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

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

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.