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Let $\mathcal{P}_\kappa(E)$ is the collection of all subsets of $E$ of cardinality $\le \kappa$, and $[E]^\kappa=\{A: A\subset E, |A|=\kappa\}.$ Then $|\mathcal{P}_\kappa(E)|=|E|^\kappa$ or only $|\mathcal{P}_\kappa(E)|\le|E|^\kappa$?

Thanks for your help.

Asaf Karagila
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  • What have you tried, for example, have you tried any basic examples? Can you relate $\mathcal{P}_2({0,1})$ and $[{0,1}]^2$? – dtldarek Mar 18 '13 at 08:19
  • This question answers your query for infinite $\kappa\leq|E|$: http://math.stackexchange.com/questions/311848/proving-that-for-infinite-kappa-kappa-lambda-kappa-lambda/ – Apostolos Mar 18 '13 at 09:18
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    Notational comment: $\mathcal P_\kappa(E)$ usually means the collection of subsets of $E$ of cardinality strictly smaller than $\kappa$. – Andreas Blass Mar 18 '13 at 13:08

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If $E$ is finite then $\mathcal P_\kappa(E)$ is finite, and has strictly more sets than $[E]^\kappa$, so the cardinality must be different.

If $E$ is infinite then $|\mathcal P_\kappa(E)|\leq|E^\kappa|=|E|^\kappa$, because the function sending $f\in E^\kappa$ to its range is surjective (sans the empty set, but one element is not important in the infinite case anyway). As Proving that for infinite $\kappa$, $|[\kappa]^\lambda|=\kappa^\lambda$ show, $[E]^\kappa$ has cardinality $|E|^\kappa$. Therefore equality ensues.

Note that it is obvious that $\kappa\leq|E|$ otherwise $\mathcal P_\kappa(E)$ is $\mathcal P(E)$, whereas $[E]^\kappa$ is empty.

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Asaf Karagila
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