Let A be a nonempty open subset of $\mathbb{C}$. Then is it possible that $A \cup {u}$ for any $u$ $\in$ $\mathbb{C}$

Question

Let A be a nonempty open subset of $\mathbb{C}$. Then is it possible that $A \cup \{u\}$ is compact for any $u$ $\in$ $\mathbb{C}$

I think it is true but how to prove it with rigor?

Answer 1

No.

It is possible to find an open set $U$ with $u\in U$ and $A-\overline U\neq\varnothing$.

Then $A-\overline U$ is open with $A-\overline U\notin\{\varnothing,\mathbb C\}$ so the set cannot be closed, hence is not compact.

Then there is an open cover $(V_i)_{i\in I}$ of $A-\overline U$ that has no finite subcover.

If $U$ is added to that cover then we are dealing with an open cover of $A\cup\{u\}$ that has no finite subcover.