Proof: For $n in mathbb{Z_+}$, the set of all functions $f: {1,...,n} to mathbb{Z_+}$ is countable

(citation: Munkres Topology 7.5 b)

Backgound (from Munkres Topology Chapter 7):

Theorem 7.1: $B$ is a nonempty countable set $LeftarrowRightarrow$ there is an injective function $g: B to mathbb{Z_+}$

Theorem 7.6: a finite product of countable sets is countable

PROOF

Let $B_n = {f | f: {1,…n} to mathbb{Z_+}}$ and let $mathbb{Z_+^n} = {(z_1, z_2,…,z_n)|z_1, z_2,…,z_nin mathbb{Z_+}}$.

Define the function $g_n: B_n to mathbb{Z_+^n}$ such that $g_n(f) = (f(1), f(2),…,f(n))$, where $fin B_n$.

Let $z = z’ in mathbb{Z_+^n}$, such that $z = g_n(f), z’ = g_n(f’)$ for $f,f’in B_n$.

Then $z = (f(1),f(2),…,f(n)) = (f'(1),f'(2),…,f'(n)) = z’$, which holds iff $f(1)=f'(1), f(2)=f'(2),…,f(n)=f'(n)$.

This implies that $f = f’$, hence $g_n$ is an injection from $B_n$ into $mathbb{Z_+^n}$.

$mathbb{Z_+^n}$ is countable by Theorem 7.6, as $mathbb{Z_+}$ is countable. Thus, by Theorem 7.1, $exists$ an injective function $h: mathbb{Z_+^n}to mathbb{Z_+}$. Then, $h circ g_n: B_nto mathbb{Z_+}$ is injective.

Therefore, by Theorem 7.1, $B_n$ is countable.

I believe this is sufficient, but please let me know, if it may be corrected or improved.