1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Lord Shark the Unknown 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 $A_k$. $\endgroup$ – Lord Shark the Unknown 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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