2

I can't seem to find a full proof that for every infinite cardinal $κ$, $k^2 = k$. Does anyone have one they could share?

Pedro
  • 122,002
Learner
  • 691
  • 1
  • 4
  • 19
  • 2
    If I recall correctly, that's equivalent to the axiom of choice. – Akiva Weinberger Feb 18 '16 at 03:40
  • @Akiva : I don't think. the axiom of choice would be to find a bijection between those, not to find one surjection $\kappa \to \kappa^2$ – reuns Feb 18 '16 at 03:53
  • 1
    See this answer for the standard proof using well-orderings, as well as a sketch of the proof that this statement implies the axiom of choice. – Eric Wofsey Feb 18 '16 at 03:58
  • @Eric : I don't understand, you think the axiom of choice is also needed to show a surjection $A \to A^2$ ? in your link they seem to talk only about a bijection – reuns Feb 18 '16 at 04:27
  • @user1952009: I suspect the axiom of choice is also needed to get such a surjection. In any case, the question here is asking about a bijection, not a surjection. – Eric Wofsey Feb 18 '16 at 04:30
  • @Eric : I don't see why, to prove $|X| = |Y|$ you only have to show a surjection $X \to Y$ and another $Y \to X$ (so that with the help of the axiom of choice we could find a bijection) – reuns Feb 18 '16 at 04:32

1 Answers1

4

Claim 1. If $X$ is infinite and $Y$ is finite, $|X|+|Y|=|X|$.

Proof Choose a countably infinite subset $A\subseteq X$. Then there is a bijection $A\to A\sqcup Y$ (by shifting). Combining this with the identity map $X-A\to X-A$ gives a bijection $X\to X\sqcup Y$.

Claim 2 If $X$ is infinite, $|X|+|X|=|X|$.

Proof Consider the collection of all bijections $A\sqcup A\to A$ where $A$ can be any subset of $X$, ordered by extension. They exist, for $X$ contains a countably infinite subset. Now use Zorn's lemma to pick a maximal element. Then $X-A$ must be finite, else we could glue up a bijection $Y\sqcup Y\to Y$ for $Y$ a countably infinite subset of $X-A$. The claim follows since then $A$ and $X$ are equinumerous by the first claim.

In particular, by Schroder-Bernstein it follows that as long as at least one of $X$ and $Y$ is infinite we have $|X|+|Y|=\max(|X|,|Y|)$.

Claim 3 If $X$ is infinite, $|X\times X|=|X|$.

Proof. Again consider bijections $A\times A\to A$, and pick a maximal one. If $|X-A|\leq |A|$, then $|X|=\max(|A|,|X-A|)=|A|$, so choosing a bijection between $A$ and $X$ we're done. Hence we can assume $|X-A|\geq |A|$. Then $X-A$ contains a subset equinumerous with $A$; call it $Y$. We can then extend the bijection $A\times A\to A$ to a bijection $(A\cup Y)\times (A\cup Y)\to A\cup Y$ (here we use that $|A\times Y\cup Y\times A\cup Y\times Y|=|A\times A|+|A\times A|+|A\times A|=|A\times A|=|Y\times Y|$, with the second equality coming from Claim 2). This contradicts the maximality of $A$.

Eric Wofsey
  • 330,363
Pedro
  • 122,002