# Family of subset of N is countable?

## Homework Statement

I am trying to show that a family F {A$$\subseteq$$$$N$$: A is finite} Show that F is countable.

I found a proof in a book but I don't understand it. Could you help me please?

## The Attempt at a Solution

If A is finite, then F is clearly also finite and the claim becomes trivially true. We therefore suppose that A is countable infinite.

To prove that F is countable it would be sufficient to show that the family of all finite subsets of "Naturals" is countable. For each n an element of "Naturals", let Mn denote the family of all subsets N of "Naturals" satisfying N$$\subseteq$$ {1,2,...,n}.
Mn is obviously finite.

If M is the family of all finite subsets of "Naturals", we see that it can be expressed as M= the union of all Mn from n=1 to infinity. From the theorem that "union of countable sets is countable" we see though that M is countable and hence see that the family F is also countable.

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My first question is as follows. In the beginning, how do they know that if A is finite, then F is clearly also finite?

Dick
Homework Helper
If A is finite and the number of elements in A is n. Then the number of subsets of A is 2^n. You can prove this be induction. But it's pretty obvious if you think about how to choose a subset. For each element a of A you can either have a in the subset or a not in the subset.

Thanks Dick,

but why, if the problem states that A is finite, must they prove the case in which A is countable infinite?

Dick
Homework Helper
The problem doesn't state A is finite. It wants you to prove that the infinite set N has a countable number of finite subsets. 'A' is just a dummy variable in the set notation. In the proof M_n are the sets that are finite.

Thanks again Dick,

Why is it that F is stated as the family of all subsets A, but the proof uses M? Could they have used F without loss?

Dick