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I've some elementary set theory problem that I came across with:

Let $S\subseteq\mathbb{R}$ be infinite set, and let $A=\{B\subseteq S: |B|<\infty \}$. I'm interested in showing that cardinality of set $A$ is equals to cardinality of $S$, i.e $|A|=|S|$.

I thought using Cantor–Bernstein–Schroeder theorem in some way.

Salech Alhasov
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Hint What is the cardinality of the set of singleton subsets of $S$?

What is the cardinality of the set of two-element subsets of $S$?

$\cdots$

What is the cardinality of the set of $n$-element subsets of $S$?

$\cdots$

Sungjin Kim
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  • Not the asker but this had bugged me: how do you prove that $|S\times S|=|S|$??? I have difficulty finding whatever theorem that is needed for this. (did not study set theory) – Gina Dec 29 '13 at 06:26
  • We need Axiom of Choice for that." $|S\times S|=|S|$ for any infinite set $S$ " is an equivalent statement to AC. – Sungjin Kim Dec 29 '13 at 06:59
  • Ah, can you tell me the name of the theorem though? Or a reference. Seems like Google only come up the the countable case. – Gina Dec 29 '13 at 07:11
  • http://en.wikipedia.org/wiki/Tarski%27s_theorem – Sungjin Kim Dec 29 '13 at 07:18
  • Thanks, but what about the other direction? (ie. AoC implies $|A|=|A\times A|$) – Gina Dec 29 '13 at 07:22
  • The other direction will be easy one: consider a well-order in a set $A$, and construct bijection between the two sets. – Sungjin Kim Dec 29 '13 at 07:24
  • Ah sorry I have not done set theory, I can't quite figure out how to do that. Can you give me a reference, if it is long to prove? – Gina Dec 29 '13 at 07:32
  • @Gina: I have written several answers about Tarski's theorem. – Asaf Karagila Dec 29 '13 at 07:59
  • @Gina See this question and other questions linked there. – Martin Sleziak Dec 29 '13 at 08:43
  • @Gina: Not every easily understandable statement has an easily understandable proof. One can prove that $|A|=|A\times A|$ using Zorn's lemma, but you still have to understand what is "transfinite induction" (on the class of cardinals), http://math.stackexchange.com/a/608555/622 – Asaf Karagila Dec 29 '13 at 08:55
  • @AsafKaragila:thanks for the link. Sorry I just thought that this is an elementary set theory question, so the answer to the question shouldn't be that hard. And the proof of $|A|=|A\times A|$ is needed to answer the question, at least for $|A|\leq 2^{\aleph_{0}}$. – Gina Dec 29 '13 at 16:51
  • @Gina: If only... one can arrange models where the axiom of choice fails, and there are infinite sets of real numbers with $|A|<|A\times A|$. – Asaf Karagila Dec 29 '13 at 17:43