Proof on why a certain topological space is compact

Question

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.

Answer 1

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.

Answer 2

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…