- #1

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## Homework Statement

Let [itex]A=\{f:\mathbb{Z}\to\mathbb{Z}: f(n)\neq 0 \text{for a finite number of n}\}[/itex], prove that [itex]A[/itex] is countable.

## Homework Equations

I'm considering using that it would be equivalent to prove that the set [itex]A'=\{f:\mathbb{N}\to\mathbb{N}: f(n)\neq 0 \text{for a finite number of n}\}[/itex] is countable.

## The Attempt at a Solution

I'm dividing this solution in two steps, first proving what I stated en (2), this is, that [itex]A[/itex] and [itex]A'[/itex] have the same cardinal. I'm still trying to prove this.

(...)

Then prove [itex]A'[/itex] is countable, I define [itex]A'_k = \{f:\mathbb{N}\to\mathbb{N}: f(n)\neq 0 \text{ for k numbers}[/itex], I want to prove that every [itex]A'_k[/itex] is countable. I define now the function [itex]g_k:A'_k\to \underbrace{\mathbb{N}\times\dots\times\mathbb{N}}_{\text{2k}}[/itex] by [itex]g_k(f)=(n_1,\dots,n_k,f(n_1),\dots,f(n_k))[/itex] where [itex]n_1,n_2,\dots[/itex] are the points where [itex]f[/itex] is not zero.

This function defines (or at least it seems to me) an injection to the set [itex]A'_k[/itex] to a countable set.

A problem is that only a finite product of countable sets is countable, can I still conclude [itex]A'[/itex] after take the union of the [itex]A'_k[/itex]?.