# Contents

## Statement

###### Proposition

(Dieudonné‘s theorem)

In particular compact Hausdorff spaces are normal.

###### Proof

Let $(X,\tau)$ be a paracompact Hausdorff space

We first show that it is regular: To that end, let $x \in X$ be a point, and let $C \subset X$ be a closed subset not containing $x$. We need to find disjoint open neighbourhoods $U_x \supset \{x\}$ and $U_C \supset C$.

First of all, by the Hausdorff property there exists for each $c \in C$ disjoint open neighbourhods $U_{x,c} \supset \{x\}$ and $U_c \supset \{c\}$. As $c$ ranges, the latter clearly form an open cover $\{U_c \subset X\}_{c \in C}$ of $C$, and so the union

$\{U_c \subset X\}_{c \in C} \,\cup\, X \backslash C$

is an open cover of $X$. By paracompactness of $(X,\tau)$, there exists a locally finite refinement, and by this lemma we may assume its elements to share the original index set and be contained in the original elements of the same index. Hence

$\{V_c \subset U_c \subset X\}_{c \in C}$

is a locally finite collection of subsets, such that

$U_C \coloneqq \underset{c \in C}{\cup} V_c$

is an open neighbourhood of $C$.

Now by definition of local finiteness there exists an open neighbourhood $W_x \supset \{x\}$ and a finite subset $K \subset C$ such that

$\underset{c \in C \backslash K}{\forall}( W_x \cap V_c = \emptyset ) \,.$

Consider then

$U_x \;\coloneqq\; W_x \cap \left( \underset{k \in K}{\cap} \left( U_{x,k} \right) \right) \,.$

which is an open neighbourhood of $x$, by the finiteness of $K$.

It thus only remains to see that

$U_x \cap U_C = \emptyset \,.$

But this holds because the only $V_{c}$ that intersect $W_x$ are the $V_{k} \subset U_{k}$ for $k \in K$ and each of these is by construction disjoint from $U_{x,k}$ and hence from $U_x$.

This establishes that $(X,\tau)$ is regular. Now we prove that it is normal. For this we use the same approach as before:

Let $C,D \subset X$ be two disjoint closed subsets. We need to produce disjoint open neighbourhoods for these.

By the previous statement of regularity, we may find for each $c \in C$ disjoint open neighbourhoods $U_c \supset \{c\}$ and $U_{D,c} \supset D$. Hence the union

$\left\{ U_c \subset X \right\}_{c \in C} \cup X \backslash C$

is an open cover of $X$, and thus by paracompactness has a locally finite refinement, whose elements we may, again by this lemma, assume to have the same index set as before and be contained in the previous elements with the same index. Hence we obtain a locally finite collection of subsets

$\{ V_c \subset U_c \subset X \}_{c \in C}$

such that

$U_{C} \coloneqq \underset{c \in C}{\cup} V_c$

is an open neighbourhood of $C$.

It is now sufficient to see that every point $d \in D$ has an open neighbourhood $U_d$ not intersecting $U_C$, for then

$U_D \coloneqq \underset{d \in D}{\cup} U_d$

is the required open neighbourhood of $D$ not intersecting $U_C$.

Now by local finiteness of $\{V_c \subset X\}_{c \in C}$, every $d \in D$ has an open neighbourhood $W_d$ such that there is a finite set $K_d \subset C$ so that

$\underset{c \in C \backslash K_d}{\forall} \left( V_c \cap W_d = \emptyset \right) \,.$

Accordingly the intersection

$U_d \coloneqq W_d \cap \left( \underset{c \in K_d \subset C}{\cap} U_{D,c} \right)$

is still open and disjoint from the remaining $V_k$, hence disjoint from all of $U_C$.

Last revised on June 24, 2017 at 16:28:10. See the history of this page for a list of all contributions to it.