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Let $S$ be a set of infinite cardinality $\kappa_1$. What is the cardinality $k$ of the set of subsets of $S$ with cardinality $\kappa\le\kappa_0<\kappa_1$? I understand that if $\kappa_0$ is finite, then $k=|S\times S\times\ldots \times S| $ (at most $\kappa_0$ factors), so $k=\kappa_1^{\kappa_0}=\kappa_1$. But does this still hold if $\kappa_0$ is infinite? Thanks for any help!

Daniel
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  • I think your argument that the result is $\kappa_1^{\kappa_0}$ still works, because if you excise a set of cardinality at most $\kappa_0$ from a set of cardinality $\kappa_1$, then the result has cardinality $\kappa_1$. (I think there may be some transfinite recursion step here in the formal proof, but the idea seems clear enough.) But $\kappa_1^{\kappa_0}$ needn't necessarily be $\kappa_1$. For instance $\aleph_1^{\aleph_0} \geq 2^{\aleph_0} = \beth_1$, which is bigger than $\aleph_1$ under the negation of the continuum hypothesis. – Ian Jun 06 '15 at 15:09
  • Thanks. Yes, this is certainly the intuition from finite $\kappa_0$. And I did assume the CH in the last step. – Daniel Jun 07 '15 at 16:42

1 Answers1

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There are exactly $\kappa_1^{\kappa_0}$ such subsets. Let $\mathcal{A} = \{ A \subseteq S : |A| = \kappa_0 \}$.

  1. $(\leqslant)$

    Let $A \in \mathcal{A}$ i.e. $A \subseteq S$ and $|A| = \kappa_0$. Then there exists a bijection $f : \kappa_0 \to A$. Hence the mapping

    $$S^{\kappa_0} \ni f \mapsto f[\kappa_0] \in \mathcal{P}(S)$$

    is onto $\mathcal{A}$, so $|\mathcal{A}| \leqslant \kappa_1^{\kappa_0}$.

  2. $(\geqslant)$

    We will use the fact that $\kappa_0 \cdot \kappa_1 = \kappa_1$ so the set $S$ can be divided into $\kappa_0$ pairwise disjoint subsets of power $\kappa_1$:

    $$S = \bigcup_{\alpha \in \kappa_0} S_{\alpha}$$

    and for each $\alpha \in \kappa_0$ there is a bijection $\varphi_{\alpha} : S \to S_{\alpha}$.

    Let every $f \in S^{\kappa_0}$ be assigned the set $A = \{ \varphi_{\alpha}( f( \alpha ) ) : \alpha \in \kappa_0 \}$. Since $S_{\alpha}$ are pairwise disjoint, the values $\varphi_{\alpha}( f(\alpha) )$ are all distinct so $|A| = \kappa_0$ hence $A \in \mathcal{A}$. The mapping $f \mapsto A$ is also injective, because given $A$ we can reproduce $f$: for each $\alpha$ there is exactly one $a \in S_{\alpha} \cap A$ so it must be $f(\alpha) = \varphi_{\alpha}^{-1}(a)$. Therefore $|\mathcal{A}| \geqslant \kappa_1^{\kappa_0}$.

Adayah
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  • Thank you very much Adayah. Sorry for the late reply --- being a humble physicist, I first had to learn some of the notations with the help of a mathematician friend, but now I understand the construction. In the original problem I had in fact $|A|\le \kappa_0$, but as my friend explained both parts easily generalize to that case by using a surjective function $f$. – Daniel Jun 15 '15 at 09:21
  • @Daniel: I'm glad I could help. :) – Adayah Jun 15 '15 at 09:40