The complement of the union of sets is compact

Question

Let $X$ be a Hausdorff space and $\hat{X}=X \cup \{\infty\}.$ Now, suppose $\hat{X} \backslash U_\alpha$ is compact in $X$ for all $\alpha \in A$, where $A$ is some index set. How do I show that $$\hat{X} \backslash \bigcup_{\alpha \in A} U_\alpha$$

is compact?

Similarly, how would I show that for $V,W$, where $\hat{X} \backslash V$ and $\hat{X} \backslash W$ are both compact, that $\hat{X} \backslash (V \cap W)$ is compact.

Answer 1

$\hat {X} \setminus \bigcup_{\alpha \in A} U_{\alpha}=\bigcap (\hat {X} \setminus {U_\alpha}) $ and intersection of compact sets in a Hausdorff space is compact. For the secodn question use the fact that union of two compact sets is compact.