# Proof on why a certain topological space is compact

I'm studying for a test on general topology and in some old exam the following question has been asked:

For any finite subset $$S\subseteq\mathbb{Z}\setminus\{0\}$$ we define $$U_S:=\mathbb{Z}\setminus S$$. Let $$\mathcal{T}$$ be the set defined by $$\mathcal{T}=\{\emptyset\}\cup\{U_S\mid S\subseteq\mathbb{Z}\setminus\{0\} \text{ finite}\}$$.

(a) Show that $$\mathcal{T}$$ defines a topology on $$\mathbb{Z}$$
(b) Show that $$(\mathbb{Z},\mathcal{T})$$ is compact

I've managed to do (a). However (b) is not really working out. Can someone give me a good tip on how to solve this, thanks in advance.

Let $$\{U_i, i \in I\}$$ be an open cover of $$X$$. Then some open set $$U_{i_0}$$ must contain $$0$$, this $$U_{i_0}$$ must be of the form $$U_S$$ for some finite $$S \subseteq \mathbb{Z}\setminus\{0\}$$ by the definition of the topology.
For each $$s \in S$$ there must be some $$U_{i_s}$$ that contains $$s$$ (as we have a cover), and then $$\{U_{i_0}, U_s: s \in S\}$$ is a finite subcover (any $$x$$ in $$\mathbb{Z}$$ is in $$S$$ and so covered or else it’s in $$U_S=U_{i_0}$$) of our cover and we are done.
Note that this topology is a weird modification of the co-finite (or finite-closed) topology on $$\mathbb{Z}$$, modified so that all non-empty open sets contain $$0$$, so $$\{0\}$$ is a dense subset of $$X$$. $$X$$ is thus only $$T_0$$, not $$T_1$$ as the co-finite topology is.
Let $$\{A_\lambda\,|\,\lambda\in\Lambda\}$$ be an open cover of $$\mathbb Z$$. If $$A_\lambda=\mathbb Z$$ for every $$\lambda\in\Lambda$$, then $$\{A_{\lambda_0}\}$$ (where $$\lambda_0$$ is some element of $$\Lambda$$) is an open subcover. Otherwise, take $$\lambda_0\in\Lambda$$ such that $$A_{\lambda_0}\neq\mathbb Z$$. Then $$\mathbb{Z}\setminus A_{\lambda_0}$$ is finite. So…