# Proof verification (sets & logic)

Question:

Let $$f$$: $$\mathbb{N} \rightarrow \mathbb{Z}$$ be a function that is eventually zero. i.e: There exists some $$N \in \mathbb{N}$$ s.t $$f(n)=0$$ for all $$n \geq \mathbb{N}$$. Prove that the set of such functions is countable.

Proof:

Define $$g$$: $$\mathbb{N} \rightarrow \mathbb{Z}$$ s.t $$g(i) = \left\{ \begin{array}{ll} f(i) & \quad i \in \{0,1…N-1\} \\ 0 & \quad , otherwise \end{array} \right.$$

Let $$B_n=\{h|h: \{0,1…N-1\}\rightarrow \mathbb{Z}$$}. Clearly, $$B_n$$ is equal to $$N$$ copies of $$\mathbb{Z}$$. Hence, it's countable.

Let $$G=\{f|f: \mathbb{N}\rightarrow \mathbb{Z}\}$$ is eventually zero}. So, $$G = \cup_{N=1}^{\infty} B_n$$. Countable union of countable sets is countable. So is $$G$$.

Is my approach correct?

The basic idea is good, but you rather want to take the union of $$G_N$$'s where $$G_N:=\{f:\Bbb N\to\Bbb Z:f(n)=0\text{ if } n>N\}$$

Your proof is correct. Here are some points of feedback.

• Your definition of $$g$$ depends on a specific $$f$$ and is actually just the same as that $$f$$. It does not really do anything in your proof, so just leave it out.
• It is not quite true that $$G = \bigcup_{N=1}^\infty B_n$$, because the maps in $$B_n$$ do not have domain $$\mathbb{N}$$. They can however easily be made into maps with domain $$\mathbb{N}$$ (which is clearly what you mean), you should say this (and how).
• A few small issues with notation, it is a bit sloppy. There is a $$\}$$ too many in the definition of $$G$$. The set $$B_n$$ depends on $$n$$ not on $$N$$ in its notation. That kind of thing. Not really a problem here, but in bigger proofs it will get confusing.
• For the first point, Think I should've used some other function, say $f'(i)$ instead of $f(i)$. Second, I think I should've defined a set $T_N$={$g: f' \in B_N$}? Then, take unions of all such $T_N$'s , which is equal to $G$? – SL_MathGuy Jan 1 at 14:39
• @SL_MathGuy That sounds about right. It would be good to make explicit that $g$ depends on $f'$ (e.g. denote $g_{f'}$). Because what you are saying is that $g$ is the extension of $f'$ to $\mathbb{N}$ by setting it to $0$ everywhere $f'$ is not defined. – Mark Kamsma Jan 1 at 14:46

I think there is a problem defining the map $$g$$...What you call $$f$$ is actually a set of functions....And also you do not use $$g$$ after you define it...

Some hint, How many maps from $$\Bbb{N}$$ to $$\Bbb{Z}$$ can you construct that they are zero after $$n=0$$ or after $$n=1$$,.....?