# Tarski's caracterisation of inaccessible cardinals

Alfred Tarski, in his paper "Ueber unerreichbare Kardinalzahlen" Fund. Math. vol 30 (1938) pp 68-89 proves the followig theorem of ZFC "If the cardinal of the set Y is equal to the cardinal of the set of the subsets of Y that are not equipotent with Y, then the cardinal of Y is (strongly) inaccessible". 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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Are you sure you quoted the theorem correctly? The continuum hypothesis implies that a set of size $\aleph_1$ has only $\aleph_1$ countable subsets. – Andreas Blass Jan 19 '11 at 16:34
I'm confused by the statement of the theorem. There are $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}=\aleph_1$ many countable subsets of $\aleph_1$, assuming $CH$, so this would imply that $\omega_1$ is strongly inaccessible. What am I missing? – Noah Schweber Jan 19 '11 at 16:38
YES,I am VERY SORRY, I did not think "inaccessible", but "regular !!! I will write the correct question; Gérard lang – Gérard Lang Jan 19 '11 at 18:58
There is no need to write another question, it was already answered here. – Andrés Caicedo Jan 19 '11 at 19:43
Is Tarski using the axiom of choice? – Andrés Caicedo Jan 19 '11 at 19:45

In modern notation, it says, "if $\kappa$ is a cardinal and $\kappa ^{< \kappa} = \kappa$, then $\kappa$ is strongly inaccessible." This isn't entirely true since the antecedent holds for $\kappa = \omega$ but $\omega$ isn't considered strongly inaccessible, but that's not a big deal. More importantly, under CH the antecedent will hold of $\aleph _1$ but $\aleph _1$ isn't a limit cardinal. So we need to add the assumptions that $\kappa$ is an uncountable limit cardinal. Given that, we can proceed:

So let's assume $\kappa ^{< \kappa} = \kappa$. First we show $\kappa$ is strong limit: $\kappa \leq 2^{< \kappa} \leq \kappa ^{< \kappa} = \kappa$. Next we show $\kappa$ is regular: Suppose not, then $\kappa ^{< \kappa} = \kappa < \kappa ^{ \mathrm{cf} ( \kappa) } \leq \kappa ^{< \kappa}$, contradiction.

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Thank you very much. This is the right answer for my very question ! Gérard Lang – Gérard Lang Jan 19 '11 at 21:48