# Tarski's caracterisation of REGULAR cardinals

Alfred Tarski, in his paper "Ueber unerreichbare Kardinalzahlen" Fund. Math. 1938 proves the following ZFC theorem "if the cardinal of the set Y is equal to the cardinal of the set of subsets of Y that are not equipotent to Y, then the cardinal of Y is REGULAR." The proof of the paper is rather long and involved. Question: Is there another known simpler proof of this theorem ? Gérard Lang

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How many theorems with rather long and involved proofs did Tarski's paper contain? –  Franz Lemmermeyer Jan 19 '11 at 19:09
This was already solved in the other question. –  Andres Caicedo Jan 19 '11 at 19:42
Which other question? –  Stefan Geschke Jan 19 '11 at 19:55
Oh I see, you mean mathoverflow.net/questions/52526/… –  Stefan Geschke Jan 19 '11 at 19:58
Hi Stefan, yes, I would have suggested to move your answer there but it doesn't matter. –  Andres Caicedo Jan 19 '11 at 20:00

Let me prove the contraposition: If $\kappa$ is not regular, then it has more than $\kappa$ subsets of size strictly less than $\kappa$.
Suppose $\kappa$ is not regular. Let $\lambda$ be the cofinality of $\kappa$. Note that the number of subsets of $\kappa$ of size $\lambda$ is $\kappa^\lambda$. It is therefore enough to show that $\kappa^\lambda$ is strictly larger than $\kappa$.
Choose a cofinal sequence $\{\alpha_\gamma:\gamma\in\lambda\}$ of ordinals in $\kappa$. Let $f:\kappa\to{}^\lambda\kappa$ be a function. We show that it is not onto. Define $g:\lambda\to\kappa$ be letting $g(\gamma)$ be the least element of $\kappa\setminus\{(f(\beta)(\gamma):\beta\in\alpha_\gamma\}.$ This exists since $\{(f(\beta)(\gamma):\beta\in\alpha_\gamma\}$ is of size striclty less than $\kappa$. Then $g$ does not appear in the range of $f$.
For suppose $g=f(\alpha)$ for some $\alpha\in\kappa$. Choose $\gamma\in\lambda$ such that $\alpha\in\alpha_\gamma$. Then $g(\gamma)\not\in\{f(\beta)(\gamma):\beta\in\alpha_\gamma\}$ and hence in particular, $g(\gamma)\not=f(\alpha)(\gamma)$, a contradiction.