When I was in my logic class I had to ask the question "What if the universe is empty?" The teacher grinned and said "We're assuming that it's not." When I simply nodded, satisfied with the answer, my classmates stared at me like I had just grown a second head. At this point, the teacher explained that this was actually a reasonable question.
Axiom of Infinity - "There exists an infinite set." Well, the langauge of Set Theory does not include the word "infinite". So we might say, the "set of all Naturals". What are Naturals? 0=∅, 1={0}, 2={0,1} n={0,1,2...n-1} "next" is the union of all previous Naturals. Now we can make the set such that it: 1.Contains 0. 2.If it contains n, it also contains the next Natural. Borcherds says all ordinals can even be defined this way. 2:45 "Is the Infinity Axiom needed?" Yes. To get Vω, you must take the Union of {V0, V1, V2...} and you can get this set by applying Replacement to an infinite set, but you first need an infinite set. 4:00 What if you replace the axiom of Infinity with an axiom saying there is no infinity? You get the models: 1.Just ∅. It satisfies all ZFC-I. 2.Hereditarily finite sets. (Finite, rooted, rigid trees.) 12:15 ZFC-Infinity + ~Infinity ↔ Peano Arithmetic, because you can encode PA in ZFC-I+~I, and you can encode ZFC-I+~I in PA. 12:45 ZFC→Con(ZFC-Infinity), so ZFC-Infinity is strictly weaker than ZFC. (It cannot prove some statements that ZFC can.) And Borcherds yet again gave no proof of it. 13:30 A third model of ZFC-I+~I, is a nonstandard (whatever that means) one: Create constants c0, c1, c2... for integers such that c0 = c1 + 1, c1 = c2 + 1... Now Borcherds says this theory is Consistent because it is finite, BUT, it is equivalent to PA, and Gödel works for PA!! It cannot prove its Consistency!!! This model of ZFC-I+~I contains these objects c0, c1, c2 etc. which I would call not-well-founded, because c0 will still be positive after subtracting 1 from it any number of times. A nonwell-founded model. c0 seems to be an infinite set. I do not see why, but Borcherds says c0 is internally finite, and only externally infinite. From inside the model, c0 supposedly appears finite. One quantifier of infinity is that an infinite set can be matched 1-1 (mapped) with its proper (smaller) subset. And such a mapping function, according to Borcherds, is not within the model. "As usual with nonstandard models, very strange freaky things happen." -_-
@9:30 looks like a bijection between the natural numbers and their powerset, which would mean they have the same cardinality, which they do not. Why does this not prove that their cardinality is the same? Is it not a bijection?
Infinity must be considered stronger than powerset in consistency power, as powerset is just an "inaccessible cardinal" axiom relative to infinity, which adds proof strength, to be sure, but it adds strength in a sort of minimal reflective way. Infinity adds an unfathomable amount of strength, without powerset, assuming some other way of defining functions, the strength of set theory with infinity should be equivalent to that of second order arithmetic, and the question is then whether the step from first order to second order arithmetic is smaller or larger than the step from second order arithmetic to ZFC. I believe the second step is tiny in comparison to the first, because inside a countable model, it is just an ordinally iterated consistency axiom, like an inaccessible axiom, and these are much weaker than true large strength axioms like measurable cardinals.
@@An-ht8so I'm not exactly sure what he meant either, but the cross product does not need power set to exist. A very weak form of replacement (more specifically, replacement on atomic formulas, ie only formulas that can be written as x = y or x in y) and union is sufficient. So the definition of functions as subsets of XxY is always well defined even in weak theories (in even weaker theories or theories of arithmetic, it may be better to take second-order/third-order conservative extensions of the theories to allow talking about functions and stuff). The proof for that goes something like this if I am not wrong. For a given x in X, the set {x} X Y exists, by atomic replacement. Then the set {{x} X Y | x in X} exists again by atomic replacement. Take the union. Note that the inner workings of this argument only assume that there are objects (x,y) such that (x,y) = (z,w) x=z and y=w, so does not depend on formulation of ordered pairs. What may not exist in these contexts is the function space between two sets X and Y. Classically, the existence of the function space implies power set if I am not mistaken (altho I encourage you to search a proof, I am not smart enough to think of one now). Intuitionistically this is false. An exemple would be constructive ZF (CZF), a theory that Aczel introduced. It has a stronger version of existence of the function space (that I will not describe without reviewing it's axiom again, cause it is harder to understand than the others) and full replacement, since replacement does not imply separation intuitionistically.
@@mariogamer1088 If I understand correctly, you first replace y by (x,y) for y in Y and a given x, but to do that you need to already have a set that contains the (x;y), i.e. a set that contains xXY, and I'm not sure where it comes from. Replacement can only be used to build subsets.
@@An-ht8so I don't think this is correct. The usual statement of replacement is as follows: Let psi be a formula taking as free parameters x1,...,xn fixed and x, y, A. Let A be any set, such that for any x in A, there is a unique y such that psi(x,y,...,A) is true. (this y need not be proved to be element of a domain or anything) Then there is a set B such that x in A implies there is y in B st. psi(x,y,...,A). This formulation then necessitates the use of separation to have only those ys that have an associated x. There are strengtenings of replacement that are used in constructive theories and that do not need this. One may also require this set be unique in the statement of replacement: Then there is B such that for all y y in B if and only if there is X such that psi(x,y,...,A) is true. This one doesn't need any sort of separation afterwards. Here is a thread on math stack about another subject where the same kind of ambiguity arose: mathoverflow.net/questions/253473/predicative-definition-and-existence-of-ordinal-numbers . In the first application, A = Y. Let y in Y. Let x0 in X. Then there is a (z,y) such that psi((z,y),x0) (def)= (z,y)=(x0,y). To see that it is unique, note that (x0,y)=(z,y) x0=z and y=y x0=z. The second application is similar except A = X and the function is equality with {x} X Y with a given x in X as "input" to the functional formula. Uniqueness is guaranteed by extensionality/replacement/separation, depending on the formulation. Edit: After a bit of a think, I realised I had forgotten that one can't speak of a set {{x},{x,y}} without implicitly using quantifier, so this isn't just atomic replacement. It's probably bounded, but I am not sure.
The video says a set in ZFC can be represented by a finite rooted rigid tree, and vice versa. I understand finite, rooted, and tree in this case, but what does "rigid" mean here?
I'm happy with the empty set, because if you can select things or not, such as for the set of all even integers then 2 in the set but 3 is not, then the answer to "is it in set X" can be "no" for everything. But then also to have to qualify vacuous truths with something to make it explicit that they are vacuous. To say "It technically works and is true but not in a way that can be built on or used for anything".
Two new solutions for the quadratic. (1) let x = y + iz. (y + iz) ^ 2 = y ^ 2 + 2iyz + z ^2. Let x ^ 2 + bx + c = 0. y ^ 2 + 2iyz + z ^ 2 + by + biz + c = 0. Equating real and imaginary parts y ^2 + z ^ 2 + by + c = 0. i(2y + b)z = 0. (2) A square is (x + A) ^ 2 = 0, but not all quadratic are squares and cannot be transformed to them. So let (x + A) ^ 2 = h, and equate to x ^ 2 + bx + c = 0.
Couldn't you drop the axiom, and use the other axioms to construct the naturals anyway? The empty set exists, union exists, powerset exists, choice exists,... doesn't the existence of N follow from the other axioms?
The existence of N does not follow from the other axioms. Only the finite subsets of N follow from the other axioms (assuming that at least one set exists).
the problem is that we cannot "count" the number of elements since the cardinal loses the ability to express said quantity in infinite sets. To count the number of elements of an infinite set we should first organize from less to greater infinite infinite sets so that when organizing the set that we want to "count" it is positioned such that, "This infinite set is smaller and larger than the infinite sets absolutely close by the difference of 1 single element"
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Do you think you'll do a video on large cardinal axioms?
When I was in my logic class I had to ask the question "What if the universe is empty?" The teacher grinned and said "We're assuming that it's not." When I simply nodded, satisfied with the answer, my classmates stared at me like I had just grown a second head. At this point, the teacher explained that this was actually a reasonable question.
Axiom of Infinity - "There exists an infinite set."
Well, the langauge of Set Theory does not include the word "infinite". So we might say, the "set of all Naturals". What are Naturals? 0=∅, 1={0}, 2={0,1} n={0,1,2...n-1}
"next" is the union of all previous Naturals.
Now we can make the set such that it:
1.Contains 0.
2.If it contains n, it also contains the next Natural.
Borcherds says all ordinals can even be defined this way.
2:45 "Is the Infinity Axiom needed?"
Yes. To get Vω, you must take the Union of {V0, V1, V2...} and you can get this set by applying Replacement to an infinite set, but you first need an infinite set.
4:00 What if you replace the axiom of Infinity with an axiom saying there is no infinity? You get the models:
1.Just ∅. It satisfies all ZFC-I.
2.Hereditarily finite sets. (Finite, rooted, rigid trees.)
12:15 ZFC-Infinity + ~Infinity ↔ Peano Arithmetic, because you can encode PA in ZFC-I+~I, and you can encode ZFC-I+~I in PA.
12:45 ZFC→Con(ZFC-Infinity), so ZFC-Infinity is strictly weaker than ZFC. (It cannot prove some statements that ZFC can.)
And Borcherds yet again gave no proof of it.
13:30 A third model of ZFC-I+~I, is a nonstandard (whatever that means) one:
Create constants c0, c1, c2... for integers such that c0 = c1 + 1, c1 = c2 + 1...
Now Borcherds says this theory is Consistent because it is finite, BUT, it is equivalent to PA, and Gödel works for PA!! It cannot prove its Consistency!!!
This model of ZFC-I+~I contains these objects c0, c1, c2 etc. which I would call not-well-founded, because c0 will still be positive after subtracting 1 from it any number of times. A nonwell-founded model.
c0 seems to be an infinite set. I do not see why, but Borcherds says c0 is internally finite, and only externally infinite. From inside the model, c0 supposedly appears finite.
One quantifier of infinity is that an infinite set can be matched 1-1 (mapped) with its proper (smaller) subset. And such a mapping function, according to Borcherds, is not within the model.
"As usual with nonstandard models, very strange freaky things happen."
-_-
@9:30 looks like a bijection between the natural numbers and their powerset, which would mean they have the same cardinality, which they do not. Why does this not prove that their cardinality is the same? Is it not a bijection?
The range is only the finite subsets of the natural numbers, not all subsets of the natural numbers.
Infinity must be considered stronger than powerset in consistency power, as powerset is just an "inaccessible cardinal" axiom relative to infinity, which adds proof strength, to be sure, but it adds strength in a sort of minimal reflective way. Infinity adds an unfathomable amount of strength, without powerset, assuming some other way of defining functions, the strength of set theory with infinity should be equivalent to that of second order arithmetic, and the question is then whether the step from first order to second order arithmetic is smaller or larger than the step from second order arithmetic to ZFC. I believe the second step is tiny in comparison to the first, because inside a countable model, it is just an ordinally iterated consistency axiom, like an inaccessible axiom, and these are much weaker than true large strength axioms like measurable cardinals.
What do you have in mind for "some other way of defining functions" ? You would need another axiom to take its place.
And by the way ZFC does have countable models, but I see what you mean (I think)
@@An-ht8so I'm not exactly sure what he meant either, but the cross product does not need power set to exist. A very weak form of replacement (more specifically, replacement on atomic formulas, ie only formulas that can be written as x = y or x in y) and union is sufficient. So the definition of functions as subsets of XxY is always well defined even in weak theories (in even weaker theories or theories of arithmetic, it may be better to take second-order/third-order conservative extensions of the theories to allow talking about functions and stuff).
The proof for that goes something like this if I am not wrong. For a given x in X, the set {x} X Y exists, by atomic replacement. Then the set {{x} X Y | x in X} exists again by atomic replacement. Take the union. Note that the inner workings of this argument only assume that there are objects (x,y) such that (x,y) = (z,w) x=z and y=w, so does not depend on formulation of ordered pairs.
What may not exist in these contexts is the function space between two sets X and Y. Classically, the existence of the function space implies power set if I am not mistaken (altho I encourage you to search a proof, I am not smart enough to think of one now). Intuitionistically this is false. An exemple would be constructive ZF (CZF), a theory that Aczel introduced. It has a stronger version of existence of the function space (that I will not describe without reviewing it's axiom again, cause it is harder to understand than the others) and full replacement, since replacement does not imply separation intuitionistically.
@@mariogamer1088 If I understand correctly, you first replace y by (x,y) for y in Y and a given x, but to do that you need to already have a set that contains the (x;y), i.e. a set that contains xXY, and I'm not sure where it comes from. Replacement can only be used to build subsets.
@@An-ht8so I don't think this is correct. The usual statement of replacement is as follows: Let psi be a formula taking as free parameters x1,...,xn fixed and x, y, A.
Let A be any set, such that for any x in A, there is a unique y such that psi(x,y,...,A) is true. (this y need not be proved to be element of a domain or anything)
Then there is a set B such that x in A implies there is y in B st. psi(x,y,...,A). This formulation then necessitates the use of separation to have only those ys that have an associated x.
There are strengtenings of replacement that are used in constructive theories and that do not need this. One may also require this set be unique in the statement of replacement: Then there is B such that for all y y in B if and only if there is X such that psi(x,y,...,A) is true. This one doesn't need any sort of separation afterwards.
Here is a thread on math stack about another subject where the same kind of ambiguity arose: mathoverflow.net/questions/253473/predicative-definition-and-existence-of-ordinal-numbers .
In the first application, A = Y. Let y in Y. Let x0 in X. Then there is a (z,y) such that psi((z,y),x0) (def)= (z,y)=(x0,y). To see that it is unique, note that (x0,y)=(z,y) x0=z and y=y x0=z.
The second application is similar except A = X and the function is equality with {x} X Y with a given x in X as "input" to the functional formula. Uniqueness is guaranteed by extensionality/replacement/separation, depending on the formulation.
Edit: After a bit of a think, I realised I had forgotten that one can't speak of a set {{x},{x,y}} without implicitly using quantifier, so this isn't just atomic replacement. It's probably bounded, but I am not sure.
Student: "Master, what is the difference between finite and infinite?"
Master: "You don't get it."
At that moment, the Student was enlightened.
I apparently have not been enlightened.
Thank you for making this video!
Is the collection of finite rigid trees actually V_w?
The video says a set in ZFC can be represented by a finite rooted rigid tree, and vice versa. I understand finite, rooted, and tree in this case, but what does "rigid" mean here?
No non-trivial automorphisms, i.e. "all branches are different". Also see the previous video on extensionality for exposition on this
I'm happy with the empty set, because if you can select things or not, such as for the set of all even integers then 2 in the set but 3 is not, then the answer to "is it in set X" can be "no" for everything. But then also to have to qualify vacuous truths with something to make it explicit that they are vacuous. To say "It technically works and is true but not in a way that can be built on or used for anything".
Two new solutions for the quadratic.
(1) let x = y + iz.
(y + iz) ^ 2 = y ^ 2 + 2iyz + z ^2.
Let x ^ 2 + bx + c = 0.
y ^ 2 + 2iyz + z ^ 2 + by + biz + c = 0.
Equating real and imaginary parts
y ^2 + z ^ 2 + by + c = 0.
i(2y + b)z = 0.
(2) A square is
(x + A) ^ 2 = 0,
but not all quadratic are squares and cannot be transformed to them. So let (x + A) ^ 2 = h, and equate to
x ^ 2 + bx + c = 0.
Couldn't you drop the axiom, and use the other axioms to construct the naturals anyway? The empty set exists, union exists, powerset exists, choice exists,... doesn't the existence of N follow from the other axioms?
The existence of N does not follow from the other axioms. Only the finite subsets of N follow from the other axioms (assuming that at least one set exists).
you could construct the naturals, but you couldn't claim that the class of all naturals is a set
Two to the zeroth power is one.
I think supercompactness is already stronger than omega.
yeeeeeeeeeeeeeeeee
the problem is that we cannot "count" the number of elements since the cardinal loses the ability to express said quantity in infinite sets.
To count the number of elements of an infinite set we should first organize from less to greater infinite infinite sets so that when organizing the set that we want to "count" it is positioned such that,
"This infinite set is smaller and larger than the infinite sets absolutely close by the difference of 1 single element"
It's True or False. It's not True, and also it is False.