Dieudonné's theorem on paracompact Hausdorff spaces

Remarks

The proof has two steps of the same shape. First, $X$ is regular: for a closed set $C$ and a point $x \notin C$, cover $C$ by open sets whose closures miss $x$, and add $X \setminus C$. Take a locally finite open refinement. The union of the closures of the members that meet $C$ is closed, by local finiteness, and it misses $x$. The second step repeats this argument with a closed set in place of the point $x$, using regularity.

Example at work: every compact Hausdorff space is paracompact, so it is normal.

The Hausdorff hypothesis cannot be dropped. Take $X = \set{a, b, c}$ with open sets $\emptyset$, $\set{a}$, $\set{a, b}$, $\set{a, c}$ and $X$. It is finite, so paracompact. The closed sets $\set{b}$ and $\set{c}$ are disjoint, but every open set containing either of them contains $a$.

Going deeper

For a student

Normality is what makes Urysohn's lemma work. So on a paracompact Hausdorff space, there are many continuous real functions. This is the first step toward partitions of unity subordinate to an open cover. In fact, a Hausdorff space is paracompact if and only if every open cover has a subordinate partition of unity.