# Is the topological space $(\mathbb{Z}, \tau)$ compact/connected?

$$\mathbf{Question:}$$ 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 $$\mathbf{VERIFY}$$.

• Is there a particular part of this proof you're worried about? – Jackson Aug 11 '20 at 19:11
• @Jackson Just the proof in general. Would it work? – Subhasis Biswas Aug 11 '20 at 19:12
• I mean, yeah. I'm trying to see if there's something you're asking besides "here's this proof from the sky, is every step valid" so I can try and give a better answer. But it sounds like the yes is all you need? – Jackson Aug 11 '20 at 19:16
• Yes, it’s correct. – Brian M. Scott Aug 11 '20 at 19:17
• @Jackson the approval is needed. In addition, if possible, a better answer. – Subhasis Biswas Aug 11 '20 at 19:18

• @SubhasisBiswas I’d just say, if $A\subseteq \Bbb Z$ has two or more points, it contains a point $u \neq 0$ and then $\{u\}$ is clopen in $A$. No case distinctions. – Henno Brandsma Aug 11 '20 at 19:59
• Let $A$ be any nonempty, not-a-one-point proper subset of $\mathbb{Z}$. The subspace topology $\tau_s=\{A\cap U: U \in \tau\}$. Case 1: $0 \in A$ Let $u ( \neq 0 ) \in A$. Now, $T=\mathbb{Z}\setminus \{u\} \in \tau$. $\ \ T_s=T \cap A \in \tau_s$ and $Q=\{u\} \in \tau_s$. $T_s \cup Q =A$ but $T_s \cap Q = \emptyset$. Case 2: $0 \notin A$ Take any $u \in A$. $A\setminus \{u\} \in \tau \implies P=(A\setminus \{u\}) \cap A=A \setminus \{u\} \in\tau_s$. Let $\{u\}=Q \in \tau_s$. $P \cup Q=A$, but $P \cap Q = \emptyset$ – Subhasis Biswas Aug 11 '20 at 20:57
• @SubhasisBiswas That'll work too. It's essentially what I did but written out longer. I wouldn't say "but" $P \cap Q=\emptyset$, just "and". It's not a contrast; we just show a disconnection. Note that in either case we just use the singleton non-zero element. This needs no case distinction. – Henno Brandsma Aug 11 '20 at 21:48