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Dedekind-Infinite Set. A set $S$ is called a Dedekind-Infinite Set if there exists a bijection $f:S\to T$ where $\emptyset \subset T\subset S$.

I was wondering if the following definition is logically equivalent to the above definition.

Descartes-Infinite Set. A set $S$ is called an Descartes-Infinite Set if $S$ is not a singleton set and there exists a bijection $f:S\to S\times S$.

My questions are the following,

  • Are the two definitions logically equivalent in any model of $\sf{ZFC}$,$\sf{TG}$ or $\sf{NBG}$?

  • Does there exist any model(s) of Set Theory in which these two definitions are not logically equivalent?

  • Suppose it is known that $S$ is a Descartes-Infinite set. So there exists a bijection between $S$ and $S\times S$. My question is, is there any way to explicitly define the bijection?

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  • They're certainly equivalent over ZFC. – goblin GONE Apr 18 '16 at 08:57
  • @goblin: But how can we prove it? –  Apr 18 '16 at 09:01
  • I think $\kappa^2=\kappa⟺\kappa = 0 \vee \kappa = 1 \vee \kappa \geq \aleph_0$ is probably a standard exercise, so have a look at your favorite introduction to set theory or cardinal numbers and you'll probably find a proof. I'm guessing the standard proof probably involves first proving $ℵ^2_α=ℵ_α$ over ZF, and then using the axiom of choice to get the full result. – goblin GONE Apr 18 '16 at 11:45

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It’s true in $\mathsf{ZF}$ that every Descartes-infinite set is Dedekind-infinite. Suppose that $S$ is Descartes-infinite; then there is a bijection $f:S\times S\to S$, and

$$g:S\to S:x\mapsto f(\langle x,x\rangle)$$

is an injection such that $g[S]\subsetneqq S$, so $S$ is Dedekind-infinite.

In this answer Asaf Karagila shows that it’s a theorem of $\mathsf{ZF}$ that the axiom of choice is equivalent to the statement that each infinite set is Descartes-infinite. (By infinite here I mean that there is no bijection between the set and any finite ordinal.) Thus, in $\mathsf{ZFC}$ the notions of infinite, Dedekind-infinite, and Descartes-infinite coincide.

In the absence of $\mathsf{AC}$, however, there can be sets that are Dedekind-infinite but Descartes-finite. Let $X$ be an infinite, Dedekind-finite set, and let $S$ be the disjoint union $X\sqcup\omega$; clearly $S$ is Dedekind-infinite. Suppose that $f:S\to S\times S$ is a bijection.

Let $N=\omega\cap f^{-1}[X\times X]$, and suppose that $N$ is infinite. If $N_x=\{n\in N:f(n)\in\{x\}\times X\}$ is infinite for some $x\in X$, it’s easy to construct an injection from $N_x$ and hence from $\omega$ into $X$, contradicting the Dedekind-finiteness of $X$, so $N_x$ is finite for each $x\in X$. Thus, if we set $Y=\{x\in X:N_x\ne\varnothing\}$, then $\{N_x:x\in Y\}$ is a partition of $N$, and therefore the map

$$g:Y\to\omega:x\mapsto\min N_x$$

is an injection. Moreover, $Y$ is infinite, so $g^{-1}$ is an injection of the infinite set $g[Y]\subseteq\omega$ into $X$, so there is an injection of $\omega$ into $X$, again contradicting the Dedekind-finiteness of $X$. It follows that $N$ must be finite and hence that $N_x=\varnothing$ for some $x\in X$. Then the function

$$h:X\to X:y\mapsto f^{-1}(\langle x,y\rangle)$$

is an injection from $X$ to a proper subset of $X$, which is impossible. Thus, there is no bijection $f:S\times S\to S$, and $S$ is Descartes-finite.

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Brian M. Scott
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  • So the two notions of infinite coincide in $\sf{TG}$ and $\sf{NBG}$, right? But what about my third question? –  Apr 19 '16 at 12:17
  • @user170039: Yes, equivalence in $\mathsf{ZFC}$ implies equivalence in the other two theories. I very much doubt that the answer to your third question is yes, but I don't know for sure. – Brian M. Scott Apr 19 '16 at 12:25
  • Actually the motivation for the third question came when I was trying to find a bijection between $\mathbb{N}$ and $\mathbb{N}\times \mathbb{N}$. So, I was wondering if there is really any general way to define bijection from $X$ to $X\times X$. –  Apr 19 '16 at 12:36
  • @user170039: For well-ordered $X$ there is, but in general I don't think so. – Brian M. Scott Apr 19 '16 at 12:45
  • Can you mention some reference(s)? –  Apr 19 '16 at 13:07
  • @user170039: My answer and the one by Asaf to which I linked contain proofs or references for everything that I've asserted here. – Brian M. Scott Apr 19 '16 at 13:14
  • Also Brian M. Scott wrote an answer about this right here. And for some reason Brian is putting my name in boldface. Is it to boldly go where several people have gone before? I don't know. But maybe Brian knows. :-) – Asaf Karagila Apr 19 '16 at 16:11
  • @Asaf: I put users' screen names in boldface because so many of them, unlike yours and mine, don't look much like names; it helps in identifying them as such. – Brian M. Scott Apr 19 '16 at 17:51
  • @AsafKaragila: The answer to which you have given a link is precisely Brain's answer to this question. –  Apr 20 '16 at 05:35
  • @user170039: Yes, this was on purpose. – Asaf Karagila Apr 20 '16 at 05:38