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)

This is a very interesting question (and I really want to see what other answers you receive). I do not know of any general metatheorems ensuring that what you ask (in particular, about consistency strength) is the case, at least under reasonable conditions. However, arguments establishing the proof theoretic ordinal of a theory $T$ usually entail this. You […]

This is false; take a look at https://en.wikipedia.org/wiki/Analytic_set for a quick introduction. For details, look at Kechris's book on Classical Descriptive Set Theory. There you will find also some information on the history of this result, how it was originally thought to be true, and how the discovery of counterexamples led to the creation of desc […]

This is open. In $L(\mathbb R)$ the answer is yes. Hugh has several proofs of this, and it remains one of the few unpublished results in the area. The latest version of the statement (that I know of) is the claim in your parenthetical remark at the end. This gives determinacy in $L(\mathbb R)$ using, for example, a reflection argument. (I mentioned this a wh […]

A classical reference is Hypothèse du Continu by Waclaw Sierpiński (1934), available through the Virtual Library of Science as part of the series Mathematical Monographs of the Institute of Mathematics of the Polish Academy of Sciences. Sierpiński discusses equivalences and consequences. The statements covered include examples from set theory, combinatorics, […]

There is a new journal of the European Mathematical Society that seems perfect for these articles: EMS Surveys in Mathematical Sciences. The description at the link reads: The EMS Surveys in Mathematical Sciences is dedicated to publishing authoritative surveys and high-level expositions in all areas of mathematical sciences. It is a peer-reviewed periodical […]

You may be interested in the following paper: Lorenz Halbeisen, and Norbert Hungerbühler. The cardinality of Hamel bases of Banach spaces, East-West Journal of Mathematics, 2, (2000) 153-159. There, Lorenz and Norbert prove a few results about the size of Hamel bases of arbitrary infinite dimensional Banach spaces. In particular, they show: Lemma 3.4. If $K\ […]

You just need to show that $\sum_{\alpha\in F}\alpha^k=0$ for $k=0,1,\dots,q-2$. This is clear for $k=0$ (understanding $0^0$ as $1$). But $\alpha^q-\alpha=0$ for all $\alpha$ so $\alpha^{q-1}-1=0$ for all $\alpha\ne0$, and the result follows from the Newton identities.

Nice question. Let me first point out that the Riemann Hypothesis and $\mathsf{P}$-vs-$\mathsf{NP}$ are much simpler than $\Pi^1_2$: The former is $\Pi^0_1$, see this MO question, and the assertion that $\mathsf{P}=\mathsf{NP}$ is a $\Pi^0_2$ statement ("for every code for a machine of such and such kind there is a code for a machine of such other kind […]

For brevity's sake, say that a theory $T$ is nice if $T$ is a consistent theory that can interpret Peano Arithmetic and admits a recursively enumerable set of axioms. For any such $T$, the statement "$T$ is consistent" can be coded as an arithmetic statement (saying that no number codes a proof of a contradiction from the axioms of $T$). What […]

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.