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Let a topology $\tau$ is defined on $\mathbb{Z}$ is as follows: $$ \tau=\{U\subset \mathbb{Z} :\ \mathbb{Z}\setminus U\ \text{is finite or}\ 0\notin U \}. $$ Then what about the space $(\mathbb{Z},\tau),$ is it connected, is it compact?

According to me, the space is neither compact nor connected. As $$ \mathbb{Z}=\mathbb{Z}\setminus \{1,2\}\cup \{1,2\} $$ And here $\mathbb{Z}\setminus \{1,2\}$ is open since its complement is finite and it is closed since its complement does not contain $0$. Hence, $\{1,2\}$ is also clopen. So these two will form a separation of the set $\mathbb{Z}$ and hence it is not connected. It is not compact as $$ \{ \{0,1,2,3,\cdots\}, \{-1\}, \{-2\},\cdots, \{-n\},\cdots \} $$ does not have a finite subcover. And hence it is not compact. Am I right?

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    $\begingroup$ $\{0,1,2,3,\cdots\}$ is not open, it contains $0$ but its complement is infinite. $\endgroup$ – Daniel Fischer Feb 5 '17 at 15:40
  • $\begingroup$ Okay that's true. I missed it. $\endgroup$ – Sachchidanand Prasad Feb 5 '17 at 15:44
  • $\begingroup$ The non-connectedness is fine, the sets are disjoint open and so are both closed as well. $X \setminus \{1\}$ and $\{1\}$ are even simpler. $\endgroup$ – Henno Brandsma Feb 5 '17 at 17:16
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If $(U_{i})_{i \in I}$ is an open cover of $\mathbb{Z}$, then there is some $i_0 \in I$ such that $0 \in U_{i_0}$. As $U_{i_0}$ is open and $0 \in U_{i_0}$, the complement of $U_{i_0}$ must be finite. For every element $x$ in the finite complement of $U_{i_0}$, pick an $i_x$ such that $x \in U_{i_x}$. Thus we obtain a finite subcover, so this space is compact.

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