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)

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 […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

The description below comes from József Beck. Combinatorial games. Tic-tac-toe theory, Encyclopedia of Mathematics and its Applications, 114. Cambridge University Press, Cambridge, 2008, MR2402857 (2009g:91038). Given a finite set $S$ of points in the plane $\mathbb R^2$, consider the following game between two players Maker and Breaker. The players alternat […]

Yes. This is a consequence of the Davis-Matiyasevich-Putnam-Robinson work on Hilbert's 10th problem, and some standard number theory. A number of papers have details of the $\Pi^0_1$ sentence. To begin with, take a look at the relevant paper in Mathematical developments arising from Hilbert's problems (Proc. Sympos. Pure Math., Northern Illinois Un […]

It is easy to see without choice that if there is a surjection from $A$ onto $B$, then there is an injection from ${\mathcal P}(B)$ into ${\mathcal P}(A)$, and the result follows from Cantor's theorem that $B

Only noticed this question today. Although the selected answer is quite nice and arguably simpler than the argument below, none of the posted answers address what appeared to be the original intent of establishing the inequality using the Arithmetic Mean-Geometric Mean Inequality. For this, simply notice that $$ 1+3+\ldots+(2n-1)=n^2, $$ which can be easily […]

First of all, $f(z)+e^z\ne 0$ by the first inequality. It follows that $e^z/(f(z)+e^z)$ is entire, and bounded above. You should be able to conclude from that.

Yes. The standard way of defining these sequences goes by assigning in an explicit fashion to each limit ordinal $\alpha$, for as long as possible, an increasing sequence $\alpha_n$ that converges to $\alpha$. Once this is done, we can define $f_\alpha$ by diagonalizing, so $f_\alpha(n)=f_{\alpha_n}(n)$ for all $n$. Of course there are many possible choices […]

I disagree with the advice of sending a paper to a journal before searching the relevant literature. It is almost guaranteed that a paper on the fundamental theorem of algebra (a very classical and well-studied topic) will be rejected if you do not include mention on previous proofs, and comparisons, explaining how your proof differs from them, etc. It is no […]

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.