$\mathbf{Question:}$ Consider the topology $\tau=\{U \subseteq \mathbb{Z}: \mathbb{Z} \setminus U$ is finite or $0\notin U \}$ on $\mathbb{Z}$. Then, the topological space $(\mathbb{Z}, \tau)$ is
(A) compact but NOT connected; (B) connected but NOT compact; (C) both compact and connected; (D) neither compact nor connected.
$\mathbf{Attempt:}$ Consider $A=\{1\} \in \tau$ and $B=\{...,-2,-1,0,2,3,4,...\} \in \tau$. Now, $A \cup B= \mathbb{Z}$ but $A \cap B= \emptyset$ where $A$ and $B$ are open sets. Therefore $(\mathbb{Z},\tau)$ is not connected.
Now, take an arbitrary family of open sets $\{G_\alpha\}_{\alpha \in I}$ where $G_\alpha \in \tau$ for all $\alpha \in I$. Further suppose that it covers $\mathbb{Z}$.
Since $\{G_\alpha\}_{\alpha \in I}$ covers $\mathbb{Z}$, there exists an $\alpha=p$ such that $0 \in G_p$. Now, by hypothesis, $\mathbb{Z}\setminus G_p$ is finite, say $| \mathbb{Z}\setminus G_p|=n$
We now require "at most" $n$ members of the family such that $(\mathbb{Z}\setminus G_p)\cap G_\alpha \neq \emptyset$. Since it is an open cover, we can find $G_{k_1},G_{k_2},...,G_{k_m}$, $m \leq n$ such that $(\mathbb{Z}\setminus G_p) \subset G_{k_1} \cup...\cup G_{k_m}$.
Therefore, for every open cover $\{G_\alpha\}_{\alpha \in I}$ of $\mathbb{Z}$, there exists a finite subcover $G_{k_1} \cup...\cup G_{k_m}\cup G_p$.
Hence, $(\mathbb{Z}, \tau)$ is compact. So, Option (A) is the right choice.
Is this correct?
Kindly $\mathbf{VERIFY}$.