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

  • $\begingroup$ Is there a particular part of this proof you're worried about? $\endgroup$
    – Jackson
    Aug 11 '20 at 19:11
  • $\begingroup$ @Jackson Just the proof in general. Would it work? $\endgroup$ Aug 11 '20 at 19:12
  • $\begingroup$ 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? $\endgroup$
    – Jackson
    Aug 11 '20 at 19:16
  • 2
    $\begingroup$ Yes, it’s correct. $\endgroup$ Aug 11 '20 at 19:17
  • $\begingroup$ @Jackson the approval is needed. In addition, if possible, a better answer. $\endgroup$ Aug 11 '20 at 19:18

Yes, both arguments are fine. Compact and disconnected ( even totally disconnected) is the right answer.

  • $\begingroup$ @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. $\endgroup$ Aug 11 '20 at 19:59
  • $\begingroup$ I tried something very elementary (being new to topology), but it uses case distinction. It would be helpful if you could verify that one. I am trying to work out what you said as well. $\endgroup$ Aug 11 '20 at 20:57
  • $\begingroup$ 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 $P \cap Q = \emptyset$ $\endgroup$ Aug 11 '20 at 20:57
  • 1
    $\begingroup$ @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. $\endgroup$ Aug 11 '20 at 21:48

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