1
$\begingroup$

Consider the smallest topology $\tau $ on $\Bbb C$ in which all the singleton sets are closed.

Is it true that $(\Bbb C,\tau)$ is compact and connected.

I thoght to derive a contradiction by using the open cover

$\Bbb C=\{\cup C_a :a\in \Bbb C\}$ where $C_a=\Bbb C \setminus \{a\} $ which is open.

But I am not getting anywhere from here .The example fails

Please help to solve the problem

$\endgroup$
7
$\begingroup$

Hint: first prove that the non-empty open sets in this topology are the complements of finite sets. Then if you have a covering $X \subseteq \bigcup_i U_i$ of a set $X$ by non-empty open sets $U_i$, pick some index $i_0$ say, and observe that $X \mathop{\backslash} U_{i_0}$ is finite. As for connectedness, note that any two non-empty open sets meet.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.