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My lecture notes state that for a set $S$, we have $|S \times S| = |S|$. Some reading on this topic here suggests that this requires the axiom of choice, which implies to me that the assumption of $S$ being infinite is necessary. This makes sense, as I can come up with a counterexample for finite $S$. If $S = \{1,2,3\}$, then $|S \times S| = 3 \cdot 3 = 9 > |S| = 3$.

Am I correct that $S$ must be infinite for this result to hold? Further, is the canonical proof an explicit bijection or using Shroder-Bernstein? Every method of trying to find a bijection led to problems with either injectivity or surjectivity.

Moni145
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JeremyS
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    You're correct that the result only holds for infinite sets. I'm pretty sure that, given a well-ordering of $S$, you can construct the bijection explicitly. – Robert Shore May 02 '21 at 06:00
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    If $S$ is infinite and if $f:S^2\to S$ is a bijection then take $s_0\in S,$ and let $g(s)=f(s,s_0)$ for each $s\in S.$ Then $g$ is injective from $S$ to a proper subset of $S$, i.e. $S$ is Dedekind-infinite. Without AC it is consistent that there is an infinite set which is NOT Dedekind-infinite. – DanielWainfleet May 03 '21 at 12:26

2 Answers2

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The result holds for finite sets of size $x$ where $x^2 = x$ (i.e. 0 or 1).

According to (Show that an infinite set $C$ is equipotent to its cartesian product $C\times C$), $|S \times S| = |S|$ for all $S$ is equivalent to the axiom of choice, so you can expect any solution to involve a well-ordering on $S$.

TomKern
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Assuming AC: To show that $|S\times S|=|S|$ when $S$ is infinite, it suffices to prove it when $S$ is an infinite cardinal ordinal.

Let $<$ be the usual order on ordinals.

Induction Hypothesis: $|T\times T|=|T|$ when $T$ is an infinite ordinal and $T<S.$

Define $<^*$ on $S\times S$ as follows. For $(x,y)$ and $(x',y')$ in $S\times S$

$(i)$ if $\max (x,y)=\max (x',y')$ then $<^*$ for $(x,y),(x',y')$ is the $<$-lexicographic order;

$(ii)$ if $\max (x,y)<\max (x',y')$ then $(x,y)<^*(x',y').$

Now show that $<^*$ is a well-order, so $(S\times S,<^*)$ is order-isomorphic to $(J,<)$ for some unique ordinal $J.$ Use the Induction Hypothesis to show that $\forall p\in S\times S \,(|\{r: r<^*p\}|<S\,),$ implying $J\le S.$ But clearly $J\ge S$ (i.e. $J\ge |J|\ge |S\times \{0\}|=|S|=S.)$

Remark: For $p=(x,y)\in S\times S,$ let $t=\max (x,y).$ Then

$\{r: r<^*p\}\subseteq$ $ \{r:r<^*(t,t)\}=[(t+1)\times (t+1)]\setminus \{(t,t)\}\subset$ $ (t+1)\times (t+1),$

and by the Induction Hypothesis, the cardinal of this last set is $<S.$

DanielWainfleet
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