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Suppose I know a result that the set of finite sets in $\mathbb{N}$ is countable.

Is there a very quick way to show that the set of finite sets in any $X$ countable is countable?

Idea...two sets have same cardinality if there is a bijection, we biject set of finite sets, bijection preserves cardinality. End of proof.

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  • $\begingroup$ I'm not sure what statement you're trying to prove. Is it the following? "If $X$ is countable, then the set of all finite subsets of $X$ is countable." $\endgroup$ – kccu Jun 15 '16 at 19:35
  • $\begingroup$ "we biject set of finite sets" You need a little more than this, because it seems to be assuming what you want to prove. For what it's worth, I think it is simpler to just prove the general result. Trying to lift a bijection from the underlying countable sets to their corresponding sets of finite subsets (and proving that the "lifting" also is a bijection) seems harder to carry out than proving the general result directly, at least to me. $\endgroup$ – Dave L. Renfro Jun 15 '16 at 19:35
  • $\begingroup$ I think you are right, but it's a matter of using language in an acceptably clear yet rigorous way. "for a finite set in N, apply the bijection to get an equivalent distinct and unique finite set in X. Likewise for any finite set in X we can apply the bijection to get an equivalent and distinct finite set in N So this is a 1-1 bijection between the set of finite sets of N and the set of finite sets of X" would make your statement clearer. Or if you want to be really pedantic you can use my answer. $\endgroup$ – fleablood Jun 15 '16 at 20:13
  • $\begingroup$ This is most certainly not a duplicate and are entirely different questions. The other is a proof that finite sets are countable. This is asking how we can transfer the result of one countable set to another countable set. Entirely different issues. $\endgroup$ – fleablood Jun 16 '16 at 20:24
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I think you have the right idea, but you need to be more specific.

Call the bijection $\mathbb N\to X$ $f$ and write down how you use that to construct a mapping from a given subset of $X$ to a subset of $\mathbb N$, and then use that you know that the set of finite subsets of $\mathbb N$ is countable.

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Mostly.

$\mathbb N$ is countable. $X$ is countable. Therefore there is an bijective $f:\mathbb N \rightarrow X$.

Let $S_N = $ the set of finite sets of $\mathbb N$. Let $S_X = $ the set of finite sets of $X$.

Claim: Let $F:S_N \rightarrow X_N$ be defined as for $A \in S_N$, $F(A) = \{f(a)|a \in A\}$. I claim (first of all that this is indeed a function between $S_N$ and $S_X$) and it is a bijection.

Pf: Left to the reader as an excercise... (c'mon... it's ... basic.)

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Oh, all right. As $A \subset \mathbb N$ and $f: \mathbb N \rightarrow X$ each $f(a) \in X$ so $F(A) \subset X$ and as $A$ is finite $F(A) = \{f(a)|a \in A\}$ is finite so $F(A) \in S_X$ so $F$ is a function from $S_N \rightarrow S_X$.

For $B \in S_X$ the set $\{f^{-1}(b)|b \in B\} \subset \mathbb N$ and is finite so so is $\in S_N$. So $F(\{f^{-1}(b)|b \in B\}) = B$ so $F$ is surjective.

If $A \ne A'$; $A \in S_N, A' \in S_N$ there must be some $a \in A; a \not \in A'$ or $a \in A'; a \not \in A$. Wolog the first. Then $f(a) \in F(A)$ but $f(a) \not \in F(A')$ (as $f$ is injective). So $F(A) \ne F(A')$ so $F$ is injective.

So $F$ is a bijection.

So $S_N$ and $S_X$ have the same cardinality.

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  • $\begingroup$ That was an insult? Aimed at whom exactly? It was a joke. I usually hate "proof is left to reader" comment but in this case the OP gave a valid argument. I just fleshed out the statement of it to make the statement more rigorous. The "proof" would be nothing more that the OP restating his/her argument. In this case the proof itself is not important compared to the stating of the claim. $\endgroup$ – fleablood Jun 15 '16 at 20:01

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