We verified the “surjective” version of Cantor’s theorem: If then is not injective. A curious weakness of the standard diagonal proof is that the argument is not entirely constructive: We considered the set and showed that there is some set such that .

Question. Can one define such a set ?

As a consequence of the results we will prove on well-orderings, we will be able to provide a different proof where the pair of distinct sets verifying that is not injective is definable. (Definability is an issue we will revisit a few times.)

We proved the trichotomy theorem for well-orderings, defined ordinals, and showed that is not a set (the Burali-Forti paradox).

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Although I understand Cantor’s theorem and the “surjective version” well enough separately, and I even see how they are essentially saying the same thing, I don’t really understand the preamble that you were providing before presenting this version.

That is, I don’t understand why we would consider this version to be a “surjective” version of Cantor’s theorem, since it’s dealing with the nonexistence of injective functions. I also don’t understand the connection to the axiom of choice, and how the fact that a surjection A –> B corresponds (with choice) to an injection B –> A plays any part in this theorem.

My guess is that there’s a really obvious surjection implicit somewhere in the two statements, which would put everything in the right context, and I’m just missing it.

The “usual” version of Cantor’s result is that no injection is onto. The “surjective” version would say that no surjection is into.

It just so happens that the argument used in the first case does not actually use that is an injection, we actually show that no such is onto. And, similarly, in the second case, we end up not needing to assume that is a surjection. I suppose one could just as well call the version the “surjective version” and the one the “injective version,” but I believe the common usage is as I have it.

At the heart of both arguments is the simple diagonalization coming from wondering whether or (for appropriate ), and it is in that sense that both arguments are more or less “the same.”

As for Choice: In Tuesday we will show the fact you mention, that if there is a surjection from A onto B then there is an injection from B into A. So, with choice, both statements “A surjects onto B” and “B injects into A” are the same, and we could use either to define the relation . Without choice, these statements are not equivalent, and we could have a different notion if there is a surjection from A onto B.

What we showed is that Cantor’s argument gives and
also , so that x is less than *no matter* which version of “less than” one chooses.

However, I would go further and say that the version one wants is the version rather than the one, since the Schröder-Bernstein theorem holds for this version.

Question. Does Schröder-Bernstein hold for ?

(Of course, yes if we assume choice, but the question is whether it holds in ZF.)

Wait, that made perfect sense the first time I read it, but then I sat down to formalize it and realized I must be missing something. I’m pretty sure that a straight translation of your first paragraph gives…

“Usual version”: f injective => neg (f surjective)
“Surjective version”: g surjective => neg(f injective)

Georgii: Let me start with some brief remarks. In a series of three papers: a. Wacław Sierpiński, "Contribution à la théorie des séries divergentes", Comp. Rend. Soc. Sci. Varsovie 3 (1910) 89–93 (in Polish). b. Wacław Sierpiński, "Remarque sur la théorème de Riemann relatif aux séries semi-convergentes", Prac. Mat. Fiz. XXI (1910) 17–20 […]

It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. (This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here (Intern […]

Stefan, "low" cardinalities do not change by passing from $L({\mathbb R})$ to $L({\mathbb R})[{\mathcal U}]$, so the answer to the second question is that the existence of a nonprincipal ultrafilter does not imply the existence of a Vitali set. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${ […]

Marginalia to a theorem of Silver (see also this link) by Keith I. Devlin and R. B. Jensen, 1975. A humble title and yet, undoubtedly, one of the most important papers of all time in set theory.

Given a positive integer $a$, the Ramsey number $R(a)$ is the least $n$ such that whenever the edges of the complete graph $K_n$ are colored using only two colors, we necessarily have a copy of $K_a$ with all its edges of the same color. For example, $R(3)= 6$, which is usually stated by saying that in a party of 6 people, necessarily there are 3 that know e […]

Equality is part of the background (first-order) logic, so it is included, but there is no need to mention it. The situation is the same in many other theories. If you want to work in a language without equality, on the other hand, then this is mentioned explicitly. It is true that from extensionality (and logical axioms), one can prove that two sets are equ […]

$L$ has such a nice canonical structure that one can use it to define a global well-ordering. That is, there is a formula $\phi(u,v)$ that (provably in $\mathsf{ZF}$) well-orders all of $L$, so that its restriction to any specific set $A$ in $L$ is a set well-ordering of $A$. The well-ordering $\varphi$ you are asking about can be obtained as the restriction […]

Gödel sentences are by construction $\Pi^0_1$ statements, that is, they have the form "for all $n$ ...", where ... is a recursive statement (think "a statement that a computer can decide"). For instance, the typical Gödel sentence for a system $T$ coming from the second incompleteness theorem says that "for all $n$ that code a proof […]

When I first saw the question, I remembered there was a proof on MO using Ramsey theory, but couldn't remember how the argument went, so I came up with the following, that I first posted as a comment: A cute proof using Schur's theorem: Fix $a$ in your semigroup $S$, and color $n$ and $m$ with the same color whenever $a^n=a^m$. By Schur's theo […]

It depends on what you are doing. I assume by lower level you really mean high level, or general, or 2-digit class. In that case, 54 is general topology, 26 is real functions, 03 is mathematical logic and foundations. "Point-set topology" most likely refers to the stuff in 54, or to the theory of Baire functions, as in 26A21, or to descriptive set […]

Although I understand Cantor’s theorem and the “surjective version” well enough separately, and I even see how they are essentially saying the same thing, I don’t really understand the preamble that you were providing before presenting this version.

That is, I don’t understand why we would consider this version to be a “surjective” version of Cantor’s theorem, since it’s dealing with the nonexistence of injective functions. I also don’t understand the connection to the axiom of choice, and how the fact that a surjection A –> B corresponds (with choice) to an injection B –> A plays any part in this theorem.

My guess is that there’s a really obvious surjection implicit somewhere in the two statements, which would put everything in the right context, and I’m just missing it.

Thanks!

The “usual” version of Cantor’s result is that no injection is onto. The “surjective” version would say that no surjection is into.

It just so happens that the argument used in the first case does not actually use that is an injection, we actually show that no such is onto. And, similarly, in the second case, we end up not needing to assume that is a surjection. I suppose one could just as well call the version the “surjective version” and the one the “injective version,” but I believe the common usage is as I have it.

At the heart of both arguments is the simple diagonalization coming from wondering whether or (for appropriate ), and it is in that sense that both arguments are more or less “the same.”

As for Choice: In Tuesday we will show the fact you mention, that if there is a surjection from A onto B then there is an injection from B into A. So, with choice, both statements “A surjects onto B” and “B injects into A” are the same, and we could use either to define the relation . Without choice, these statements are not equivalent, and we could have a different notion if there is a surjection from A onto B.

What we showed is that Cantor’s argument gives and

also , so that x is less than *no matter* which version of “less than” one chooses.

However, I would go further and say that the version one wants is the version rather than the one, since the Schröder-Bernstein theorem holds for this version.

Question. Does Schröder-Bernstein hold for ?

(Of course, yes if we assume choice, but the question is whether it holds in ZF.)

Wait, that made perfect sense the first time I read it, but then I sat down to formalize it and realized I must be missing something. I’m pretty sure that a straight translation of your first paragraph gives…

“Usual version”: f injective => neg (f surjective)

“Surjective version”: g surjective => neg(f injective)

Um, aren’t these just contrapositives?

I fear I failed to explain myself.

We proved two different theorems. Theorem 1 says that no map is surjective. Theorem 2 says that no map is injective.

And we shouldn’t call them `injective’ or `surjective’ versions of Cantor’s result any longer, as this seems to be misleading.

I agree with you, the first paragraph says the same thing twice. But we actually obtained two theorems from our analysis.