I have considered as following:

Consider any infinite set $A$ with cardinality $m$ for some infinite cardinal $m$, for each natural number $n$, denote the set of subsets of $A$ with exactly $n$ elements as $A_n$.

Denote the set of finite subsets of $A$ by $F$, thus $F=\lbrace \emptyset \rbrace \cup A_1\cup A_2 \cup...=\lbrace \emptyset\rbrace \oplus A_1\oplus A_2\oplus...$.

There is the missing argument to conclude that for each $n\in \Bbb N$, $A_n\approx A$

Hence $\#F=\#(\lbrace \emptyset\rbrace \oplus A_1\oplus A_2\oplus...)=1+m+m+... =m+m+...=(1+1+1+...)m=\aleph_0\cdot m=m$

Is that correct? I fell I am stuck on proving that each $A_n$ has the same cardinality as $A$. It seems to be related to the product set but I can only find an injection from each $A_n$ to $A^n$. So could someone please help? Thanks so much!


The collection $A_k$ of $k$-element subsets of $A$ has cardinality at most $|A^k|$, since a subset $A^{(k)}$ of $A^k$ maps onto $A_k$ by $(a_1,\ldots,a_k)\mapsto\{a_1,\ldots,a_k\}$. ($A^{(k)}$ will be the collection of $k$-tuples of distinct elements) By standard cardinal arithmetic $|A^k|=|A|^k =|A|$ when $A$ is infinite. Therefore $|A_k|\le |A|$. That's all you need, equality is true, but $\le$ is good enough.

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  • $\begingroup$ Is that the fact that as we have already know that $|A|\le |A_k|$, then as we have both $|A|\le |A_k|$ and $|A_k|\le |A|$ so the cardinality must be equal? May I please ask if the rest part of my argument is correct? $\endgroup$ – PropositionX May 5 '17 at 15:40
  • $\begingroup$ @PropositionX The rest ($|F|\le\aleph_0|A|=|A|$) is fine! $\endgroup$ – Angina Seng May 5 '17 at 15:42
  • $\begingroup$ May I please ask for an argument to prove that $|A_k|\le |A|$? I stucked when I tried to write it out formally. $\endgroup$ – PropositionX May 5 '17 at 16:23
  • $\begingroup$ @PropositionX $|A^{(k)}|\le|A^k|$ as $A^{(k)}$ is a subset of $ A^k$, $|A_k|\le |A^{(k)}|$ as $A^{(k)}$ maps onto $A_k$. $\endgroup$ – Angina Seng May 5 '17 at 16:25
  • $\begingroup$ Oh, sorry. I am now asking about $|A|\le |A_k|$. Apologize for the typo. $\endgroup$ – PropositionX May 5 '17 at 16:30

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