Other examples of a "standard compactness argument"?

Question

By a "standard compactness argument" I mean a proof where one has two families $(U_i)_{i \in I}$ and $(V_i)_{i \in I}$ in a topological space, obtains finite subfamilies $(U_j)_{j \in J}$ and $(V_j)_{j \in J}$ of those families, and finally takes $U = \bigcup_{j \in J} U_j$ and $V = \bigcap_{j \in J} V_j$.

That argument is used in the proofs of these two well-know results:

1. A compact subset of a Hausdorff space is closed.
2. The product of two compact spaces is itself compact.

What other results use the same argument? (Well-known results would be preferable.)

Note: I am not asking for the actual proofs — just mention of results whose proofs use that argument.

Answer 1

More very useful facts proved by such standard compactness arguments:

The closed subspace of a compact space is compact, is another type of argument (namely easier: we just add $X\setminus A$ to a cover of $A$), but what is an argument as you mean is the partial converse: if $A$ is a subspace of a Hausdorff space and $A$ is compact (in the subspace topology) then $A$ is closed in $X$.

Let $f: X \to Y$ be a perfect map (i.e. surjective, continuous, closed and for every $y \in Y$, $f^{-1}[\{y\}]$ is compact.). Then for every $K \subseteq Y$ compact, $f^{-1}[K]$ is compact.

If $Y$ is compact then, for every space $X$, the projection function $\pi_X : X \times Y \to X$ is a closed map.

Answer 2

Every compact (Hausdorff) topological space $K$ is regular.

Proof: Let $p\in K$ and let $C$ be a closed subset of $K$. For each $c\in C$, there are open sets $U_c$ and $V_c$ such that:

• $p\in U_c$;
• $c\in V_c$;
• $U_c\cap V_c=\emptyset$.

Since $(V_c)_{c\in C}$ is an open cover of $C$ and $C$ is compact (since it is closed and $K$ is compact), there are $c_1,\ldots,c_n\in C$ such that $C\subset\bigcup_{k=1}^nV_{c_k}$. Since$$\left(\bigcap_{k=1}^n U_{c_k}\right)\cap\left(\bigcup_{k=1}^nV_{c_k}\right)=\emptyset$$we have two open sets ($\bigcap_{k=1}^n U_{c_k}$ and $\bigcup_{k=1}^nV_{c_k}$) such that $p$ belongs to the first one, $C$ is a subset of the second one and whose intersection is empty.

It is now easy to prove that $K$ is in fact normal.

Answer 3

A compact Hausdorff space is locally compact, i.e. every neighbourhood of a point contains a compact neighbourhood of that point.

We can prove this with your "standard compactness argument" (Note that, as it was stated in the comments, authors can mean very different things when they say "standard compactness argument")