Easy proof the set of finite Set in countable is countable

Question

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.

Answer 1

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.

Answer 2

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.