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.

Advertisements

Like this:

LikeLoading...

Related

This entry was posted on Wednesday, April 9th, 2008 at 2:09 am and is filed under 116c: Set theory. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

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)

The technique of almost disjoint forcing was introduced in MR0289291 (44 #6482). Jensen, R. B.; Solovay, R. M. Some applications of almost disjoint sets. In Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968), pp. 84–104, North-Holland, Amsterdam, 1970. Fix an almost disjoint family $X=(x_\alpha:\alpha

At the moment most of those decisions come from me, at least for computer science papers (those with a 68 class as primary). The practice of having proceedings and final versions of papers is not exclusive to computer science, but this is where it is most common. I've found more often than not that the journal version is significantly different from the […]

The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

I assume by $\aleph$ you mean $\mathfrak c$, the cardinality of the continuum. You can build $D$ by transfinite recursion: Well-order the continuum in type $\mathfrak c$. At stage $\alpha$ you add a point of $A_\alpha$ to your set, and one to its complement. You can always do this because at each stage fewer than $\mathfrak c$ many points have been selected. […]

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 negative. More precisely: Assume determinacy in $L({\mathbb R})$. Then $2^\omega/E_0$ is a successor cardinal to ${\mathfrak c}$ (This doesn't matter, all we need is that it is strictly larger. T […]

R. Solovay proved that the provably $\mathbf\Delta^1_2$ sets are Lebesgue measurable (and have the property of Baire). A set $A$ is provably $\mathbf\Delta^1_2$ iff there is a real $a$, a $\Sigma^1_2$ formula $\phi(x,y)$ and a $\Pi^1_2$ formula $\psi(x,y)$ such that $A=\{t\mid \phi(t,a)\}=\{t\mid\psi(t,a)\}$, and $\mathsf{ZFC}$ proves that $\phi$ and $\psi$ […]

Yes, the suggested rearrangement converges to 0. This is a particular case of a result of Martin Ohm: For $p$ and $q$ positive integers rearrange the sequence $$\left(\frac{(−1)^{n-1}} n\right)_{n\ge 1} $$ by taking the ﬁrst $p$ positive terms, then the ﬁrst $q$ negative terms, then the next $p$ positive terms, then the next $q$ negative terms, and so on. Th […]

Yes, by the incompleteness theorem. An easy argument is to enumerate the sentences in the language of arithmetic. Assign to each node $\sigma $ of the tree $2^{

A simple example is the permutation $\pi$ given by $\pi(n)=n+2$ if $n$ is even, $\pi(1)=0$, and otherwise $\pi(n)=n−2$. It should be clear that $\pi$ is computable and has the desired property. By the way, regarding the footnote: if a bijection is computable, so is its inverse, so $\pi^{-1}$ is computable as well. In general, given a computable bijection $\s […]

The question is asking to find all polynomials $f$ for which you can find $a,b\in\mathbb R$ with $a\ne b$ such that the displayed identity holds. The concrete numbers $a,b$ may very well depend on $f$. A priori, it may be that for some $f$ there is only one pair for which the identity holds, it may be that for some $f$ there are many such pairs, and it may a […]

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.