I know many of you might be wondering where I've been the last several months, and why I made a random video on Zorn's lemma rather than continuing From Zero to Geo. I made a community post about it here: ua-cam.com/users/postUgkxa6bKAm4cOHqySeTehTAnKvJOBXFv0lZD. The short answer is that I will be continuing From Zero to Geo soon, and hopefully as time progresses, videos should start coming out at a faster rate.
This is maybe the best presentation of ZL that I have seen. I wish I had seen this video in 1986 when I was trying to get my head around Rings & Modules at Cambridge. As often, there, there was a lack of motivating examples. I just wanted to come back on your comment that "the Axion of Choise is obviously true". I can't remember who said it, but one mathematician claimed that AC was obviously true, the Well Ordering Theorem was obviously false, and ZL was so arcane that it was neither obviously true or false. And I was hoping that you'd throw in a "wait a minute" or two somewhere 😉. Look forward to seeing the continuation of your GA series.
Axiom of Choice gets invoked in the proof of Zorn when you define f(S)=*an* upper bound of S It is a choice function that chooses one of the elements of the set of upper bounds and it requires AoC since Zorn can be shown to be equivalent and the fact that there are models of ZF not(C) I agree that I see AoC as "obviously true", but I find the implications of it's existence and negations fascinating
Nah, AoC is absolutely not obviously true when it comes to uncountable infinite sets. Like let's say you want to choose a vector for every real number. You start with zero, and then what? I just don't understand even how you COULD do that in the abstract
@@tylerfusco7495 I know what you are getting at, I have had this discussion before, but just FYI for a lot of common vector spaces I can think of a choice function with that requirement, let V be a non-trivial, real or complex valued vector space, let e be a non-zero vector in V, define our choice function f:R->V as f(x)=x*e My go-to example that choice lets us make weird sets/choices is choosing a representative for each equivalence class of the relation x~y iff x-y is rational for x and y being real numbers The set of representatives is non-measurable
@@TheLuckySpades The function g(x)=x*e isn't a choice function, though. (I'm using g instead of f to distinguish it from f(S) = a strict upper bound of S.) A choice function takes non-empty subsets of the set A and maps those to elements of A contained in the input subset. In other words, c:𝔓(A)\{∅}→A s.t. c(S)∈S ∀S∈𝔓(A)\{∅}. (𝔓(A) being the power set of A.) With a choice function, it's easy enough to construct a function from chains with a strict upper bound to such an upper bound: f(S) = c({b∈A:a
@@mjeffery if we are going to be picky about language then it is important to note that the choice function does not require the whole powerset of some set as it's dlmain, it only needs to be a non-empty subset of the powerset (rather in set theory since there are only sets as elements any set has a choice function with it as it's domain), else the function in the video is not a choice function (upper bounds need not be parts of the chains), nor would a function that chooses representatives of equivalence classes be one, nor would the selection of an element in an infinite Cartesian product (these are all equivalent to choice functions) And my vagueness with the usage was because they asked how we could assign a vector to each real, if you need it rephrased to the strict version, consider the set \{\{x\}×V|x\in \mathbb{R}|}, then my choice function defined on that set could be is c(\{x\}×V)=(x,x*e), which is indeed an element of \{x\}×V, you can then use the map that send x to \{x\}×V, plug that into the choice function, then apply the map that sends (x,v) to v to get the function I described That seems a bit very tedious and overly technical for a youtube comment I wrote while in a bus, so I shortcut a bunch and was maybe a bit vague and could have used air quotes for "choice function" in the first case
@@TheLuckySpades I'm using a different definition of choice function than you just described, but the distinction between them is technical, not substantial. Either way, the function in the video isn't a choice function. It can be constructed using a choice function, and so its existence is guaranteed by AC, but it isn't one itself. OTOH, a function selecting a representative element from equivalence classes would be one. But the issue with g(x)=x*e isn't technical, it's just that it has no relationship to a choice function at all. It's just a regular function. Calling it a choice function just confuses the matter. Throwing in the trivial choice function mapping a set of singletons to their elements doesn't really change that, since you can do that with any function in existence. To be fair, you were responding to a question that already conflated the existence of a function mapping ℝ→V with AC. As you demonstrated, such a function doesn't require AC at all.
I don't have an issue with doing a process infinitely , but using this to define the Real numbers feels very dubious since even if you _do_ add a missing irrational number infinitely many times, you will still be missing _100% of the irrational numbers!_ I do not see how this would work with uncountably infinite sets. Countably infinite sets, yes, but not UNcountably infinite ones. (After watching, I'm more willing to accept this, though it still feels somewhat dubious.) Maximality was defined as for all a, M is not less than a. What if a set has multiple such objects? If I have a set {x, y, z} and define a partial ordering where x < y and x < z, then we have two chains that both have different upper bounds, and two elements that are not less than other element, including each other, while also being different elements. My best guess is that it's really saying that every partially ordered set has _at least_ one maximal element, and not that it has _exactly_ one maximal element. Ok, this was touched on near the end with m = f(A_
That's right, Zorn's lemma does not give the uniqueness of Maximal element. This can be easily shown from the problem "every set can be well-ordered". The ordering 1≤2≤3≤4≤5≤6 is a well-order, and 2≤1≤4≤3≤6≤5 is also a well-order.
@@tylerfusco7495 you don't "believe" anything in maths, you just derive conclusions from your assumptions (following the rules of logic you decided to use). If you don't want to assume AC, that's completely fine, but it isn't a question of whether you believe in it or not.
@@enderstephSo, kinda yes, but also, eh, I think that overstates things slightly? It’s common to take a mathematical Platonism type view, and, I think for many people it would be reasonable to describe them as believing or disbelieving some axiom. Especially things like Con(PA)? Like, yes, assuming that Peano Arithmetic is consistent, then both PA+Con(PA) and PA+not(Con(PA)) are self-consistent axiom schemas, but I think most people could be reasonably said to “believe Con(PA)”.
I've JUST about finished my own deep dive into Zorn's Lemma, AoC, orderings, and cardinality, starting from "f is a surjection iff there exists right inverse," a statement I find more intuitive than either Zorn's Lemma or the traditional AoC (it's also more categorical which is a nice bonus!). Glad-- well, as glad as you can be-- I'm not the only one who's been dissatisfied with the way Zorn's lemma/AoC is normally introduced.
I don't know. I don't find it intuitive, as it is a synonym of 'The preimage mapping of every function has a selection' and thus much less transparent than 'Every multifunction has a selection' (the AoC).
@@pedroteran5885 Totally fair. I find intuition is pretty dependent on context, and I haven't approached many issues where a choice funciton is the natural object to define (not that it isn't helpful, I just don't find it to be obvious). I'm definitely more algebraically minded, and I think the form "surjective iff right inverse" not only looks more algebraically clean, being phrased in terms of (partial) inverses, but is also motivated well by the dual statement "injective iff left inverse," which is easily provable. Thinking just in terms of function behaviour, I also think it's obvious that if you map fully onto a target set that you should be able to (partially) work that process backwards. But again, intuition isn't uniform across the board. I'm unsure of where the form "every multifunction has a selection" shows up most naturally, but I'm sure I'd happily work with that perspective in the appropriate context.
@@sudgylacmoe "Implications are right-associative" This is a convention not a rule. And it is a point on which mathematicians and logicians often differ. Most logicians (especially those working in type theory) take A→B→C to mean A→(B→C) which as you say is equivalent to (A∧B)→C, while most mathematicians take A→B→C to mean (A→B)∧(B→C) which is different. Whenever there are multiple interpretations possible, it is always best to use parentheses for clarity.
@@twwc960 I don't think so. _Those_ mathematicians usually don't use → for such notion, but ⇒. And it is not a connective (operator on formulas giving formulas), but a symbol of a relation between formulas. Think + vs
This video is pretty repetitive, even infinitely so, you can object to this by saying you can't repeat anything in the video infinitely times, and well, you're right, but the video must eventually end, and we'll need a Zorns lemma to prove it!
at some point it's just word salad; "every set endomorphism from a nonisometric vector space to a splitting field over ℚ is an endoisododecamorphism of functors" theorems dreamed up by the utterly deranged
@@Ganerrrit's like they want their stuff to be as non-practical and obfuscated as possible. Mathematicians self-gatekeep their stuff but want to be recognized so bad. Make up your goddamn minds.
@@macchiato_1881 unfortunately it would be immeasurably less practical to use simpler words to describe these concepts every time. nobody in their right mind, when referring to an archimedean ordered field, would want to say "a set equipped with two operations which each take two inputs chosen from that set and return an output also from that set, such that for either operation swapping the inputs for each other does not affect the output; such that for either operation and for any three inputs, applying that operation on the first two and then applying it again to the output and the third, gives the same result as applying that operation to the last two and then applying it again to the output and the first; such that for each operation there exists a special element of the set which when operating with that element and any other element results in that other element; such that for each operation and for each element of the set, there exists another element for which applying that operation to those two elements results in the aforementioned special element for the operation used, with a single exception in one of the operations if the element in question is the aforementioned special element of the other operation; such that for any three elements of the set, applying the operation which does not have such an exception to the latter two and then applying the operation that does have such an exception to the output and the first, gives the same result as first applying the latter operation to the first and second elements and to the first and third elements separately, then applying the former operation to those two outputs; equipped with a relation under which every pair of elements is labeled either "the first element of the pair is greater" or "the first element of the pair is not greater", such that every element is not labeled as "greater" than itself; such that for any three elements of the set, whenever the relation labels the first element as "not greater" than the second and it labels the second element as "not greater" than the third, it also labels the first element as "not greater" than the third; such that for any two elements, if the relation labels the first as "not greater" than the second but also labels the second as "not greater" than the first, it is the case that the two are the same element; such that for any two elements of the set, the relation does not label both as "greater" than each other; such that for any two elements of the set, when first applying the former operation to the special element of the latter as both inputs and then continuing by repeatedly applying the former operation to the output and the special element of the latter, you will always come across an element as a result of that process such that the relation labels it "greater" than the result of the operation with the exception when applied to the two given elements" when they could just say "an archimedean ordered field". and, yes, the definitions have to go that deep and specific, because the entire point is to be able to use simple logical deductions to show exactly what conclusions follow when you're working with something that can be modeled in terms of a given concept.
@@macchiato_1881yeah it’s impractical because set theory wasn’t made to be “practical” it was made to prove all of maths with the least amount of axioms (basically things that are true because we say so) that’s why set theory has EXACT (and arguably hard to understand) definitions and dense notation so that it can be as unambiguous as possible
@@macchiato_1881 You received a good answer from rarebeeph1783, though I don't know if you can see it (since UA-cam seems to be hiding that response unless you go to "Newest First"). But to reiterate rarebeeph's point, let's say you're in a geometry class. Would it get annoying to you to constantly have to write phrases like "the collection of all points that are a fixed distance from a given point". After a while, you might want to just give that concept a name, such as "circle". Because it's much easier to repeatedly use the word "circle" than repeatedly use the phrase "the collection of all points that are a fixed distance from a given point". Now, just take this to its natural conclusion when continuing to be precise about mathematics.
I wonder what the maximal of the Zorn's colour palette is. I guess that it could refer to the set of pigments used for the basis colours, which are a basis for the space of colours that one can obtain using Zorn's colour palette.
Great video! I'm always glad whenever someone takes the time to examine Zorn's lemma. It's honestly very intuitive when you dig into the details, but people just like to joke about it. As a mild critique, I think this video would have been improved if you had made a supporting video specifically about transfinite recursion. Each time you point at an infinite procedure and claim that they are "not possible", I kept thinking "of course it's possible! That's exactly what transfinite recursion is for". Especially when you get around to 28:09, that's a perfect place to invoke transfinite recursion. You already set up the choice function, and outlined the exact mechanism of the recursion, so you could just invoke TR and stop the proof right there. The proof you offer is almost exactly how TR is proven, too, but I feel the demonstration is weakened and cluttered due to the surrounding context of the video. What I'm getting at is that you can't do anything interesting in set theory without occasionally stubbing your toe on the ordinals. Might as well bite the bullet and address them directly.
13:10 for some reason the font you use causes a regular M (look at Maximum and Maximal) to have big spikes as if two of the vertices defining the letter are shifted up by like half a letter height. This is not the case for the cursive green Ms, just for the three regular white Ms.
That's not the font's fault, it's manim's fault and I have no idea how to fix it. Hopefully this is my last video made with manim so I won't have issues with this in the future. I call that manim's devil horns.
28:39 : I’m a little unclear on what the order relation here is? What you had before at 28:15 looks to me like a_\alpha = f({a_\beta | \beta < \alpha }) or something like that (where \alpha and \beta are ordinals). Is that right? 30:17 : ohh, right, these a_i are all coming from a partially ordered set A, and the order relation being used is the one on A. I should have gotten that.
I really like the PDFs you crafted. What references did you use for the PDF you wrote constructing the real numbers? I think it will be a fun exercise for me to phrase it with the language of category theory. My mind is still blown by the fact that Archimedian fields are always embeddable in P(N). Great video
I came up with the real number construction on my own. I've always been interested in constructions of the real numbers (especially because the construction is one of the compilation bottlenecks in my Coq code so I have a practical reason to want to make it better), and I got to thinking if you could construct them by starting with the rationals and doing a bunch of field extensions. Over time I developed that idea into that construction. Sadly it ended up not speeding things up though because like all constructions of the reals, it gets way more complicated than you think it should.
@@sudgylacmoe Wow, it is impressive that you came up with a construction of the reals yourself. Because you said you are interested on such constructions, what other ways do you know to define R besides [equivalence classes of Cauchy sequences of rationals] or through [Dedekind cuts] (or the one you devised here)?
Those are the three good ones. I'm also interested in the Eudoxus reals, and in constructing them as a quotient of the hyperrational numbers. I found a document containing a bunch of constructions here: arxiv.org/pdf/1506.03467
hey nice video! you've been doing a pretty good job at these explainers! I'm a little worried that some of the informal exposition here may be running together the principle of transfinite induction and the axiom of choice? it's transfinite induction that allows us to do those infinitary constructions, but it's choice that tells us they "conclude" in the right way. this framing of the issue also makes the passage to explicitly using Zorn's lemma a little simpler: you don't have to change the flow of information in the proofs, you just have to add on a little bit showing that each stage (successor and limit) in the construction gives a bounded chain. (the tricky part here, of course, is that transfinite induction as it's often employed invokes choice at some point, but technically it works on any set you already know to be well-ordered) also the bit about the construction of R seems a little confusing to me. like, you never said why you need the completeness of R. because the reals are the unique complete ordered (Archimedean) field, right? or is there something more subtle here that I'm missing? (also also, glancing at your writeup, it seems like the chain construction is "just" a colimit in the category of archimedean ordered fields, so there may be some overpowered proof in the vicinity which proceeds like: the category of AOF has all colimits of some nice shape (that is, chains), and so by Zorn's lemma applied to this posetal category, it has some "maximal" object. so you might also consider applying this kind of procedure to similar-seeming categories.)
Hello. I think it would have been better to dive into one specific example and to be more thorough with it. In my opinion the whole video felt a little bit rushed because of this. For a video that was meant to Demistify Zorn's lemma, I feel like this video left me more confused than enlightened. I also think the fornat of explaning the three problems, and only then "solving" the three problems is not very good, since you can forget details/get overwhelmed. For reference, I have a master's degree in physics and I am generally curious about math, and do some independent learning,so I feel like I'm probably more familiar with these topics than most.
This is one of the issues that I mentioned at the beginning of the section that I said I would pretend didn't exist. There is no set of all Archimedean ordered fields. Thankfully the solution here is simple. You can prove that all Archimedean ordered fields have cardinality of at most 2^|ℕ|, so you can restrict yourself to Archimedean ordered fields made on subsets of the power set of the natural numbers. This does form a set.
copied directly from Wikipedia In mathematics, informal logic and argument mapping, a lemma (pl.: lemmas or lemmata) is a generally minor, proven proposition which is used as a stepping stone to a larger result. For that reason, it is also known as a "helping theorem" or an "auxiliary theorem". In many cases, a lemma derives its importance from the theorem it aims to prove; however, a lemma can also turn out to be more important than originally thought
Max Zorn was a mathematician who worked at UCLA and Indiana State University. He was not the first person to come up with his eponymous lemma ("lemma" having been explained in a previous reply), but apparently a lot of people associated the result with him, so it became known as "Zorn's Lemma". I had a logic professor in grad school who met Max Zorn, and my professor remarked that Zorn disliked the result being called "Zorn's Lemma". Apparently, Zorn would retort, "It's not mine, and it's not a lemma."
"Relegating to obscure jokes" is an oxymoron. A good joke is not a relegation, it is an elevation. A good joke (even if "so bad it is funny") is what is missing, a crucial missing ingredient, in all known mathematics texts that are for sale.
You should at least mention why Zorn's lemma works with the contruction of the real numbers, it not being a proper class, I don't t refering to a paper in the description is appropriate here.
You were doing great but failed to show (in the vector space example) that the union of the chain is linearly independent. If it's not, then there's a dependence relation that must already be true in one of the elements of the chain, contradicting the linear independence of each member of the chain. Without that detail, your proof fails. Good exposition overall though.
I explained at 17:26 that you do need to prove that, but I skipped it for the sake of time. I generally skimmed the three proofs because as I said in the video, the overall structure of the proof is what's important here, not the particular details.
does this show that the upper bound of every infinite chain must be a maximal element? The choice function f should be able to generate any chain you choose, right?
Great video! I wish you had explained why the maximal well-order was still a well-order. I guess this is one of the details that you didn't include in the video. Oh well, you know what they say: "When Zorn gives you lemmas, …"
Even the document that I got the proof from didn't explain that step since it's "trivial". Here's a quick proof: Let A be a nonempty subset of C. Because A is nonempty, there exists some x ∈ A ⊆ C, and because x ∈ C, that means there is some conforming set S with x ∈ S. Then A ∩ S is a nonempty subset of S, and because S is well-ordered, it has some least element a. If there was some y ∈ A with y < a, we would have y ∈ S as well because all conforming sets start with the same elements. Thus, y would be in A ∩ S, contradicting a being the least element of A ∩ S. Thus, a is the least element of A.
I'm reading old comments and I just realized that I think I misunderstood your question! Were you asking about the step in the proof of the well-ordering theorem? The maximal well order is still a well order by definition. Our partially ordered set there was the set of well orders on subsets, so any element of it will be a well order. Zorn's lemma gives us a maximal element of that set of well orders, so the maximal well order is a well order.
When pure mathematicians say that Zorn's lemma is „practical“, what they really mean is that it’s a convenient way not to think deeply about minor details like actually constructing something algorithmically so we might use it in the real world. The best arguments I’ve seen so far for why we might want to do such a thing is bc it allows us to „construct“ stuff like non-measurable subsets of R and thus prove that we need to be a bit thoughtful in defining measures. But if we had cared about actually algorithmically computing measures from the beginning, we just wouldn’t have run into difficulties like banach-tarski. I for one don’t really care for the axiom of choice, because it implies excluded middle which… ewww, you know? It doesn’t help me in the slightest to know abstractly that a Turing machine must either stop or loop indefinitely, I wanna know which of the two options it is!
"But if we had cared about actually algorithmically computing measures from the beginning, we just wouldn’t have run into difficulties like banach-tarski." This result is strictly weaker than ZL. It's provable from, say, Hahn Banach. And you _can't_ do functional analysis without Hahn Banach. The mere need to have Hahn Banach implies this "difficulty". And there's just no _reason_ to compute measures in the first place. Requiring that axiom leads to horrific results. If you do it, deny AC, accept LM instead, guess what happens? Then you can partition the real numbers into _strictly_ more non empty parts than there are real numbers. A result far worse than BT. Like, sorry that non constructive proofs exist. But they do. It's impossible to do like half of analysis without them.
@@najawin8348 maybe like half of analysis is useless for practical purposes then? Constructive practically useful. Also, no one’s saying we should require eg all measures to be computable. What I am saying is: let’s just not talk about non-computable ones. Let’s just not use „theorems“ that depend on choice. You may not get all the results that look so neat, but at least you’re not lying to the engineers who actually try to solve real world problems using mathematical CONSTRUCTIONS. Or here’s an idea: at the very least state in those theorems that you need a computable choice function of appropriate type. That immediately makes the theorem constructively valid. You just have to do some extra work to be able to apply it. And while proving other theorems based on that, you can just pass that burden on to the next guy until someone actually wants to apply it to something concrete to solve a real world problem. There’s really zero excuse not to do it other than laziness.
@@methatis3013 because when you solve real world problems, you actually care about finding the solution rather than being happy that it exists in the platonic sphere of ideas?
I prefer a good rhyme. What's brown and can be tattooed on your forehead in case you forget you can pick an element of a set? Zorn's henna. Or: What's weird and goes in an espresso of choice suitable for drinking with the experimental film by Hollis Frampton? Zorn's crema. ua-cam.com/video/zDQfAUxQdvI/v-deo.html
In the proof of Zorn's Lemma, when you say the "set" C is the set of all conforming sets, I conclude C must contain itself. This fact reminds me strongly of Russell's Paradox, although it is not entirely related. I wonder if we are even allowed to call C a set then.
C is not the set of all conforming sets, it's the union of all conforming sets. That's a big difference. The set of all conforming sets does exist but it's not conforming.
Ok, so I saw the thumbnail and it concerned me that you seemed to suggest that Zorn’s Lemma is unrelated to the Axiom of Choice. I’m glad that by the end of the video you pointed out that there’s a usage of the Axiom of choice in your proof; while you let the viewer try to figure that out for themselves, the main problem I have with your presentation is that this “correction” of your headline on the thumbnail comes way too late.
so hm, that's all good but what's the justification for why zorn's lemma should be intuitive enough that it essentially can be taken as an axiom? (this is particularly important if you do not consider the well-ordering theorem nor the axiom of choice as intuitive). why when studying for example vector spaces, do we assume all vector spaces must have a basis, instead of being satisfied with the fact that most of our theorems are about vector spaces that happen to have a basis?
for some time, people thought the axiom of choice (equivalently Zorn's lemma) was a theorem that could be proved using the other axioms of set theory. it is an axiom because it can or cannot be used in your reasoning, not because it is intuitive enough that it doesn't necessitate being proved. the thing is, there is just a significant enough number of results that rely on AoC for it to be commonly accepted as true.
I don't understand the confusion. partially ordered - like the y component of a parabola of the form, y = ax^2 +bx +c not all pairs of elements will be ordered, but some will. for instance, the roots of this polynomial are not ordered, but if there are two roots, then they are ordered with the vertex. every chain has an upper bound - given any ordered subset, there is a maximal element within it the set has a maximal element - if all ordered subsets have a maximal element, then one of those must be the maximal element of the whole set what can possibly be confusing about that unless you're illiterate or very stupid? and in either of those cases, what are you doing at this level of mathematics? you should be in a kindergarten class to learn the names of colors and not to eat paste.
I guess I shouldn't be surprised, though, since none of you idiots understand how exponents work. at best you claim that a^b is b copies of a multiplied together, but this doesn't explain what happens when b = 0 at all, so you make up some extra rule about that. but then that rule doesn't seem to work when a = 0, so it's an open question what 0^0 is. and all of this is separate from what happens when b is Complex. not to mention the fact that if b is Rational then you predict there are as many solutions as the denominator of the reduced Rational expression of b, but this has absolutely no extension to irrational values for b, where you only recognize 1 solution. note that the 'a^b is b copies of a multiplied together' is the basis of the definition of all hyperoperations, but it's not even possible to take fractional hyperoperations under this definition. so... yeah, you're necessarily just very stupid.
after watching your video it's clear that you use lots of terms you do not understand. you should probably stop doing that. to start, let's consider something that you definitely take for granted that you understand, but in fact do not, numbers. you consider things like 7, -7, 7i, -7 +7i, 3/7, etc. to be numbers. you also consider |-7| = 7, and this 7 is identical to +7. however, you don't consider 7 elephants, 7 ounces of water, or 7 poems to be numbers. with these, only the 7 part is a number. so now let's ask the question why does the unit sometimes count as being part of a number and sometimes not? -7 +7i is not only considered to be a number, but also a vector, because it is a combination of magnitudes and units. those magnitudes being 7 and 7, and the units being -1 and i. but now, if this -7 is a combination of unit and magnitude, and thus a vector, then isn't -7 always a vector? and if this is the case, then isn't 7 elephants a vector, and that's why it's not a number? so why are vectors sometimes numbers and sometimes not? are you beginning to see that you have no idea wtf you're talking about? now, if |-7| = 7, and the absolute value bars are isolating the magnitude of the vector -7, then clearly -7 is not a number, because numbers are vector magnitudes, not whole vectors, but to get 7 from |-7| we specifically removed the negative unit to isolate the magnitude. but it's worse than that, because we didn't really remove the negative unit, we just converted it to a positive unit, because by definition the Naturals, like 7, are identical to the positive Integers, like +7. so |-7| doesn't isolate the magnitude of the vector -7 at all. and this then plays into Godel Incompleteness being a property of all mathematical systems which incorporate the Naturals, since Godel Incompleteness arises specifically from a condition where the Naturals can be randomly converted between an ordinal, like +7, and a nominal, like the bare unsigned number 7. this is occurring due to an inconsistency in the definition of the Naturals, and yet everyone agrees that systems which build on the Naturals, like Peano Arithmetic, are consistent... that's obviously wrong. and you call this rigor? if you're still lost, consider why addition is Abelian: 3 +4 = 4 +3 = 7; this is true because there is an unstated grammar rule in mathematical orthography which renders expression-initial + optional we could instead write this as: +3 +4 = +4 +3 = +7 but now that we're doing this, we can see why subtraction is non-Abelian: 3 -4 = -4 +3 = -1; unlike the + unit specification, - is always obligatory we could instead write this as: +3 -4 = -4 +3 = -1 this means that neither addition nor subtraction are actually operations at all. the symbols we're writing and calling 'addition' and 'subtraction' are always unit specifiers, and things prefixed with them are simply listed together, in an unordered list. like a shopping list, it doesn't matter if you get milk before you get oranges, it just matters that you get them both. thus it is this hidden operation of listing vectors which is Abelian, not addition, because addition is not an operation at all. similarly, subtraction is not Abelian, but the reason is that it's not an operation, not because it's a non-Abelian operation. we can rewrite all of these in set notation, for clarity: 3 +4 = 4 +3 = {+3, +4} = {+4, +3} = {+7} 3 -4 = -4 +3 = {+3, -4} = {-4, +3} = {-1} and while you'll probably attempt to object, that is precisely what those set definitions mean. which means that the Axiom of Extension is wrong... but we already knew that from our previous analysis of the Naturals, since for Godel Incompleteness to work it must be the case that the Successor Function both generates the Naturals, and does not generate the Naturals, which is a blatant violation of the Axiom of Extension that is canonically accepted as perfectly acceptable by all who mean to accept the Axiom of Extension as inviolable. now take this, and reconsider your discussion of vector spaces... you have absolutely no idea wtf you're talking about, at all.
oh yeah, why do you specify 'a random non-zero vector'? by definition, zero is not a vector, since it has neither magnitude nor unit. every dimension has zero, but zero has no dimension. this is why most proofs have to use special pleading for the 'trivial case' of zero... because it should never have been included in most of the things that it's included in. for instance, how can you claim that the Rationals are Archimedean when zero is Rational? by definition, no set including zero can be Archimedean. when you're this dumb, you shouldn't be trying to teach people.
you think there are Transcendental Reals, don't you? take a look at the definition of e: e = lim n->inf (1 +1/n)^n now, since n is approaching infinity, the 1/n part can be replaced with an infinitesimal, c, yielding: e = lim n->inf (1 +c)^n if we now grow n via incrementation n must always be an Integer, which allows us to apply the binomial theorem, giving: e = lim n->inf 1 +Ac +Bc^2... +Wc^(n -1) +Xc^n since this limit will only be infinitesimally far away from e, we can remove the limit condition by simply adding c: e = 1 +(A +1)c +Bc^2... +Wc^(n -1) +Xc^n all of the coefficients and exponents in this expansion of e are Integers, which means that this is a non-Real Levi-Civita series. that is, e is not a member of the set of Reals because its definition includes irreducible infinitesimal terms. because infinitesimals are non-Real, the Levi-Civita field is a field extension of the Reals which adds the infinitesimals back in. but something that nobody noticed is that the definition of a Transcendental is that it includes an irreducible infinitesimal component, which means that absolutely no Transcendentals can be Real by definition. oops. frankly this should be obvious, since if the Reals are Complete as Dedekind claimed... what does the Levi-Civita field include that the Reals don't already? under the assumed definition of what a Dedekind Cut does, there shouldn't be any way for the Levi-Civita field to extend the Reals. or, for that matter, for the surreals, or hyperreals to extend the Reals. and we definitely shouldn't be finding familiar values, like e, which are naively defined to be Reals, within these extensions. the definitions of everything you work with are complete nonsense. if I were you I'd be having a panic attack right about now, since all this stuff you think you understand is actually just mythology and fairy tales.
@@sumdumbmick слушай, не хочешь в мой клуб альтернативно одаренных? я знаю некого рамзеса который с радостью обсудит с тобой эти твои выкрутасы с числами
I find something nostalgic in hearing someone disregard the axiom of choice as "obvious", and sparring no "qualms" over it. Studying advanced set theory is a sobering experience, I don't recommend it 🥲
I'll admit that I used to have the "default" opinion of "I don't really know about the axiom of choice and I'll just use it only when I need it". However, as I have learned more math and studied some foundations, I've come to the conclusion that the axiom of choice belongs in logic as a fundamental law, on the same level as the law of noncontradiction and the law of the excluded middle. Basically, I don't hold my current opinion because I haven't studied enough. I hold this opinion because I have studied enough.
@@sudgylacmoe I appreciate the response (and I apologize if I implied that you didn't know enough). But my comment was about advanced set theory specifically, things like determinacy, V=L, Solovay’s Model, countable/dependent choice… (there’s a lot weirder but you might know these). The short of it is you can find a continuum of statements in set theory and how they relate to the AoC. Deciding what “obviously” is true is a gruesome task to torture the most ardent platonists. Not to mention the Suslin line, diamond principle, Martin axioms and much more. I think anyone subjected to this will come out marked by a strong sense of relativism, like “well… I don’t think some axioms are really true or false, adding them leads to different theories and that’s it.”. To save the best for last, logic is the worst place to argue about AoC, logic thrives in the paradigm of constructivism (in particular proof theory). I can’t imagine any natural motivation from logic for the axiom of choice (I am eager to be proven wrong).
@@alexismiller2349 Type theory provides the simplest statement of choice I've ever seen: ∀ A : Type, inhabited A → A. inhabited A is just an inductive type that has a single constructor that takes in a value in A, so to me this statement is incredibly obvious, since you needed a value in A to get inhabited A to begin with. I see this formulation of choice as much more logical, intuitive, simple, and obvious.
Thanks for the answer... unfortunately I'm not so familiar with type theory, and I don't see the link between what you wrote and AoC. Also I couldn't find anything similar on nLab and mathstackexchange, do you have a link to anything that explains your answer ?
I came up with the formulation myself for use in Coq, but I have since learned that it's what Lean uses for choice as well, so you could look into people describing Lean's axioms. The idea behind choice is extending when exactly you can get a value satisfying a predicate. In normal logic you can only do that with a finite number of existential statements, and choice is what you use to extend it to an infinite number of such statements. But people obfuscate the axiom by describing it in terms of sets when the fundamental idea behind it has nothing to do with sets.
I know many of you might be wondering where I've been the last several months, and why I made a random video on Zorn's lemma rather than continuing From Zero to Geo. I made a community post about it here: ua-cam.com/users/postUgkxa6bKAm4cOHqySeTehTAnKvJOBXFv0lZD. The short answer is that I will be continuing From Zero to Geo soon, and hopefully as time progresses, videos should start coming out at a faster rate.
Good luck, glad to have the shorts in the mean time too
Ah! I missed that. Thanks for linking it!
just discovered this channel, great content
I have never seen a wolfram wiki page that unserious. My god lol
The great thing about this video is that it teaches exactly enough for you to understand the entire video
WHO LET ZORN COOK AGAIN???
This is maybe the best presentation of ZL that I have seen. I wish I had seen this video in 1986 when I was trying to get my head around Rings & Modules at Cambridge. As often, there, there was a lack of motivating examples.
I just wanted to come back on your comment that "the Axion of Choise is obviously true". I can't remember who said it, but one mathematician claimed that AC was obviously true, the Well Ordering Theorem was obviously false, and ZL was so arcane that it was neither obviously true or false.
And I was hoping that you'd throw in a "wait a minute" or two somewhere 😉. Look forward to seeing the continuation of your GA series.
Take a look at 0:30 (upper right quote). It doesn't identify the original source, but it is a statement of the same thing.
3:27 - pi is less than tau, fact!
Watched the whole zero to geo now I hate the fact that I found this channel before the zero to geo playlist was completely uploaded
Axiom of Choice gets invoked in the proof of Zorn when you define f(S)=*an* upper bound of S
It is a choice function that chooses one of the elements of the set of upper bounds and it requires AoC since Zorn can be shown to be equivalent and the fact that there are models of ZF not(C)
I agree that I see AoC as "obviously true", but I find the implications of it's existence and negations fascinating
Nah, AoC is absolutely not obviously true when it comes to uncountable infinite sets. Like let's say you want to choose a vector for every real number. You start with zero, and then what? I just don't understand even how you COULD do that in the abstract
@@tylerfusco7495 I know what you are getting at, I have had this discussion before, but just FYI for a lot of common vector spaces I can think of a choice function with that requirement, let V be a non-trivial, real or complex valued vector space, let e be a non-zero vector in V, define our choice function
f:R->V as f(x)=x*e
My go-to example that choice lets us make weird sets/choices is choosing a representative for each equivalence class of the relation x~y iff x-y is rational for x and y being real numbers
The set of representatives is non-measurable
@@TheLuckySpades The function g(x)=x*e isn't a choice function, though. (I'm using g instead of f to distinguish it from f(S) = a strict upper bound of S.) A choice function takes non-empty subsets of the set A and maps those to elements of A contained in the input subset. In other words, c:𝔓(A)\{∅}→A s.t. c(S)∈S ∀S∈𝔓(A)\{∅}. (𝔓(A) being the power set of A.)
With a choice function, it's easy enough to construct a function from chains with a strict upper bound to such an upper bound: f(S) = c({b∈A:a
@@mjeffery if we are going to be picky about language then it is important to note that the choice function does not require the whole powerset of some set as it's dlmain, it only needs to be a non-empty subset of the powerset (rather in set theory since there are only sets as elements any set has a choice function with it as it's domain), else the function in the video is not a choice function (upper bounds need not be parts of the chains), nor would a function that chooses representatives of equivalence classes be one, nor would the selection of an element in an infinite Cartesian product (these are all equivalent to choice functions)
And my vagueness with the usage was because they asked how we could assign a vector to each real, if you need it rephrased to the strict version, consider the set \{\{x\}×V|x\in \mathbb{R}|}, then my choice function defined on that set could be is c(\{x\}×V)=(x,x*e), which is indeed an element of \{x\}×V, you can then use the map that send x to \{x\}×V, plug that into the choice function, then apply the map that sends (x,v) to v to get the function I described
That seems a bit very tedious and overly technical for a youtube comment I wrote while in a bus, so I shortcut a bunch and was maybe a bit vague and could have used air quotes for "choice function" in the first case
@@TheLuckySpades I'm using a different definition of choice function than you just described, but the distinction between them is technical, not substantial.
Either way, the function in the video isn't a choice function. It can be constructed using a choice function, and so its existence is guaranteed by AC, but it isn't one itself. OTOH, a function selecting a representative element from equivalence classes would be one.
But the issue with g(x)=x*e isn't technical, it's just that it has no relationship to a choice function at all. It's just a regular function. Calling it a choice function just confuses the matter. Throwing in the trivial choice function mapping a set of singletons to their elements doesn't really change that, since you can do that with any function in existence.
To be fair, you were responding to a question that already conflated the existence of a function mapping ℝ→V with AC. As you demonstrated, such a function doesn't require AC at all.
Dr. Zorn's first name was exactly what it had to have been: Max.
The Axiom of Choice is the maximal element of Mathematics.
I don't have an issue with doing a process infinitely , but using this to define the Real numbers feels very dubious since even if you _do_ add a missing irrational number infinitely many times, you will still be missing _100% of the irrational numbers!_ I do not see how this would work with uncountably infinite sets. Countably infinite sets, yes, but not UNcountably infinite ones. (After watching, I'm more willing to accept this, though it still feels somewhat dubious.)
Maximality was defined as for all a, M is not less than a. What if a set has multiple such objects? If I have a set {x, y, z} and define a partial ordering where x < y and x < z, then we have two chains that both have different upper bounds, and two elements that are not less than other element, including each other, while also being different elements. My best guess is that it's really saying that every partially ordered set has _at least_ one maximal element, and not that it has _exactly_ one maximal element.
Ok, this was touched on near the end with m = f(A_
That's right, Zorn's lemma does not give the uniqueness of Maximal element. This can be easily shown from the problem "every set can be well-ordered".
The ordering
1≤2≤3≤4≤5≤6
is a well-order, and
2≤1≤4≤3≤6≤5
is also a well-order.
The existence of the function f is given by The Axiom of Choice, which means we don't know the exact outputs of f. The important part is that f exist.
@@user-oe5eg5qx4chow am i supposed to just blindly believe that a function like that exists when i don't even know any of its outputs?
@@tylerfusco7495 you don't "believe" anything in maths, you just derive conclusions from your assumptions (following the rules of logic you decided to use). If you don't want to assume AC, that's completely fine, but it isn't a question of whether you believe in it or not.
@@enderstephSo, kinda yes, but also, eh, I think that overstates things slightly? It’s common to take a mathematical Platonism type view, and, I think for many people it would be reasonable to describe them as believing or disbelieving some axiom.
Especially things like Con(PA)?
Like, yes, assuming that Peano Arithmetic is consistent, then both PA+Con(PA) and PA+not(Con(PA)) are self-consistent axiom schemas,
but I think most people could be reasonably said to “believe Con(PA)”.
This is the best presentation of Zorn's lemma that I've ever seen. Congrats on this video -- it's seriously great :)
looking forward to watch new zero to geo episodes 😢
I've JUST about finished my own deep dive into Zorn's Lemma, AoC, orderings, and cardinality, starting from "f is a surjection iff there exists right inverse," a statement I find more intuitive than either Zorn's Lemma or the traditional AoC (it's also more categorical which is a nice bonus!). Glad-- well, as glad as you can be-- I'm not the only one who's been dissatisfied with the way Zorn's lemma/AoC is normally introduced.
"Every epi splits" is a nice AoC expressible in every category
I don't know. I don't find it intuitive, as it is a synonym of 'The preimage mapping of every function has a selection' and thus much less transparent than 'Every multifunction has a selection' (the AoC).
@@pedroteran5885 Totally fair. I find intuition is pretty dependent on context, and I haven't approached many issues where a choice funciton is the natural object to define (not that it isn't helpful, I just don't find it to be obvious). I'm definitely more algebraically minded, and I think the form "surjective iff right inverse" not only looks more algebraically clean, being phrased in terms of (partial) inverses, but is also motivated well by the dual statement "injective iff left inverse," which is easily provable.
Thinking just in terms of function behaviour, I also think it's obvious that if you map fully onto a target set that you should be able to (partially) work that process backwards. But again, intuition isn't uniform across the board. I'm unsure of where the form "every multifunction has a selection" shows up most naturally, but I'm sure I'd happily work with that perspective in the appropriate context.
This is an incredibly lucid presentation
Cinema right here ❤
4:43 are those supposed to be conjunctions? They don't make any sense to me as implications.
Implications are right-associative, so A -> B -> C means "If A, then if B, then C" which is equivalent to "If A and B, then C".
@@sudgylacmoe "Implications are right-associative" This is a convention not a rule. And it is a point on which mathematicians and logicians often differ. Most logicians (especially those working in type theory) take A→B→C to mean A→(B→C) which as you say is equivalent to (A∧B)→C, while most mathematicians take A→B→C to mean (A→B)∧(B→C) which is different. Whenever there are multiple interpretations possible, it is always best to use parentheses for clarity.
@@twwc960 I don't think so. _Those_ mathematicians usually don't use → for such notion, but ⇒. And it is not a connective (operator on formulas giving formulas), but a symbol of a relation between formulas. Think + vs
This video is pretty repetitive, even infinitely so, you can object to this by saying you can't repeat anything in the video infinitely times, and well, you're right, but the video must eventually end, and we'll need a Zorns lemma to prove it!
This video is pure gold. Thank you so much !
Mathematicians just like making words up dont they
at some point it's just word salad; "every set endomorphism from a nonisometric vector space to a splitting field over ℚ is an endoisododecamorphism of functors" theorems dreamed up by the utterly deranged
@@Ganerrrit's like they want their stuff to be as non-practical and obfuscated as possible. Mathematicians self-gatekeep their stuff but want to be recognized so bad. Make up your goddamn minds.
@@macchiato_1881 unfortunately it would be immeasurably less practical to use simpler words to describe these concepts every time.
nobody in their right mind, when referring to an archimedean ordered field, would want to say
"a set equipped with two operations which each take two inputs chosen from that set and return an output also from that set, such that for either operation swapping the inputs for each other does not affect the output; such that for either operation and for any three inputs, applying that operation on the first two and then applying it again to the output and the third, gives the same result as applying that operation to the last two and then applying it again to the output and the first; such that for each operation there exists a special element of the set which when operating with that element and any other element results in that other element; such that for each operation and for each element of the set, there exists another element for which applying that operation to those two elements results in the aforementioned special element for the operation used, with a single exception in one of the operations if the element in question is the aforementioned special element of the other operation; such that for any three elements of the set, applying the operation which does not have such an exception to the latter two and then applying the operation that does have such an exception to the output and the first, gives the same result as first applying the latter operation to the first and second elements and to the first and third elements separately, then applying the former operation to those two outputs; equipped with a relation under which every pair of elements is labeled either "the first element of the pair is greater" or "the first element of the pair is not greater", such that every element is not labeled as "greater" than itself; such that for any three elements of the set, whenever the relation labels the first element as "not greater" than the second and it labels the second element as "not greater" than the third, it also labels the first element as "not greater" than the third; such that for any two elements, if the relation labels the first as "not greater" than the second but also labels the second as "not greater" than the first, it is the case that the two are the same element; such that for any two elements of the set, the relation does not label both as "greater" than each other; such that for any two elements of the set, when first applying the former operation to the special element of the latter as both inputs and then continuing by repeatedly applying the former operation to the output and the special element of the latter, you will always come across an element as a result of that process such that the relation labels it "greater" than the result of the operation with the exception when applied to the two given elements"
when they could just say "an archimedean ordered field".
and, yes, the definitions have to go that deep and specific, because the entire point is to be able to use simple logical deductions to show exactly what conclusions follow when you're working with something that can be modeled in terms of a given concept.
@@macchiato_1881yeah it’s impractical because set theory wasn’t made to be “practical” it was made to prove all of maths with the least amount of axioms (basically things that are true because we say so) that’s why set theory has EXACT (and arguably hard to understand) definitions and dense notation so that it can be as unambiguous as possible
@@macchiato_1881 You received a good answer from rarebeeph1783, though I don't know if you can see it (since UA-cam seems to be hiding that response unless you go to "Newest First").
But to reiterate rarebeeph's point, let's say you're in a geometry class. Would it get annoying to you to constantly have to write phrases like "the collection of all points that are a fixed distance from a given point". After a while, you might want to just give that concept a name, such as "circle". Because it's much easier to repeatedly use the word "circle" than repeatedly use the phrase "the collection of all points that are a fixed distance from a given point".
Now, just take this to its natural conclusion when continuing to be precise about mathematics.
I wonder what the maximal of the Zorn's colour palette is. I guess that it could refer to the set of pigments used for the basis colours, which are a basis for the space of colours that one can obtain using Zorn's colour palette.
At 19:26 I suddenly needed to stop the video and have a little lie down. I am waking to a new world here.
Great video! I'm always glad whenever someone takes the time to examine Zorn's lemma. It's honestly very intuitive when you dig into the details, but people just like to joke about it.
As a mild critique, I think this video would have been improved if you had made a supporting video specifically about transfinite recursion. Each time you point at an infinite procedure and claim that they are "not possible", I kept thinking "of course it's possible! That's exactly what transfinite recursion is for". Especially when you get around to 28:09, that's a perfect place to invoke transfinite recursion. You already set up the choice function, and outlined the exact mechanism of the recursion, so you could just invoke TR and stop the proof right there. The proof you offer is almost exactly how TR is proven, too, but I feel the demonstration is weakened and cluttered due to the surrounding context of the video.
What I'm getting at is that you can't do anything interesting in set theory without occasionally stubbing your toe on the ordinals. Might as well bite the bullet and address them directly.
13:10 for some reason the font you use causes a regular M (look at Maximum and Maximal) to have big spikes as if two of the vertices defining the letter are shifted up by like half a letter height. This is not the case for the cursive green Ms, just for the three regular white Ms.
That's not the font's fault, it's manim's fault and I have no idea how to fix it. Hopefully this is my last video made with manim so I won't have issues with this in the future. I call that manim's devil horns.
@@sudgylacmoe 😈
28:39 : I’m a little unclear on what the order relation here is?
What you had before at 28:15 looks to me like a_\alpha = f({a_\beta | \beta < \alpha }) or something like that (where \alpha and \beta are ordinals). Is that right?
30:17 : ohh, right, these a_i are all coming from a partially ordered set A, and the order relation being used is the one on A. I should have gotten that.
I really like the PDFs you crafted. What references did you use for the PDF you wrote constructing the real numbers? I think it will be a fun exercise for me to phrase it with the language of category theory. My mind is still blown by the fact that Archimedian fields are always embeddable in P(N). Great video
I came up with the real number construction on my own. I've always been interested in constructions of the real numbers (especially because the construction is one of the compilation bottlenecks in my Coq code so I have a practical reason to want to make it better), and I got to thinking if you could construct them by starting with the rationals and doing a bunch of field extensions. Over time I developed that idea into that construction. Sadly it ended up not speeding things up though because like all constructions of the reals, it gets way more complicated than you think it should.
@@sudgylacmoe Wow, it is impressive that you came up with a construction of the reals yourself.
Because you said you are interested on such constructions, what other ways do you know to define R besides [equivalence classes of Cauchy sequences of rationals] or through [Dedekind cuts] (or the one you devised here)?
Those are the three good ones. I'm also interested in the Eudoxus reals, and in constructing them as a quotient of the hyperrational numbers. I found a document containing a bunch of constructions here: arxiv.org/pdf/1506.03467
@@sudgylacmoe nice, thank you
hey nice video! you've been doing a pretty good job at these explainers!
I'm a little worried that some of the informal exposition here may be running together the principle of transfinite induction and the axiom of choice? it's transfinite induction that allows us to do those infinitary constructions, but it's choice that tells us they "conclude" in the right way. this framing of the issue also makes the passage to explicitly using Zorn's lemma a little simpler: you don't have to change the flow of information in the proofs, you just have to add on a little bit showing that each stage (successor and limit) in the construction gives a bounded chain. (the tricky part here, of course, is that transfinite induction as it's often employed invokes choice at some point, but technically it works on any set you already know to be well-ordered)
also the bit about the construction of R seems a little confusing to me. like, you never said why you need the completeness of R. because the reals are the unique complete ordered (Archimedean) field, right? or is there something more subtle here that I'm missing? (also also, glancing at your writeup, it seems like the chain construction is "just" a colimit in the category of archimedean ordered fields, so there may be some overpowered proof in the vicinity which proceeds like: the category of AOF has all colimits of some nice shape (that is, chains), and so by Zorn's lemma applied to this posetal category, it has some "maximal" object. so you might also consider applying this kind of procedure to similar-seeming categories.)
Hello. I think it would have been better to dive into one specific example and to be more thorough with it. In my opinion the whole video felt a little bit rushed because of this. For a video that was meant to Demistify Zorn's lemma, I feel like this video left me more confused than enlightened.
I also think the fornat of explaning the three problems, and only then "solving" the three problems is not very good, since you can forget details/get overwhelmed.
For reference, I have a master's degree in physics and I am generally curious about math, and do some independent learning,so I feel like I'm probably more familiar with these topics than most.
Why is it a lemma instead of a theorem?
Because it is a general solution to a common problem.
Here from SOME Voting
in the "Real Number Proof" section, do we assume that there exists a set of all archimedian ordered fields, and is that true?
This is one of the issues that I mentioned at the beginning of the section that I said I would pretend didn't exist. There is no set of all Archimedean ordered fields. Thankfully the solution here is simple. You can prove that all Archimedean ordered fields have cardinality of at most 2^|ℕ|, so you can restrict yourself to Archimedean ordered fields made on subsets of the power set of the natural numbers. This does form a set.
The first what for me was much earlier than the end of that statement. It was who/what is Zorn, and what is a Lemma?
copied directly from Wikipedia
In mathematics, informal logic and argument mapping, a lemma (pl.: lemmas or lemmata) is a generally minor, proven proposition which is used as a stepping stone to a larger result. For that reason, it is also known as a "helping theorem" or an "auxiliary theorem". In many cases, a lemma derives its importance from the theorem it aims to prove; however, a lemma can also turn out to be more important than originally thought
Max Zorn was a mathematician who worked at UCLA and Indiana State University. He was not the first person to come up with his eponymous lemma ("lemma" having been explained in a previous reply), but apparently a lot of people associated the result with him, so it became known as "Zorn's Lemma".
I had a logic professor in grad school who met Max Zorn, and my professor remarked that Zorn disliked the result being called "Zorn's Lemma". Apparently, Zorn would retort, "It's not mine, and it's not a lemma."
"Relegating to obscure jokes" is an oxymoron. A good joke is not a relegation, it is an elevation. A good joke (even if "so bad it is funny") is what is missing, a crucial missing ingredient, in all known mathematics texts that are for sale.
You should at least mention why Zorn's lemma works with the contruction of the real numbers, it not being a proper class, I don't t refering to a paper in the description is appropriate here.
Why do you say in the thumbnail that it's not about the axiom of choice when both statements are equivalent (over ZF)?
Because it's just not the point of the lemma, and I see that equivalence as purely coincidental.
You were doing great but failed to show (in the vector space example) that the union of the chain is linearly independent. If it's not, then there's a dependence relation that must already be true in one of the elements of the chain, contradicting the linear independence of each member of the chain. Without that detail, your proof fails. Good exposition overall though.
I explained at 17:26 that you do need to prove that, but I skipped it for the sake of time. I generally skimmed the three proofs because as I said in the video, the overall structure of the proof is what's important here, not the particular details.
does this show that the upper bound of every infinite chain must be a maximal element? The choice function f should be able to generate any chain you choose, right?
Think about the integers. The set {..., -3, -2, -1} is infinite and has an upper bound, but any upper bound is not maximal.
But that chain is not generateable by the choice function, as (as you prove) it must have some starting point f({})
@@minerharry Well even simpler example, just consider the natural numbers. It's well ordered but has no maximal element for any infinite chain.
Great video! I wish you had explained why the maximal well-order was still a well-order. I guess this is one of the details that you didn't include in the video.
Oh well, you know what they say: "When Zorn gives you lemmas, …"
Even the document that I got the proof from didn't explain that step since it's "trivial". Here's a quick proof: Let A be a nonempty subset of C. Because A is nonempty, there exists some x ∈ A ⊆ C, and because x ∈ C, that means there is some conforming set S with x ∈ S. Then A ∩ S is a nonempty subset of S, and because S is well-ordered, it has some least element a. If there was some y ∈ A with y < a, we would have y ∈ S as well because all conforming sets start with the same elements. Thus, y would be in A ∩ S, contradicting a being the least element of A ∩ S. Thus, a is the least element of A.
I'm reading old comments and I just realized that I think I misunderstood your question! Were you asking about the step in the proof of the well-ordering theorem? The maximal well order is still a well order by definition. Our partially ordered set there was the set of well orders on subsets, so any element of it will be a well order. Zorn's lemma gives us a maximal element of that set of well orders, so the maximal well order is a well order.
When pure mathematicians say that Zorn's lemma is „practical“, what they really mean is that it’s a convenient way not to think deeply about minor details like actually constructing something algorithmically so we might use it in the real world. The best arguments I’ve seen so far for why we might want to do such a thing is bc it allows us to „construct“ stuff like non-measurable subsets of R and thus prove that we need to be a bit thoughtful in defining measures. But if we had cared about actually algorithmically computing measures from the beginning, we just wouldn’t have run into difficulties like banach-tarski.
I for one don’t really care for the axiom of choice, because it implies excluded middle which… ewww, you know? It doesn’t help me in the slightest to know abstractly that a Turing machine must either stop or loop indefinitely, I wanna know which of the two options it is!
"But if we had cared about actually algorithmically computing measures from the beginning, we just wouldn’t have run into difficulties like banach-tarski."
This result is strictly weaker than ZL. It's provable from, say, Hahn Banach. And you _can't_ do functional analysis without Hahn Banach. The mere need to have Hahn Banach implies this "difficulty".
And there's just no _reason_ to compute measures in the first place. Requiring that axiom leads to horrific results. If you do it, deny AC, accept LM instead, guess what happens? Then you can partition the real numbers into _strictly_ more non empty parts than there are real numbers. A result far worse than BT.
Like, sorry that non constructive proofs exist. But they do. It's impossible to do like half of analysis without them.
@@najawin8348 what do you mean by LM?
@@najawin8348 maybe like half of analysis is useless for practical purposes then? Constructive practically useful. Also, no one’s saying we should require eg all measures to be computable. What I am saying is: let’s just not talk about non-computable ones. Let’s just not use „theorems“ that depend on choice. You may not get all the results that look so neat, but at least you’re not lying to the engineers who actually try to solve real world problems using mathematical CONSTRUCTIONS.
Or here’s an idea: at the very least state in those theorems that you need a computable choice function of appropriate type. That immediately makes the theorem constructively valid. You just have to do some extra work to be able to apply it. And while proving other theorems based on that, you can just pass that burden on to the next guy until someone actually wants to apply it to something concrete to solve a real world problem. There’s really zero excuse not to do it other than laziness.
@@markuspfeifer8473 why should "constructive practically useful" be true?
@@methatis3013 because when you solve real world problems, you actually care about finding the solution rather than being happy that it exists in the platonic sphere of ideas?
After studying measure theory my complete acceptance of the axiom of choice was shattered hahahahaha
Is this what they meant by jerma watches zorn?
I prefer a good rhyme. What's brown and can be tattooed on your forehead in case you forget you can pick an element of a set? Zorn's henna.
Or: What's weird and goes in an espresso of choice suitable for drinking with the experimental film by Hollis Frampton? Zorn's crema. ua-cam.com/video/zDQfAUxQdvI/v-deo.html
In the proof of Zorn's Lemma, when you say the "set" C is the set of all conforming sets, I conclude C must contain itself. This fact reminds me strongly of Russell's Paradox, although it is not entirely related. I wonder if we are even allowed to call C a set then.
C is not the set of all conforming sets, it's the union of all conforming sets. That's a big difference. The set of all conforming sets does exist but it's not conforming.
@@sudgylacmoe Ah thank you! That makes much more sense.
Thanks,
3:36 a few too many arrows there, eh?
❤
Ok, so I saw the thumbnail and it concerned me that you seemed to suggest that Zorn’s Lemma is unrelated to the Axiom of Choice. I’m glad that by the end of the video you pointed out that there’s a usage of the Axiom of choice in your proof; while you let the viewer try to figure that out for themselves, the main problem I have with your presentation is that this “correction” of your headline on the thumbnail comes way too late.
All questions have maximum answers. There is always a more maximal answer. I hope I get unbanned from your bivector community one day. Nice video.
so hm, that's all good but what's the justification for why zorn's lemma should be intuitive enough that it essentially can be taken as an axiom? (this is particularly important if you do not consider the well-ordering theorem nor the axiom of choice as intuitive). why when studying for example vector spaces, do we assume all vector spaces must have a basis, instead of being satisfied with the fact that most of our theorems are about vector spaces that happen to have a basis?
for some time, people thought the axiom of choice (equivalently Zorn's lemma) was a theorem that could be proved using the other axioms of set theory. it is an axiom because it can or cannot be used in your reasoning, not because it is intuitive enough that it doesn't necessitate being proved. the thing is, there is just a significant enough number of results that rely on AoC for it to be commonly accepted as true.
I'm afraid the era of indisputably intuitive axioms was left behind long ago.
Meow's lemma: If S is any nonempty partially ordered set in which every chain has a lower bound, then S has a minimal element. No one can prove this.
Second
I don't understand the confusion.
partially ordered - like the y component of a parabola of the form, y = ax^2 +bx +c
not all pairs of elements will be ordered, but some will. for instance, the roots of this polynomial are not ordered, but if there are two roots, then they are ordered with the vertex.
every chain has an upper bound - given any ordered subset, there is a maximal element within it
the set has a maximal element - if all ordered subsets have a maximal element, then one of those must be the maximal element of the whole set
what can possibly be confusing about that unless you're illiterate or very stupid? and in either of those cases, what are you doing at this level of mathematics? you should be in a kindergarten class to learn the names of colors and not to eat paste.
I guess I shouldn't be surprised, though, since none of you idiots understand how exponents work.
at best you claim that a^b is b copies of a multiplied together, but this doesn't explain what happens when b = 0 at all, so you make up some extra rule about that. but then that rule doesn't seem to work when a = 0, so it's an open question what 0^0 is. and all of this is separate from what happens when b is Complex. not to mention the fact that if b is Rational then you predict there are as many solutions as the denominator of the reduced Rational expression of b, but this has absolutely no extension to irrational values for b, where you only recognize 1 solution.
note that the 'a^b is b copies of a multiplied together' is the basis of the definition of all hyperoperations, but it's not even possible to take fractional hyperoperations under this definition.
so... yeah, you're necessarily just very stupid.
after watching your video it's clear that you use lots of terms you do not understand. you should probably stop doing that.
to start, let's consider something that you definitely take for granted that you understand, but in fact do not, numbers. you consider things like 7, -7, 7i, -7 +7i, 3/7, etc. to be numbers. you also consider |-7| = 7, and this 7 is identical to +7. however, you don't consider 7 elephants, 7 ounces of water, or 7 poems to be numbers. with these, only the 7 part is a number. so now let's ask the question why does the unit sometimes count as being part of a number and sometimes not?
-7 +7i is not only considered to be a number, but also a vector, because it is a combination of magnitudes and units. those magnitudes being 7 and 7, and the units being -1 and i. but now, if this -7 is a combination of unit and magnitude, and thus a vector, then isn't -7 always a vector? and if this is the case, then isn't 7 elephants a vector, and that's why it's not a number? so why are vectors sometimes numbers and sometimes not?
are you beginning to see that you have no idea wtf you're talking about?
now, if |-7| = 7, and the absolute value bars are isolating the magnitude of the vector -7, then clearly -7 is not a number, because numbers are vector magnitudes, not whole vectors, but to get 7 from |-7| we specifically removed the negative unit to isolate the magnitude. but it's worse than that, because we didn't really remove the negative unit, we just converted it to a positive unit, because by definition the Naturals, like 7, are identical to the positive Integers, like +7. so |-7| doesn't isolate the magnitude of the vector -7 at all.
and this then plays into Godel Incompleteness being a property of all mathematical systems which incorporate the Naturals, since Godel Incompleteness arises specifically from a condition where the Naturals can be randomly converted between an ordinal, like +7, and a nominal, like the bare unsigned number 7. this is occurring due to an inconsistency in the definition of the Naturals, and yet everyone agrees that systems which build on the Naturals, like Peano Arithmetic, are consistent... that's obviously wrong.
and you call this rigor?
if you're still lost, consider why addition is Abelian:
3 +4 = 4 +3 = 7; this is true because there is an unstated grammar rule in mathematical orthography which renders expression-initial + optional
we could instead write this as:
+3 +4 = +4 +3 = +7
but now that we're doing this, we can see why subtraction is non-Abelian:
3 -4 = -4 +3 = -1; unlike the + unit specification, - is always obligatory
we could instead write this as:
+3 -4 = -4 +3 = -1
this means that neither addition nor subtraction are actually operations at all. the symbols we're writing and calling 'addition' and 'subtraction' are always unit specifiers, and things prefixed with them are simply listed together, in an unordered list. like a shopping list, it doesn't matter if you get milk before you get oranges, it just matters that you get them both. thus it is this hidden operation of listing vectors which is Abelian, not addition, because addition is not an operation at all. similarly, subtraction is not Abelian, but the reason is that it's not an operation, not because it's a non-Abelian operation.
we can rewrite all of these in set notation, for clarity:
3 +4 = 4 +3 = {+3, +4} = {+4, +3} = {+7}
3 -4 = -4 +3 = {+3, -4} = {-4, +3} = {-1}
and while you'll probably attempt to object, that is precisely what those set definitions mean. which means that the Axiom of Extension is wrong... but we already knew that from our previous analysis of the Naturals, since for Godel Incompleteness to work it must be the case that the Successor Function both generates the Naturals, and does not generate the Naturals, which is a blatant violation of the Axiom of Extension that is canonically accepted as perfectly acceptable by all who mean to accept the Axiom of Extension as inviolable.
now take this, and reconsider your discussion of vector spaces... you have absolutely no idea wtf you're talking about, at all.
oh yeah, why do you specify 'a random non-zero vector'?
by definition, zero is not a vector, since it has neither magnitude nor unit.
every dimension has zero, but zero has no dimension.
this is why most proofs have to use special pleading for the 'trivial case' of zero... because it should never have been included in most of the things that it's included in. for instance, how can you claim that the Rationals are Archimedean when zero is Rational? by definition, no set including zero can be Archimedean.
when you're this dumb, you shouldn't be trying to teach people.
you think there are Transcendental Reals, don't you?
take a look at the definition of e:
e = lim n->inf (1 +1/n)^n
now, since n is approaching infinity, the 1/n part can be replaced with an infinitesimal, c, yielding:
e = lim n->inf (1 +c)^n
if we now grow n via incrementation n must always be an Integer, which allows us to apply the binomial theorem, giving:
e = lim n->inf 1 +Ac +Bc^2... +Wc^(n -1) +Xc^n
since this limit will only be infinitesimally far away from e, we can remove the limit condition by simply adding c:
e = 1 +(A +1)c +Bc^2... +Wc^(n -1) +Xc^n
all of the coefficients and exponents in this expansion of e are Integers, which means that this is a non-Real Levi-Civita series. that is, e is not a member of the set of Reals because its definition includes irreducible infinitesimal terms. because infinitesimals are non-Real, the Levi-Civita field is a field extension of the Reals which adds the infinitesimals back in. but something that nobody noticed is that the definition of a Transcendental is that it includes an irreducible infinitesimal component, which means that absolutely no Transcendentals can be Real by definition.
oops.
frankly this should be obvious, since if the Reals are Complete as Dedekind claimed... what does the Levi-Civita field include that the Reals don't already? under the assumed definition of what a Dedekind Cut does, there shouldn't be any way for the Levi-Civita field to extend the Reals. or, for that matter, for the surreals, or hyperreals to extend the Reals. and we definitely shouldn't be finding familiar values, like e, which are naively defined to be Reals, within these extensions. the definitions of everything you work with are complete nonsense.
if I were you I'd be having a panic attack right about now, since all this stuff you think you understand is actually just mythology and fairy tales.
@@sumdumbmick слушай, не хочешь в мой клуб альтернативно одаренных? я знаю некого рамзеса который с радостью обсудит с тобой эти твои выкрутасы с числами
I find something nostalgic in hearing someone disregard the axiom of choice as "obvious", and sparring no "qualms" over it.
Studying advanced set theory is a sobering experience, I don't recommend it 🥲
I'll admit that I used to have the "default" opinion of "I don't really know about the axiom of choice and I'll just use it only when I need it". However, as I have learned more math and studied some foundations, I've come to the conclusion that the axiom of choice belongs in logic as a fundamental law, on the same level as the law of noncontradiction and the law of the excluded middle. Basically, I don't hold my current opinion because I haven't studied enough. I hold this opinion because I have studied enough.
@@sudgylacmoe
I appreciate the response (and I apologize if I implied that you didn't know enough).
But my comment was about advanced set theory specifically, things like determinacy, V=L, Solovay’s Model, countable/dependent choice… (there’s a lot weirder but you might know these).
The short of it is you can find a continuum of statements in set theory and how they relate to the AoC. Deciding what “obviously” is true is a gruesome task to torture the most ardent platonists. Not to mention the Suslin line, diamond principle, Martin axioms and much more.
I think anyone subjected to this will come out marked by a strong sense of relativism, like “well… I don’t think some axioms are really true or false, adding them leads to different theories and that’s it.”.
To save the best for last, logic is the worst place to argue about AoC, logic thrives in the paradigm of constructivism (in particular proof theory). I can’t imagine any natural motivation from logic for the axiom of choice (I am eager to be proven wrong).
@@alexismiller2349 Type theory provides the simplest statement of choice I've ever seen: ∀ A : Type, inhabited A → A. inhabited A is just an inductive type that has a single constructor that takes in a value in A, so to me this statement is incredibly obvious, since you needed a value in A to get inhabited A to begin with. I see this formulation of choice as much more logical, intuitive, simple, and obvious.
Thanks for the answer... unfortunately I'm not so familiar with type theory, and I don't see the link between what you wrote and AoC. Also I couldn't find anything similar on nLab and mathstackexchange, do you have a link to anything that explains your answer ?
I came up with the formulation myself for use in Coq, but I have since learned that it's what Lean uses for choice as well, so you could look into people describing Lean's axioms. The idea behind choice is extending when exactly you can get a value satisfying a predicate. In normal logic you can only do that with a finite number of existential statements, and choice is what you use to extend it to an infinite number of such statements. But people obfuscate the axiom by describing it in terms of sets when the fundamental idea behind it has nothing to do with sets.