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.

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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)

(As I pointed out in a comment) yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). To illustrate this, let me "randomly" highlight two examples: See here for $\Sigma^1_2$-Woodin cardinals and, more generally, the noti […]

I am not sure which statement you heard as the "Ultimate $L$ axiom," but I will assume it is the following version: There is a proper class of Woodin cardinals, and for all sentences $\varphi$ that hold in $V$, there is a universally Baire set $A\subseteq{\mathbb R}$ such that, letting $\theta=\Theta^{L(A,{\mathbb R})}$, we have that $HOD^{L(A,{\ma […]

A Wadge initial segment (of $\mathcal P(\mathbb R)$) is a subset $\Gamma$ of $\mathcal P(\mathbb R)$ such that whenever $A\in\Gamma$ and $B\le_W A$, where $\le_W$ denotes Wadge reducibility, then $B\in\Gamma$. Note that if $\Gamma\subseteq\mathcal P(\mathbb R)$ and $L(\Gamma,\mathbb R)\models \Gamma=\mathcal P(\mathbb R)$, then $\Gamma$ is a Wadge initial se […]

Craig: For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. Perhaps the first pape […]

Yes. Consider, for instance, Conway's base 13 function $c$, or any function that is everywhere discontinuous and has range $\mathbb R$ in every interval. Pick continuous bijections $f_n:\mathbb R\to(-1/n,1/n)$ for $n\in\mathbb N^+$. Pick a strictly decreasing sequence $(x_n)_{n\ge1}$ converging to $0$. Define $f$ by setting $f(x)=0$ if $x=0$ or $\pm x_n […]

All proofs of the Bernstein-Cantor-Schroeder theorem that I know either directly or with very little work produce an explicit bijection from any given pair of injections. There is an obvious injection from $[0,1]$ to $C[0,1]$ mapping each $t$ to the function constantly equal to $t$, so the question reduces to finding an explicit injection from $C[0,1]$ to $[ […]

One way we formalize this "limitation" idea is via interpretative power. John Steel describes this approach carefully in several places, so you may want to read what he says, in particular at Solomon Feferman, Harvey M. Friedman, Penelope Maddy, and John R. Steel. Does mathematics need new axioms?, The Bulletin of Symbolic Logic, 6 (4), (2000), 401 […]

"There are" examples of discontinuous homomorphisms between Banach algebras. However, the quotes are there because the question is independent of the usual axioms of set theory. I quote from the introduction to W. Hugh Woodin, "A discontinuous homomorphism from $C(X)$ without CH", J. London Math. Soc. (2) 48 (1993), no. 2, 299-315, MR1231 […]

This is Hausdorff's formula. Recall that $\tau^\lambda$ is the cardinality of the set ${}^\lambda\tau$ of functions $f\!:\lambda\to\tau$, and that $\kappa^+$ is regular for all $\kappa$. Now, there are two possibilities: If $\alpha\ge\tau$, then $2^\alpha\le\tau^\alpha\le(2^\alpha)^\alpha=2^\alpha$, so $\tau^\alpha=2^\alpha$. In particular, if $\alpha\g […]

Fix a model $M$ of a theory for which it makes sense to talk about $\omega$ ($M$ does not need to be a model of set theory, it could even be simply an ordered set with a minimum in which every element has an immediate successor and every element other than the minimum has an immediate predecessor; in this case we could identify $\omega^M$ with $M$ itself). W […]

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.