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Lets say that A is a countable set. How would I go about showing that A has countably many finite subsets?

I need to show that for every $n \in N$, the set $P^n(A)$ of finite subsets - with exactly $n$ elements - of A is countable. How would I then show how the union $U_{n \in N} P^n(A)$ of all these sets is countable?

P. J.
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Iver Finne
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    "The union of a countable number of countable sets is itself countable" should be a theorem you have access to. – Greg Martin Feb 04 '21 at 05:58
  • Suppose $A=\mathbb{N}$. Let $p_1,\ldots,p_n$ be distinct primes. Consider the function $\phi : P^n(A)\longrightarrow \mathbb{N}$ given by $$\phi\Big({a_1,\ldots, a_n}\Big)=p_1^{a_1}\times \dots \times p_n^{a_n}$$ Show $\phi$ is injective. – Matthew H. Feb 04 '21 at 06:07
  • @MatthewPilling With $a_1 < a_2 < \cdots < a_n$? – Brian Moehring Feb 04 '21 at 06:14
  • @Brian Moehring Yes. You beat me to it. $\phi$ wouldn't be a function otherwise. – Matthew H. Feb 04 '21 at 06:23

1 Answers1

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Hints:

By the definition of a countable set, there exists an injection $f: A \to \Bbb{N}$.

Let $\mathcal{G}$ denote the set of all finite subsets of $A$.

Define $\xi: \mathcal{G} \to \Bbb{N}$ by $\xi(G) = \prod_{k \mathop \in f(G)} p_{k + 1}$

where $p_n$ denotes the $n$th prime number.

Unknown
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