Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 8:19
This question answers your query for infinite $\kappa\leq|E|$:… – Apostolos Mar 18 '13 at 9:18
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
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.