S sums to zero (and the coefficients needed to make the elements sum to zero aren’t themselves zero) we can make the function be literally anything at any given value, so f(x) wouldn’t be well defined. But then again clearly the set S could be defined without those properties so I’ve said nothing interesting 🙃
If it only attains a single value, then the argument that intermediate value theorem fails and so it isn't continuous doesn't apply. As soon as f(x) attains two values, then you can use that argument. Clearly the f(x) given achieve many values.
One of my favorite things I studied were the bizzaro function. Such as those that are continuous everywhere but differential nowhere. The real fun isn’t the functions themselves. But to go back in history and read how angry they made (some) mathematicians is fascinating.
Maybe the Weierstrass Function?
en.m.wikipedia.org/wiki/Weierstrass_function
bizarre?? you mean weistrass probably
Bolanzo Weierstrass?
bolzano function
I would argue that the Weierstrass function is not as bizarre as it sounds. When you look at what continuity really is, you know that f : U -> R is continuous at p iff for all real ε > 0, there exists some δ > 0 (which may depend on p and ε), such that for all x in U, if |x - p| < δ, then |f(x) - f(p)| < ε. Meanwhile, f is differentiable at p if lim (f(x) - f(p))/(x - p) (x -> p) exists, which translates to saying there exists some real k such that for all real ε > 0, there exists some δ > 0 (which may depend on p and ε), such that for all x in U, if |x - p| < δ, then |(f(x) - f(p))/(x - p) - k| < ε, but the latter is equivalent to |f(x) - f(p) - k·(x - p)| < ε·|x - p|, which is to say (|k| + ε)·|x - p| < |f(x) - f(p)| < (|k| + ε)·|x - p|. This clearly looks like it should be a stronger condition than the condition for continuity, which is simply that |f(x) - f(p)| < ε. Now, of course, that does not prove it is stronger, but it should not be unexpected at all if it were for some functions at some p. And it being the case for some function at all p is not such a stretch from there. So, I would argue that the existence of such functions conceptually is actually sort of intuitive. The only reason people call it bizarre is because people think of continuity and differentiability in terms of graphs, rather than in terms of what they actually are.
It's bad enough that there is always a real number between any two rationals, but it's far worse that there's always a rational number between any two reals.
What’s interesting is that rationals are countable and irrational aren’t countable.
@@egberthuizinga4408 Agreed. It's easy to have a mental model of the rationals and irrationals zipping together alternately - but that's so misleading. The uncountability of the irrationals is just so big we have the logical tools for it but not the mental models or analogies.
@@andrewharrison8436 Which logical tools do you use?
@@egberthuizinga4408 I was thinking of the techniques that Cantor used in defining a hierachy of infinities or the logic within this video. Careful reasoning from axioms, construction of examples or counter examples that sort of thing.
So logically we prove the reals are uncountable (a bigger infinity) but at the same time any 2 rationals have an irrational between them and vica versa - but at this point I have conclusions and no mental model of where those "extra" irrationals are hiding.
@@andrewharrison8436 The extra irrational is hiding in the infinities. There is a countably infinite number of rationals between 0 and 1. There is also countably infinite number of rational numbers between any two rational interval. So in conclusion there is infinity in between any pair of rational number. The number line is a bad model of understanding this thing because in reality real numbers are define as an infinite set of rationals. By definition it follows that for every real number there is a rational below it and a rational above it otherwise two real numbers without a unique rational between them is define as equivalent.
Lol this is bizarre! If you asked me whether linearity implied a continuous function, I would have said yes and had been 100% confident and 100% wrong.
Right?! You wouldn't have batted an eye! The tricky bit is your intuition is correct for "R-linear" functions, it is a standard proof in any intro real analysis class. But not for "Q-linear".
@@DrTrefor This mind-blowing! I remember proving “linearity implies continuity” using the epsilon-delta definition, could you shed some light on this? Or is there any implicit assumptions I missed?
@@mingmiao364 Yes! You would have used that the coefficients are real numbers. That whole epsilon-delta proof falls apart when you restrict to only rationals.
@@DrTrefor Thanks for pointing it out! Is is true that such example motivates the definition for topological vector spaces? As linearity and continuity really are seperate concepts
@@mingmiao364 The important hypothesis to say linearity implies continuity is finite dimension. Even with R Vectorial spaces, if you have an infinite basis, some function can be linear but not continous.
Some follow-ups to common questions!
1) The set S is certainly infinite, in fact it is uncountable. Nevertheless, the definition of a basis being used here (and guaranteed from axiom of choice) is we can write any x as a FINITE linear combination of things from S. So each choice is finite even if the set S is infinite. This means that f(x) always does give a rational number not an infinite series.
2) The big thing we relaxed from the "usual" notion of linearity is that we are only using rational coefficients, what I called "Q-Linear". For an "R-linear" function it is a standard theorem in any analysis course to prove that this does imply continuity. You might have initially guessed that since Q is dense in R that only Q-linear was needed, but no!
3) A number of people have suggested alternatives along the lines of f(x)=x for x irrational and f(x)=0 for x rational. It's tempting, but this actually fails the additivity property of linearity. Consider for instance x=1-pi and y=pi. So f(x+y)=0 which is not equal to f(x)+f(y=)=1-pi +pi=1.
W.r.t (1) can you link to a video or proof that explains why every x in reals is a *finite* linear combination of q in rationals? My intuition would have expected each x in irrationals to be only the limit of a series, i.e. an infinite linear combination.
@@drewtmacha34 I thought so too
@@drewtmacha34 Your intuition is correct, but that is not what the bit at 10:40 is saying. It is claiming that every real number x is a finite Q-linear combination of stuff from a set S. The (uncountable) set S may consist of rationals, irrationals, etc.
.
The finiteness of the linear combinations comes from the definition of a Hamel basis (as opposed to Schauder basis, which talks about infinite linear combis)
You got were my mind was thinking with Number 1, and then I was thinking is there any x in R, such that there was an infinite number of basis used to create x, and thus have an infinite summation that may be divergent. I quickly got that S was uncountable, if S was countable, then Q x S is countable, but we defined Q x S to cover R, which is uncountable, therefore either Q or S is uncountable, and we already know Q is countable.
@@strikeemblem2886 To be fair, the basis exists only if we accept the Axiom of Choice. We cannot ever fully construct a complete set, but any finite independent set will extend if AC is accepted. But once we choose a starting sequence of basis elements and locked them in, we can't t know what else the AC picked for us. The end result is not very concrete, but the AC itself sounds believable.
When I saw the thumbnail, It did seem to me you would need to restrict c to the rationals for a discontinuous function to be possible, and that you would use a basis for R as a vector space over Q.
Interestingly, when you restrict c to be rational, the condition f(cx)=cf(x) becomes redundant; i.e. f(x+y)=f(x)+f(y)⇒f(cx)=cf(x) for c∈Q.
For c=0:
Setting x=y=0, we get f(0)=f(0)+f(0), so f(0)=0, and f(0x)=f(0)=0=0f(x).
For c∈Z⁺, a simple inductive argument does the trick. The base case c=1 is trivial.
For c=-1:
Seting y=-x, we get f(0)=f(x)+f(-x). Since f(0)=0, we get f(x)=-f(-x).
For c∈Z⁻, c≤-2, we have f(cx)=f(-|c|x)=-f(|c|x) (using the result for c=-1) =-|c|f(x) (using the result for c∈Z⁺) =cf(x).
So the result is true for c∈Z.
If c∈Q, we can write c=p/q where p∈Z and q∈Z⁺. So pf(x)=f(px)=f(qcx)=qf(cx) and f(cx)=p/q f(x) (as q≠0) =cf(x).
So the result is true for c∈Q. Done.
For the same reason, all field morphisms starting from a field of characteristics zero are Q-linear.
Oh wow so the axiom of choice is the math equivalent of the "tree falls in the woods" question. If there's no way for us to explicitly construct this 'linear' but discontinuous function, does it still exist? Of course it does!
I work on discrete dynamical models in biology (esp. gene regulatory networks), and Boolean linear functions have been getting some attention lately. The classic example is the "exclusive or" function xor(x,y)=x+y mod 2. This function is linear (mod 2), but is not monotonic! Note that because ordered fields can't be finite, the definition of monotonic is somewhat different* in this context than in, e.g., the reals, but basically the observation is that "increasing" x from 0 to 1 can either "increase" xor(x,y) form 0 to 1 (if y=0) or "decrease" it from 1 to 0 (if y=1).
Boolean linear functions are, according to multiple formal definitions, the most "sensitive" Boolean functions. Their non-monotonic nature, however, makes them unlikely candidates for describing how the presence or absence of transcription factors impacts gene expression. This has some interesting implications for a bunch of concepts related to the robustness of genetic circuitry, such as canalization, damage spread, and criticality.
* For completeness, a Boolean function f is monotonic if for all input variables x, across all input configurations, changing x from 0 to 1 never changes f from 1 to 0 OR changing x never changes f from 0 to 1. (Some authors, e.g., on Wikipedia, use a stricter definition that leaves off the part after the "OR", but I don't like that because it's more analogous to the definition of non-decreasing functions.)
Biology is complex enough without shoving Computer Science into it. Your field must be endlessly fascinating lol
2:44 , I'm currently taking real analysis now and it still baffles me that numbers behave so nicely in this way. So much that one of the definitions of a real number is similar to this (ie, letting S be all the non-positive powers of 10, or 2, or ...whatever base you want).
Can I ask what your major is?
@@ZinkZoodles I'm in my senior year of a math degree.
@@zachpence8283 Thank you for sharing! Number theory and mathematics intrigue me as theoretical ideas, but I am struggling to choose a major as a high school student applying to university (looking for majors more like applied math/engineering fields). If I may, what do you plan to do after you get your degree? Thank you for your time.
Are you missing a condition on S?
By your definition why can't you just choose S=R? Then there's only one coefficient (1) and your function f is just 1 everywhere
Edit: nevermind, did not catch that the decomposition must be *unique*
First I thought: uh? Then I remembered: linear functions over finite dimensional (normed) R or C vector spaces are continuous, so one has to go into infinite dimension or change the underlying number field. So your answer is correct. The thing is: you play with the ambiguity of not naming the underlying number field.
Ya exactly. Effectively I'm saying "A Q-linear function is not continuous" which is much more reasonable sounding.
@@DrTreforYou can also take a more general definition of continuity and then say that your function returns to being continuous - just choose the right topology and everything will work out fine...
@@eofirdavid Does it? If you take the definition based on neighbourhoods it's not clear to me why the function would be continuous. Image you construct your set such that both 1 and pi are in the basis, now all rational numbers just map to themselves, but 10pi maps to 10, so all neighbourhoods around 10pi will always map to a set containing at least 10pi and 10, so you can never find one that maps to a (9;11) despite that being a neighbourhood of f(10pi)
@@chalkchalkson5639 You can define a metric on R (in fact on any set), with d(x,y)=1 if x!=y and d(x,x)=0 (this induces the discrete topology). Using that metric every function from R to any topologic space is trivially continuous.
@@chalkchalkson5639 Whenever you have a function f:A->B (and in this case both A and B were the set of real numbers), you can always make it continuous by either giving B the trivial topology (the only open sets are the empty sets and B) or giving A the discrete topology (all the sets are open. This topology also comes from the discrete metric).
You can then try to create some more interesting topologies by looking for stronger topologies on B or weaker topologies on A. Not sure how many of them will be "interesting" but there sure are many of them.
2 missleading aspects:
1:
Not saying that S has to be infinite.
2:
Drawing f as a line, like we are used from linear functions.
Let 2 basevectors be e1=1 and e2=pi. Then f(3)=3, f(pi)=1, f(4)=4.
It is mathematically linear, but looks not like a line.
For the people confused with the set S. It is called a hamel basis or algebraic basis. You can look it up in Wikipedia.
The second property of linearity you left up to us can be found through factoring by grouping , no? Awesome video, I’m taking college classes but still a high school junior and find your videos quite impressive and intuitive. Thank you!
Yup! And thank you!
I sort of imagine constructing S as follows:
Start with S_0 = Ø, repeat ad infinitum the following procedure: pick any nonzero real number and check if it can be written as a finite linear combination (with rational coefficients) of elements that are already in S_n thus far. If not, then add it to S_n, to make S_n+1. This way we construct our basis S (in principle).
Although f is linear as proved, it may be a bit misleading to depict it as a line. My intuition tells me it comes arbitrarily close to any point on the plane.
Edit to backup my intuition a bit: Suppose a and b are different elements of S.
Then, for any point (x,y) on the R2- plane, we can choose q1 in Q arbitrarily close to (x-by)/(a-b) and q2 in Q arbitrarily close to (-x+ay)/(a-b).
Then q1 a + q2 b is arbitrarily close to x, whereas f(q1 a + q2 b) = q1+q2 is arbitrarily close to y.
And guess the missing part is: pick your favorite well-ordering of real numbers, and construct the basis by transfinite recursion in the order of well-ordering. The only part left is to well-order the reals, and while the existence of well-ordering is a consequence of the axiom of choice, no formula can be proven in ZFC to actually define one (in the sense that 1) there is exactly one set satisfying the formula, and 2) that set is a well-ordering relation on reals).
And if you don't accept the axiom of choice, it can be shown that the existence of such a function cannot be proved. This is even crazier to me that you can construct a setting in which you can prove that every Q-linear real function is continuous!
Did you claim that you did not accept the axiom of choice? I think you meant you accept that the axiom of choice is false. It’s different, because if you don’t accept the axiom of choice is true then you can’t imply the axiom of choice is false. To sum it up, there are 3 possibilities: we can claim axiom of choice is true, we can claim axiom of choice is false, we can neither claim it’s true neither claim it’s false.
@@becomepostal I do not claim anything about me. I personally think it's very interesting to consider theories with the axiom of choice as well as theories where it is false and to compare them.
To answer your question, I chose not to get into details but you're absolutely right that working without the axiom of choice doesn't mean it's false. What I meant is that there exists a coherent theory containing ZF in which you can prove that every Q-linear real function is continuous (and therefore it is the multiplication by a real number). Such a theory must contradict the axiom of choice. This is due to Cohen and it uses the method of forcing.
To clarify, I'm not saying that refusing the axiom of choice would imply anything. However even if you consider a theory in which the axiom of choice is false, you could very well have non continuous Q-linear real functions. It is not sufficient to say that the axiom of choice is false to get to the result I stated.
In other words, what Cohen proved is that "every Q-linear real function is continuous" could itself be taken as an axiom instead of the axiom of choice and it would be exactly as coherent. In a way, both possibilities coexist (if either one actually exists, which is another question...)!
@@swenji9113 Thank you for your answer and thank you for your time.
I think a much easier to understand example is for example the differential operator from and to the set of smooth functions on an interval [a,b] with the supremum norm,
D_x : C_inf[a,b]-->C_inf[a,b].
It is obviously linear and well defined. But you can find a sequence like
f_n(x)=sin(n²x)/n, with
D(f_n)(x) = cos(n²x)*n, and
||f_n|| --> 0, but ||D(f_n)|| --> inf.
In fact, the "additive linearity" implies the rational multiplication linearity, because you can reproduce all the process, i.e: f(2x)=f(x+x)= f(x)+f(x)= 2f(x)... f(x+(-x))=f(x)+f(-x)=f(0)=0 and therefore f(-x)=(-1)f(x)...f(x)=f(x/3+x/3+x/3)=3f(x/3)... This extension of linearity from addition to multiplication is automatic for the rational field of the "big field".
I think, if I learn this in the university, it won't be half as interesting as you explain it. I love it ❤️
As a math educator myself, I appreciate the love you have for the subject :)
Me an intellectual changing the topology on R where the image lives to the discrete topology: The power I have scares me
Nice video. I'm glad you took the time out to describe the Axiom of Choice, and how this is one of the many non-constructive examples that falls out of assuming it (AoC) to be true. I might rank the Cantor function above this one, simply because it's constructive and is the first time I bumped into a function that was continuous but couldn't be drawn with an "unbroken stroke of pencil on paper" (the way continuity is usually described, pre-calculus). Its differentiablility almost everywhere with derivative zero, despite the ascent from 0 to 1 was equally baffling. That was not the first time my measure theory course broke my brain for a moment or three.
Maybe I misunderstood something but: if your function was R-linear, i.e. f(cx)=cf(x) for all real x and c, then f would be onto. This is a contradiction since f(R)=Q. What do I miss?
If I'm not wrong what you say is that R is a Q-vector space, and f is a Q-linear map. The map you described is not R-linear. I wonder if there is a discontinuous additive real function f, such that f(cx)=cf(x) for all real x,c
0:30 - 0:38 There is something that needs to be cleared up. In general, when non-mathematicians talk about linear functions, they are imagining an equation of the form y = m·x + b (which partially stems from the misconception that a function is a kind of equation, which is not true at all). Those who are more careful will instead say that f is a linear function if it satisfies f(x) = a·x + b everywhere for some a, b, and the inspiration in calling this linear is that if you graph f on a Cartesian coordinate system, the set of points is a line. However, this is still incorrect. Such a function f is not linear unless b = 0. This is confusing, because the functions that we use to graph lines on a plane are not themselves called linear functions, they are called affine functions. If b = 0, then the affine function is also linear. But what is the difference? Well, as stated in the video, the actual definition of a function being linear is that it satisfies the property in the video. To be more precise: it only makes sense to talk about linear functions in the context of vector spaces (or if you want to get more general, modules). If you have two vector spaces V, W over the same field scalar field F, then f : V -> W is called linear iff f(v0 + v1) = f(v0) + f(v1) for all v0, v1 in V, and f(α·v) = α·f(v) for all α in F and v in V. In this context, though, since we are dealing with calculus, linearity means that the vector space in question is R, the field of real numbers. In particular, we are taking R to be a field over itself. R acts as both the vector space and the scalar field, and it is a 1-dimensional vector space. Applying the definition of linearity, then, looks like this: f : R -> R iff f(A·x + y) = A·f(x) + f(y) = x·f(A) + f(y) for all real A, x, y. Understanding that linearity is about functions of vector spaces specifically (and not just arbitrary sets) is key to understand the results in the video.
2:30 - 3:12 By the way, the proof of the existence of S usually uses Zorn's lemma (though you can get it from weaker assumptions), and Zorn's lemma is actually equivalent to the axiom of choice. Also, S is not unique, generally speaking. Also, I want everyone to pay close attention to what is being said here. S is some set of real numbers, an uncountably infinite set, and if you take some finite number of these real numbers, scale each of them *by a rational number,* and add them, you get a real number x, and this is possible for *all real numbers* x. What this is really saying is that if you take the real numbers R to be a vector space over the scalar field Q, the field of rational numbers, then S is a basis for this vector space. Notice that this is different from how we usually handle R as a 1-dimensional vector space over itself. Here, R is a Beth(1)-dimensional vector space over Q. This is very important.
6:14 - 6:31 In other words: it is very important that R is being considered a vector space over Q, rather than over R itself. In fact, you could not define f otherwise, to begin with, but even if you could, it would not be linear.
Here is the point: usually, when we talk about linear functions, we are talking about one vector space, and one vector space only: R as a vector space over itself. We talk about functions f : R -> R being linear, and because this vector space is 1-dimensional, it works out to also be continuous. However, when we switch vector spaces, say R with R' (this is just me abusing notation: the set of real numbers is the same set, but now it is being considered as a vector space over Q, so I am denoting it as R'), and we talk about linear functions g : R' -> R', since R' is Beth(1)-dimensional, and yet the scalar field, Q, only has cardinality Beth(0) = Aleph(0), it is impossible for it to be continuous. The more general lesson here is that if you have a vector space V over a field F and you consider linear functions h : V -> V in general, whether these functions are continuous or not depends largely on the properties of V and F and how they interact together. Continuity is guaranteed if V is finite dimensional, but not so if V is infinite dimensional. Intuitively speaking: If F is "too small" compared to V, the topologies may be "incompatible," in a certain sense, resulting in these linear functions being not continuous.
And if you are familiar with some basic category theory, you can think of it this way: we are talking about endormorphisms in the category of topological vector spaces over a field F. There is a relationship between these endomorphisms and the endomorphisms in the category of *topological* vector spaces over a field F: there is a functor between these categories. If an endomorphism in the category of vector spaces is also a monomorphism (generalizing the idea of an injective function), then it will be an isomorphism, and thus the functor maps this endomophism to a homeomorphism. However, the endomorphism f that was "constructed" in the video (I am using the word "constructed" very loosely here, you guys should know what I mean) is not a monomorphism, clearly, since its range is Q, of cardinality smaller than R. So it also cannot be a homeomorphism in the category of topological vector spaces over F. In other words, it cannot be continuous.
7:30 - 8:15 For those who are curious what an exact statement of the axiom of choice looks like, it is like this: consider a set S, and consider all the sets X that are elements of S. We are gonna to index the sets of S with some other set I, and to index them, you have a surjective (onto) function I -> S. This indexing allows me to talk about each set X in S as being labeled by some i, so we can denote as X(i). Why do I care about this labeling? Because now I can state the axiom of choice: the axiom of choice is simply saying that there is some set T such that for all X(i), there is some x(i) in X(i) such that x(i) in T, and this is true for all S, and regardless of the indexing chosen. This is more or less how you make formal the intuitive idea explained in the video.
8:50 - 9:15 Without the axiom of choice, you can still prove all *finite dimensional* vector spaces have a basis. But remember: the vector space defined in this video is Beth(1)-dimensional (the dimension is uncountably infinite). To prove that infinite dimensional vector spaces always have a basis, you need something like Zorn's lemma, which is equivalent to the axiom of choice.
11:30 - 11:48 Once again, this comes down to the fact that Q is countable, but R is not. Q is dense in R *even though* it is countable and not uncountable. As a result, the order topology of Q is the discrete topology, but the order topology of R is not the discrete topology. These are "incompatible" in some sense, which is why you get this discontinuity.
for Q-linearity, I would just chose two linear functions, one defined for x in Q, and the other for x in R, like f(x) = x for Q, & f(x)=2x for R.
This doesn't work. It doesn't always satisfy f(x+y) = f(x) + f(y)
In particular, (2-sqrt(2)) + (sqrt(2)) = 2 so the sum of 2 irrationals is rational
Although axiom of choice is not constructive, I would like to know if my understanding is correct: is that S = {1, 2^(1/2), 2^(1/3), ..., 3^(1/2), 3^(1/3), ..., pi, e, ...} that includes all the "irrationality" basis and then f(x) is kind of "counting" the "irrationality". For example f(pi+1/3*e) = 1+1/3=4/3, f(sqrt(2) - 1/2*sqrt(3)) = 1-1/2=1/2. f(c) = c if c is rational.
If that is the case it is easy to see it is only Q-linear but not R-linear: f(sqrt(2)*sqrt(2))=f(2)=2 but sqrt(2)*f(sqrt(2))=sqrt(2)*1=sqrt(2)
A good corollary from this example is that the linear space R over Q is infinite dimensional.
At 5:00… does not satisfying the intermediate value theorem for every interval mean it’s discontinuous? I’m really confused because it seems like it could still satisfy the delta epsilon definition of continuity
Answer: yeah not satisfying IVT does mean it’s discontinuous, which I think means the delta epsilon definition isn’t satisfied but the internet says the explanation is complicated
Exactly! IVT says "if it is continuous, then it must hit all intermediate values". As it fails to do so, it is not continuous. We don't need to open up delta epsilon definition for this.
I understand that the axiom of choice implies that such set S exists, because R is a vector space, and Q is the field of scalars over that vector space. However, You assumed that this set S is finite, and I don't quite understand why that should be the case. And S must be finite, or alternatively all the possible series sum_of{q_i} must be convergent for f to be well-defined, and I don't see the reason why either of those must be true
A basis means any x can be written as a FINITE linear combination from the infinite set S
I encountered a similar reasoning when a friend challenged me to write the identity function f(x)=x as the sum of two periodic functions.
Find a Q-base of R, define g(x) as the projection of x on b_1, the first element of the base and h(x) as the projection of x on span(all the others elements of the base). Then you have naturally that f(x)=g(x)+h(x) by definition of the projection. Moreover, g(x) is periodic of period b_1 while h(x) is periodic of any period b_i where i=/=1. That’s crazy.
How is h periodic?
I have one confusion, why is f not continuous if its range is cut by irrational numbers? By definition of continuous, (1) f is defined on R, (2) f(x) -> f(c) when x -> c for every c on R. I can choose rational numbers q = f(x) that are infinitely close to f(c) when x is close to c. So the second condition is also satisfied.
Continuity means more than just that it gets close in the limit, it actually has to have a value c so f(c) equals any number, and that won't work for irrationals.
If we’re just looking for a Q-linear function that isn’t continuous, can’t we make a much simpler function? Just say f(x)=x for rational numbers and f(x)=0 for irrationals. Then f is Q-linear since for all rational c, f(cx)=cx=cf(x) if x is rational and f(cx)=0=cf(x) of x is irrational. Furthermore f isn’t continuous for the same reason mentioned in the video: it can’t output irrational numbers.
Sadly this is not linear as it fails additivity
I don’t understand the proof for discontinuity. Why does the fact that f only takes values in Q implies discontinuity? Counter example: suppose f is the constant function that is always equal to 1. This function is continuous and only takes value in Q.
I was using the intermediate value theorem which says for continuous functions it attains all values between any two given values. But your example only attains one value so I can’t reject continuity via contradicting IVT
sounds like the domain needs a different topology to make this continuous.
the main propose of this video is to show you a change of paradigm... the linearity or continuity is only an abstract way from our language and how this lead in your maths... so its okay to make clear the idea .... at all not everyone gives importance to this because it seems not relevant in the diary life... but is thoutght involving itself in a infinite cycle jeje
Not only is there at least one irrational between any two rationals, there are infinite rationals and irrationals between any two rationals. Any interval of nonzero length on the real line contains "a whole real line" of its own, because there are bijections that can be defined between the open interval we get by excluding the endpoints, no matter how small it is, and the entire real line.
i don't know much math but your video was very entertaining
I don't know that much about core mathematics so maybe the 2 conditions are the definition of linearity but I would expect the second one (f(cx) = cf(x)) to use c real (even irrational) because for rational c's it easily follows from f(x+y) = f(x) + f(y).
That absolutely is the ‘normal’ expectation. In almost all cases, it is going to work out the same so we don’t talk about it much, but this example really shows we do have to be careful about whether the c is real or rational.
how can we know that f(x) converges ? since S is infinite we are adding up an infinite amount of terms, it is not clear at all if f(x) should converge or not.
While S is indeed uncountably infinite, the definition of basis used here (and guaranteed by AOC) is that any x an be written as a FINITE linear combination
@@DrTrefor It has cardinality of the continuum, right? (Assuming axiom of choice.)
* Consider the set of all finite subsets of the set S. This set has the same cardinality as S.
* For each set consider the set of all sequences of rational numbers with the same number of elements. Of these there are countably many.
* Every such sequence defines a real number, and by assumption every real number can be obtained as a linear combination of finitely many elements of S, with rational coefficients.
* It follows: The set of all linear combinations is a union of S-many countable sets, which has the same cardinality as S.
* Therefore: S has cardinality of the continuum.
(I don't think that the argument would work without axiom of choice, or at least some weaker consequence thereof - even assuming that the basis exists. Without choice, I can't even be sure that a union of countably many countable sets is countable.)
The big "cheat" is that the function is only linear over the rational field, which is only mentioned in the middle of the video. The existence of such a function is not surprising at all. There are very few rational numbers compared to reals. So linearity over the rationals is a quite weak condition on a real function.
In contrast, rationals are dense in R, so one might have guessed that this was sufficient. Indeed, we can’t construct such a function so not so easy to do, need non-constructive proofs via axiom of choice to make it work
@@DrTrefor Density is a topological notion. We did not make any assumptions that takes into consideration the topology or the metric of the reals (e.g., that the function be bounded, or upper semi-continuous, or open, or whatever). So we are simply looking for a linear operator on the continuum dimensional Q-vector space R. It is no surprise that there are a lot of them. Discontinuity of most of these functions is again not surprising for the same reason: the vector space structure is oblivious to the proximity of real numbers. Basically, in order to make this construction sound interesting, you have to bring up concepts that were never mentioned in the problem, and point out that those concepts are violated. Nobody expected them to hold to begin with. It's like saying "I'm, gonna mix water and salt, and believe it or not, the mixture is not gonna sing Beethoven's fifth symphony."
A common knowledge says that any linear functional on finite dimensional vector space is continuous. In infinite dimensions it is a very common thing to have discontinuous linear functional.
This function you provided is not R linear (because if you multiply x by irrational number you may have a whole new set of coefficients q), but it is Q linear, and R is infinite dimensional space over Q, so there is nothing strange in your result:)
0:30 By that definition, f(x)=x+1 isn't linear, because c*f(x) =/= f(c*x).
Example:
c=2, x=0
f(x)=x+1
2*(0+1)=(2*0)+1
2=/=1
I suppose it's only linear additively, not linear multiplicitatively.
This is an annoying terminology thing. I’d call x+1 an “affine transformation”
@@DrTrefor Ah, so f(x)=x+1 is considered an affine transformation of the base function f(x)=x?
How do you know the basis (set S) isnt just the whole R. Such that the coeficients are always 0 except from 1, which equals 1. Hence the function would be just the constant function 1 which is continous?
Then it would no longer have a unique decomposition for each x because you could always write x=1x and x=1(2x)-1(x).
@@DrTrefor nice. Thanks!
the moment you mentioned a vector space over Q, I stopped the video, typed this. if you're allowed to use a nonstandard VP, I would also be allowed to make up some topology in the codomain for any function y=ax such that it's not linear. for example, discrete topology
The functions "f(x)= p(x)×n^kx, f(x)= p(x)×n^|kx| and f(x)= p(x)×n^-kx" with n as negative real number, k a non negative and nor zero real number and p(x) a polynomial of x fascinates me. The functions can be linear or curved and all real set are included but the numbers alternate between positive. But this occurs in Rational numbers. But I thought a little about irrational numbers in this case and those are perfectly possible to apply but we don't know if f(x) for a irrational x would be positive or negative. That's a in indetermination? The domain must be just the rationals? If irrational is really included the function is a non continuous but delimited in two curves well described
I don't speak English so I'm sorry. Hahaha.
If 1_Q is the characteristic function over the rationals (1 if x in Q, 0 if not), then f(x) = x * 1_Q(x) is also Q-linear and not continuous over R.
EDIT: it is continuous at 0.
Counter-example: let a=pi, b=4-pi.
No? For example,
f(1)=1
f(0)=1
Thus:
f(1+0)=f(1)=1
However,
f(1)+f(0)=1+1=2≠f(1+0)
@@methatis3013 that would be the case for 1_Q(x) but we're multiplying by x to ensure the linearity.
@@estebanmarco8755 ah, my bad, I missed that part
It still fails though because of the reasons person before me said. If you have 2 irrational numbers x and y that add up to a rational number q different than 0, you will have
f(x)+f(y)=0+0=0
but
f(x+y) = f(q) = q = x+y ≠ 0
Ayy, I knew that it had to be some conversion from R to Q!
This seems like a bit of slight of hand. You've made a linear function that maps real numbers to rational numbers. The rational numbers by definition are not continuous, so of course your function is not continuous.
Also, the way you define x seems to me a type of field extension with regard to Q? I'm still in my Abstract Algebra course, but isn't each of the s_a going to either be an element within Q are an irrational element that acts as a field extension?
I think you can think of it as a field extension, but this is not the canonical structure to consider (i.e as a vector space over Q). The video is a bit slight of handy, but only a bit. The wierd property is that axiom of choice gives us this funky basis. The function itself is just repackaging that funky basis into a "clickbait" kind of way, but the existence of the funky basis is the real meat of the problem.
Choiciness feels like diet "False".
In a similar way to how False implies anything in logic, choice implies choiciness. There's like a bajillion unintuitive things as a consequence of choice, I'm not surprised that there are other things that are unintuitive and choicy, if you assume choice. :P
OK. Would you prefer it that a set can be split into more subsets - disjoint and non-empty - than it has elements? Let's visualize this: imagine Cantor's Hotel (a competitor of Hilbert's Hotel), where rooms are indexed by real numbers. Of course, the hotel has infinitely many floors, and infinitely many rooms on each floor: on each floor there are numbers which differ from each other by a rational number. In absence of axiom of choice, it's consistent that the hotel has more floors than rooms. ("Say what? Of course it has no more floors than rooms." Okay, what does that mean? "It means that the set of floors can be mapped one-to-one with a subset of rooms." And can you do that? "Sure, just pick a single room from each floor..." Oops. That's precisely what axiom of choice says.) On the other hand, it can be proven even without axiom of choice that the hotel has no less floors than rooms.
And that's not going to Feferman-Levy model of set theory, where real numbers (obviously an uncountable set) are a union of countably many countable sets. (Really, don't go there, because for obvious reasons in this model Lebesgue measure is not countably additive, with a number of other unfortunate consequences for calculus.)
Thank you, i was actually searching an example like this last year. I was wondering if f(x+y)=f(x)+f(y) could imply linearity and I easily proved that Q-linearity is indeed implied. Then I showed that if f is also continuos, than it's linear. I did guess that some kind of "loose continuity" property was indeed needed, but I failed to find an example where f failed to be linear. Your example works very nicely, even though it requires uncountable AC.
You're aware that your set S could just be R itself? Every x can be written as "linear combination" x = 1x
Yes, but not uniquely. The claim of a basis is a unique linear combination.
@@DrTrefor Yes I remembered and deleted the post, or at least it looked deleted xD
I was thinking this would be smth like f(x) = (x^2)/x
I would say this is continuous on its domain of R\{0}, in contrast to my example where the domain is all of R
if I take (f(x+h)-f(x))/h, I get
(f(x) + f(h) - f(x)) / h
Which gives f(h)/h
And, since f(cx) = cf(x),
f(h)/h = h f(1)/h
= f(1)
So it seems a linear function is differentiable, so long as f(1) exists, and that implies continuity.
Am I wrong?
I don't think you "know" a function that is defined via a Hamel Basis.
why are you using y as in input variable? Why wouldn't you instead use x_1 and x_2?
You can't use y as both an independent variable and dependent variable (as measured along the vertical axis) in 2 space.
Thinking of R as a vector space over Q, the existence of this functional actually shows that R is infinite dimensional over Q because if R were finite dimensional then all functional would be continuous, right?. Also, since R is a banach space then by the Baire Category theorem the hamel basis is uncountable. This is really cool. Is there any other way to show that R is infinite dimensional without constructing the discontinuous functional? And do you know any book where i can learn more about similar results? I mean books that collect results related to the axiom of choice.
By the way. This is a great video, don't be discouraged by the negative comments. This piece of mathematics is really beautiful and it also confused me when i first learned it. Another similar beautiful theorem is that if you have an additive funtion over the reals that is also lebesgue measurable then it is continuous. This can be proven by luzin theorem or using mollifications. It's very cool because luzin theorem intuitively tells us that measurable functions are almost continuous and if you add additivity you obtain continuity! That could be great topic for a video.
@12:18 This crazy set sounds like one associated with the Riemann-Zeta function.
I'm confused, why is it guaranteed that the sum of coefficients of the basis vectors for any real number is finite? or does that not matter?
Is the existance of this function completely dependent on the axiom of choice? Or is it actually imposible to construct the set? I'm always unconfortable with theorems that are undecidable without this axiom, as you can have objects that exist, but are "unfindable". Like, is pi on the set? can we prove it?
My understanding is that it is known this is not constructable.
@@DrTrefor So it isn't posible to know the value of any value other than 0?
@@bautibunge737 Or more like: there are uncountably many such functions, so you can pick the values rather arbitrarily. For example, it may be the case that pi is an element of the basis, so pi f(pi)=1. Or 2*pi is an element of the basis, in which case f(pi)=1/2. Or no rational multiple of pi is an element of the basis, instead 1 and pi-1 is. (In that case f(pi) is 2.)
Linear operators in infinite dimensions are kinda discontinuous by default (continuous = bounded for those). This baffled me when I first learned about it. For instance, the derivative operator is discontinuous because functions of fixed norm can still have arbitrarily large derivatives.
P.S. Note that the reals are an infinitely-dimensional vector space over the rationals.
First of all, the set S has to be infinite or we've just proven that real numbers are countable. And if S is infinite, then vectors over it have infinite number of coefficients as well. And there's even no guarantee their sum is a rational number. Pi is sum of countable number of rational numbers too.
It is true that S is infinite. Actually it is necessarily uncountably infinite! However, the fact that we have a basis means every x can be written as a FINITE linear combination from the the set S. So the sum is a finite sum.
@@DrTrefor That one got me as well, upon checking, yes "finite linear combination" is part of basically every standard definition, but I guess it's just a part used so rarely that it's easy to forget :D Same with the finite part of span.
@@chalkchalkson5639 ya you probably never explicitly state it if you are working in something like R^n which is already finite dimensional and has a finite basis.
i may be stupid but does not
f(x)={
x ∈ Q: x
otherwise: 0
}
work for this?
Sadly not. This function fails the additivity property of linearity (ex take (pi+1) and (-pi)
@@DrTrefor what about the alternate case being x+1
How does the plate on your shirt turn into a torus? It doesn't have any holes to begin with
Ha, I think it is meant to be a shadow:D
So only the restriction to the rationals is linear, but isn't that map from Q->Q also continuous?
Implicitly I'm banning restricting the domain as that is somehow unsurprising.
You already implicitly restricted the domain to declare linearity, didn't you,@@DrTrefor? That it's somehow unsurprising in this light was kinda my point.
You eventually clarify with the weaker statement that the original R->Q function is "Q-linear," though this just amounts to restricting the linearity condition rather than restricting the function itself--which _is_ different but _not really_ .
If it's unsurprising that the function is discontinuous while its restriction to Q is linear, isn't it equally unsurprising that the function is discontinuous while merely Q-linear?
My first thought when I saw your thumbnail was the Dirac delta but then I remembered you're a math professor and you guys tend to insist that the Dirac delta function doesn't exist.
I know info was cut to make a more accessible video, but where can I find a more rigorous proof?
The main thing left out is the proof that Axiom of Choice implies that every vector space has a basis, a theorem that can be found in plenty of places. Everything else is more or less rigorous even if stated casually.
@@DrTrefor Thanks, but that actually wasn't the point I initially ran into confusion from. The part I wanted more rigor was in the assumption that there must be an irrational number between any two values of f(x). This proof that f(x) is not continuous relies on the fact that f(x) has at least two distinct elements in its image, how do we know that f(x) is not a constant function? Intuitively, it's obvious, but that isn't enough.
@@varmituofm By construction of the function, f(0) must be equal 0. Now let x be some number which belongs to the basis; then by construction f(x)=1 (and by definition of a basis, any real number can be produced as a linear combination of elements of the basis in only one way).
@@MikeRosoftJH perfect, thank you!
Is the function discontinuous even when viewed as a function from Q^|S| -> Q, with the usual Euclidean metrics?
I'm fairly certain it would be continuous in that sense. I feel like the linearity condition treating R as the multidimensional Q^|S| space while continuity condition framed as treating R as the one-dimensional R space is the crux of the clash here. "R as a vector space over Q" is fundamentally different from R as we normally view it, and the metric we normally use for R doesn't map well onto the structure of R as a vector space over Q.
Here's my attempt at a proof (which includes an explicit hand-wave) using the preimage definition of continuity.
Let I = (a, b) be any open interval in R. We want to show that the preimage f^-1(I) is open in Q^|S|. This preimage is precisely {real x: sum of coefficients of S in presentation of x is strictly between a and b}. It seems clear to me that the boundary points of the preimage is precisely {real x: sum of coefficients of S in presentation of x is equal to a or equal to b}, and the preimage contains none of these, and thus must be open.
Is the set [0, 1] not sufficient for this?
What blew my mind was to learn that f(x)=mx+b is not linear, assuming b is not 0.
This is more a definitional thing, we call such transformations when b is nonzero affine transformations.
Can S be constructed without the axiom of choice? Possibly even explicitly?
I don’t believe so.
@@DrTrefor Them I don't understand why not just use a representant system for R/Q
Wait, so "f(x) = x+1" is NOT linear, since f(2x) = 2x+1 != 2(x+1) = 2f(x) ?
I call this an affine transformation, but it’s just terminology and mostly these distinctions are helpful in higher dimensions.
What are the indices alpha in s sub alpha sub 1 ... s sub alpha sub k at 10:27 ? I don't think these can be integers if S is uncountable but I'm not aware how you'd index an uncountable set. Unless this is just some abstract shorthand for "some member of the set S, exactly which is undefined"
There is a small gap in the proof that the function is not linear. If the function outputs two different values for two different inputs, then it cannot be continuous. But perhaps the function always outputs the same number for any input. It's not hard to show that this cannot be, but without that step the proof is incomplete.
Isnt the numberline f(x)=0 also linear? (funny question to ask if a line is linear lol). So if you have f(x)=0 and x ∈ Q, it is going to be discontinous, and the R numbers is indeed linear combinations of Q. I dont get how this is so exciting
Sadly not continuous! It fails additivity.
@@DrTrefor do you mean «sadly not linear»? Of course it is not continous
f(x)=0 should be linear and not continous
A physicist, an engineer, and a mathematician are all staying in the same hotel. By sheer coincidence, a fire breaks out in each of their rooms.
The physicist wakes up and sees the fire. He assumes the fire is perfectly ideal, calculates the amount of water needed to put it out, measures that much water, puts it out, and goes back to bed.
The engineer wakes up and sees the fire. He grabs the ice bucket, fills it with water, and dumps it on the fire. He fills the bucket again, and gives it another dousing. Now it's definitely out.
The mathematician wakes up, sees the fire and the ice bucket, exclaims, "there is a solution!" and goes back to bed.
I don't like the arguing for it not being continuous.
I'd state any sequence which basically is an function from the natural numbers into some set, or let's say some metric space is continuous. However there are "holes" between 1 and 2 in the identity-function of the natural numbers...
OK, I have watched the whole video - the next week's insomnia is about rationalising this proof or is it irrationalising this proof?
Every irrational number has a countably infinite set of friends formed by multiplying by a rational number - pick one.
There must be an uncountably infinite number of such sets, however we have combinations of these.
Suppose we pick pi and e as representatives of their sets then we now have all rational combinations of pi and e for example pi*2 + e*13/27 (Excel convention of * for multiplication) - but that's still only a countable number of irrational numbers.
If we only pick a finite number of irrationals and take linear combinations then we still only have a countable number of irrationals. So we must use the axiom of choice an infinite number of times - is that a countably infinite number of times or an uncountably infinite number of times? If it's uncountable has it the same cardinality as the reals (surely it should be smaller)?
I fully accept the overall logic but the consequences feel more like Lovecraft than mathematics.
Thank you?
I'm not sure if I understand your question right, I think what you're asking is, is a countably infinite set of real numbers big enough for the set of all their linear combinations (over Q) to be uncountably infinite?
I think the answer is yes. Take such a countably infinite set M={r1,r2,r3,...} of reals. Now lets look at one particular linear combination x=q1*r1+q2*r2+q3*r3+... of that set. We will now construct a subset P of M where r1 is in P if and only if q1 is equal to 1, r2 is in P if and only if q2 is equal to 1 and so on. (Suppose for example that M={e,pi,sqrt(2),...} and we take the linear combination x=1*e+2*pi-1.3*sqrt(2)+... Then P contains e, but not pi or sqrt(2).)
You can get all subsets of M by setting the coefficients of all numbers from M that you want to be in the subset to 1 and the rest to something else, for example 0. So the cardinality of the set of linear combinations of the numbers from M is at least the cardinality of subsets of M. But because M is countably infinite, that cardinality is the same as the number of subsets of natural numbers and thus the cardinality of real numbers, i.e. uncountably infinite.
@@dorisch8038 Yes, I agree we have some countable and uncountable sets - and that the set of all subsets of a set of size Q is uncountable.
What I was thinking about is the unresolved question: are there sets of size > the size of the integers and less than the size of the reals? The question can be phrased as: is Aleph1 the cardinality of the real numbers?
Aleph1 is definitely uncountable as it is greater than Aleph0. The reals are uncountable but are they the same uncountable?
Better minds than mine have shown that either would be consistent with set theory so don't take my question too seriously.
However the sets involved in this video have such strange properties that it is difficult to think about sizes hence my reference to Lovecraft and his eldritch horrors.
The video shows that the function is not continuous, gaps at all the irrationals - that's a lot of gaps.
It gets worse, the function takes every rational value in every interval.
Outline of the proof:
Take 2 irrationals in different sets that map to none zero rationals a & b
For any rational c there are an infinite number of ways of combining rational multiples of a & b to give c
Each combination goes back to a rational multiple of our original irrationals which gives an infinite number of irrationals that map to c
Every interval will contain some of those irrationals.
Thats hideous, plot it out and you are shading the whole plane but somehow with zero density.
Is S an infinite set? If so, what's stopping me from saying S = R - {pi + 1}, so S is a proper subset of the reals. Now the statement is that x = f(q) covers all reals, so for any rational number in S we can construct it by just picking a single q_i=1 corresponding to that number and then nothing else. Then add in pi + 1, the last real we need to cover R and define it with q_i=1 for pi and q_j = 1 for 1. I'm not seeing what's stopping us from just assigning q's and s's so we can do the trivial thing I just outlined. Perhaps there is a constraint I'm not seeing? Again it would help to know if this property is true when S is a finite set.
The basis isn't finite; it's uncountably infinite. But R\{pi+1} obviously isn't a basis, because it isn't linearly independent.
if f(x) = sum of q_{i} doesnt that mean it's a linear form, ( constant) so f(cx) again equals sum q_{i} ( cuz there is no x )
I must be missing something in my proof, some underling assumption that i am not taking into account
If you use the definition for checking if a function is continuous using only the first two properties you end up showing that any linear function must be continuous.
so let's show that: for all epsilon > 0 exist a delta > 0 st: || f(x+delta) - f(x) || < epsilon
|| f(x+delta) - f(x) || = (for property [a] ) = || f(x) + f(delta) - f(x) || = ||f(delta) ||
Now if we take delta = epsilon^b
|| f(x+delta) - f(x) || = epsilon^b ||f(1)|| and taking a appropriate b we can show that 0 < || f(x+delta) - f(x) || = epsilon^b ||f(1)|| < epsilon
Creating a angry mathematician ( i am not a mathematician)
Topologically speaking, coffee = doughnuts lol.
Oh, Vitalli sets how you haunt my darkest nightmares. You just couldn't leave me alone after you first traumatized me in measure theory, and now look what you have done: I like my vectorspaces to have bases, but I like my linear functions continuous, too. Ah, but boundedness shall save me.
When I clicked on the video I was expecting something about the differentiation operator and was thinking "okay this is gonna be harsh to explain in a YT video". Then, during the first half of the video I was like "wtf isn't he taking a finite dimensional vector space his linear function can't be discontinuous" and then you went "we only consider Q-linearity" and I was like "Aaaaaaah of course"...
About that set S, if we remove the uniqueness assumption and only ask for existence, it's quite obvious that such a set exists, for example take the set of irrational numbers together with {1}; but this would only make S a spanning set and not a basis, so it's not suitable to define your function f. However, from this S we could think of removing elements to get a basis, couldn't we? Wouldn't some sort of quotient do the trick? Or is that so that we can't possibly construct any such S by removing elements from the irrational numbers?
Why is it assumed that the set {si} is finite? For the infinite set f(x) could be infinite at all xs
is there any way to choose this magic set such that this function always converges? and also, it wasn't said explicitly, but in order to have discontinuity you must also have that the function takes on multiple values, which isn't obvious...
Assuming axiom of choice, there is a basis of real numbers over rational numbers; that is, there is a set of real numbers with the following properties: 1) Every real number can be expressed as a linear combination of finitely many elements of the set: x = a1*x1 + a2*x2 + x3*x3 + ... + an*xn, where a1, a2, ..., an are rational numbers and x1, x2, ..., xn are elements of the basis; and 2) the set is linearly independent - no element of the set can be expressed as a linear combination of finitely many of the other elements (as above; otherwise, you could have trivially picked the set of all real numbers as a basis). So every real number x can be expressed as a sum of finitely many terms, and therefore the value of f(x) is a sum of finitely many rational numbers. That the way to express a real number as a linear combination of the elements of the basis is unique follows from the linear independence.
And what do you mean by "in order to have discontinuity you must also have that the function takes on multiple values"? This is a function in one variable, and for a given x the function has only one value f(x) (that's by definition of a function). Really, a function is a set of ordered pairs: if [a,b] is an element of the set, then we say that f(a)=b.
@@MikeRosoftJH firstly: is it per definition of a basis that every real is a linear combination of finitely many elements of the set?
Secondly, by takes on multiple values, I mean that the set of outputs of the function (the values that the function takes on) is not a singleton
@@blizzard_inc Yes, the usual notion of a basis of a vector space is that every element of the space can be expressed as a linear combination of finitely many elements of the basis. For precision, this notion of a basis can be called 'Hamel basis', as opposed to 'Schauder basis', which allows a combination of infinitely many elements (see the corresponding Wikipedia articles). This has an interesting consequence: Take the set of all infinite sequences of all real numbers (R^N). If we pick a countably infinite basis [1], [0,1], [0,0,1], ... (by convention, I omit a trailing infinite sequence of numbers 0), the resulting vector space (over reals) is the set of all sequences of real numbers which have finitely many non-zero elements. The space R^N as a whole doesn't have an easily defined basis (it's uncountably infinite, and its existence is a consequence of the axiom of choice - similar to the space of real numbers over rational numbers).
That the range of the function has more than one element is fairly obvious: by construction (and by linear independence of the basis), f(0)=0; and if x belongs to the basis, then f(x)=1. (Similarly, if y=a*x, where x is an element of the basis, and a is a rational number, then f(y)=a.)
What if the basis isn't finite? If the sum of the q's diverges, can we still say it's rational?
I wish we received a bit more information about this set S. It's not actually surprising at all that there is such a set, and one can be constructed easily right away, say, the real numbers. Or, if you want it to be a proper subset, just remove tau. Of course, what would make this interesting is just how much you can cut out of S. What seems obvious to me is that you could just take the real numbers modulo multiplication by rationals, so just pick one element from each unique set that looks like: {y: y = q*x, q \in Q}.
Well, here's my problem, suddenly you are writing a finite sum! Now *that's* a real shocker! There is such a set S that x can be written as a *finite* superposition (over the rational numbers) of the elements of S and also *S is finite* ??? Hoyly moly
nevermind it's clear now that was just a typo
But is such set really finite? That would imply, that there is not infinitely many "independent" real numbers. But just take sqrt(n) for every integer n. There's infinitely many of these that aren't rational multiples of any combination of the previous ones, just take n to be a prime number. How would you accomplish such decomposition with a finite basis?
If it is finite, what is the smallest such basis?
If it is infinite, then your function may contain infinitely many terms as the value, does it converge?
Can you sketch the function for some choice of the basis?
The set obviously is not finite; it's uncountably infinite. But by definition of a basis, every real number can be expressed as a linear combination of finitely many elements of the basis. And the basis can't really be constructed in the usual sense; its existence follows from the axiom of choice. (Well, you can do the following: pick your favorite well-ordering of real numbers, and then construct the basis by transfinite recursion, with the following recursion step: if the current element is linearly independent with the set constructed so far, then add it to the set, otherwise, do nothing. However, that real numbers can be well-ordered is a consequence of the axiom of choice.)
Well the basis is of infinite length so there is no quarantee that the sum of the q factors doesn't have to converge => the function isn't necessarily defined. It is also clearly not linear with irrational factors as that would imply an irrational output ov the function.
The basis is uncountably infinite; but by definition of a (Hamel) basis every real number can be expressed as a linear combination, with rational coefficients, of finitely many elements of the basis. The other condition is that the basis must be linearly independent; from this follows that any real number can be expressed as a linear combination of finitely many elements of the basis in only one way. Therefore: the function is well-defined, and is finite and rational, for every real number.
Why does axiom of choice imply there exists a basis for any vector space?
EDIT: I ask because it's not trivial to me that there would be a finite set of bases to satisfy the property you outline. But I want to understand!
Oh it is not trivial at all, it is its whole own proof I'd have to do a video on or your could look up:)
Without going into details, AC is equivalent to Zorn’s Lemma:
“a partially ordered set containing upper bounds for every chain (that is, every totally ordered subset) necessarily contains at least one maximal element.”
From here you can take all linearly independent sets of vectors, and order them by inclusion.
For example, in R3 you could take
A= {(1,0,0)}
B={(1,0,0), (2,3,0)}
C={(2,3,0)}
D={(0,0,9), (2,3,0)}
…
Then, A
AC implies that the real numbers are "well ordered" (actually they could be well ordered even without AC...)
Start the basis with, say B = {1}. Then take S = R \ span B and use well ordering to pick a maximal element, and add it to B. Continue this process until R = span B. (the rigorous way to do this is 'transfinite induction', but it gets the idea across).
I read about the axiom of choice a long time ago but I didn't get the point of it, like yea of course you can choose any number you want and why did this only come up so late? After watching this it's quite obvious that it was invented around the time people started to play with infinity more formally and it makes so much more sense now. The name is so misleading
One mathematician has explained the axiom of choice as follows: axiom of choice is not required to pick one of each from infinitely many pairs of shoes (you can always pick the left shoe); but it is required to pick one of each from infinitely many pairs of socks (there is no obvious way to distinguish between socks in a pair). So for example consider the following relation on reals: x~y, if x-y is a rational number. This is an equivalence relation, which splits the real numbers into equivalence classes. Does there exist a set which contains exactly one element from each equivalence class? That it does is a consequence of the axiom of choice. Going back to the "shoes/socks" example: what if you have a collection of infinitely many pairs of real numbers? Obviously, a choice set exists; just pick a minimum from each set. But what if you have an arbitrary infinite collection of two-element sets (without telling what are these elements)? Then it's consistent without axiom of choice that there is such a collection for which no choice set exists. So: axiom of choice is important if you need to pick one of each from infinitely many sets (even infinitely many finite sets). Conversely, for a collection of finitely many sets (even finitely many infinite sets) it is trivial. In particular, for one set, it becomes a truism: a non-empty set has an element.
@@MikeRosoftJH Thank you for taking your time to explain it (to me or to anyone else reading). I say the naming of the axiom of choice is a little misleading because it is not just about the ability to make a choice, but about being able to make infinitely many arbitrary choices.
Isn‘t existence of Hamel Basis different to existence of basis (which is what‘s said in the video)?
I want to watch more closely when I have time, but I think the property f(cx)=cf(x) is for c rational only? Otherwise, f(x)=mx, where m=f(1)
Yes exactly
@@DrTrefor I didn't see the pinned comment right away, as phones tend to hide important information. (otherwise, I would've seen my question was already answered)
If I understand, the basis of the reals over the rations is (uncountably) infinite and so the function f is generally an infinite sum. Do we have to worry about this sum always converging? (ie is f always well defined)?
Cool video! Thanks for taking the time to present this idea
Each vector is a combination of only a finite number of basis elements, so the sum is always finite, no worries there :)
Even when an infinite base is generating a vector space, the meaning is still that every vector in the vector space can be written as a linear combination of a finite number of vectors from that infinite base.
The sum has a finite number of terms.
Interesting to notice that, in general linearity implies continuity in any finite-dimensional vector space. The real numbers are infinite-dimensional over Q, which is why this fails!
If you allow infinite sets, you can build R with coefficients from Z and elements of S from Q via the diagonal argument. The most obvious set has Sa=10^(-a), zero indexed. I'm not really qualified to know, but my intuition is that this means this is one of those cases where the axiom of choice breaks math.
I thought it would be 2x if x is rational, and x if x is irrational
Hold up, 5:10, what if f(x) is constant in some interval?? I don’t see why that’s not possible?
Actually, if we choose a set S such that one of its elements is 0 OR where some linear combination of elements of S sums to zero (and the coefficients needed to make the elements sum to zero aren’t themselves zero) we can make the function be literally anything at any given value, so f(x) wouldn’t be well defined. But then again clearly the set S could be defined without those properties so I’ve said nothing interesting 🙃
04:15 so... if you have a constant function f(x)=0, so it takes only rational values, so this function isn't continuous?
If it only attains a single value, then the argument that intermediate value theorem fails and so it isn't continuous doesn't apply. As soon as f(x) attains two values, then you can use that argument. Clearly the f(x) given achieve many values.