Contents

# Contents

## Statement

###### Proposition

In fact a stronger statement holds: paracompact Hausdorff spaces are normal.

To prove this, consider the following lemma:

###### Lemma

(separation by neighbourhoods of points from compact subspaces in Hausdorff spaces)

Let

1. $(X,\tau)$ be a Hausdorff topological space;

2. $Y \subset X$ a compact subspace.

Then for every $x \in X \backslash Y$ there exists

1. an open neighbourhood $U_x \supset \{x\}$;

2. an open neighbourhood $U_Y \supset Y$

such that

• they are still disjoint: $U_x \cap U_Y = \emptyset$.
###### Proof

of lemma

By the assumption that $(X,\tau)$ is Hausdorff, we find for every point $y \in Y$ disjoint open neighbourhoods $U_{x,y} \supset \{x\}$ and $U_y \supset \{y\}$. By the nature of the subspace topology of $Y$, the restriction of all the $U_y$ to $Y$ is an open cover of $Y$:

$\left\{ (U_y \cap Y) \subset Y \right\}_{y \in Y} \,.$

Now by the assumption that $Y$ is compact, there exists a finite subcover, hence a finite set $S \subset Y$ such that

$\left\{ (U_y \cap Y) \subset Y \right\}_{y \in S \subset Y}$

is still a cover.

But the finite intersection

$U_x \coloneqq \underset{s \in S \subset Y}{\bigcap} U_{x,s}$

of the corresponding open neighbourhoods of $x$ is still open, and by construction it is disjoint from all the $U_{s}$, hence in particular from their union

$U_Y \coloneqq \underset{s \in S \subset Y}{\bigcup} U_s \,.$

Therefore $U_x$ and $U_Y$ are two open subsets as required.

###### Proof

of prop.

First we claim that $(X,\tau)$ is regular. To show this, we need to find for each point $x \in X$ and each disjoint closed subset $Y \in X$ dijoint open neighbourhoods $U_x \supset \{x\}$ and $U_Y \supset Y$. But since closed subspaces of compact spaces are compact, the subset $Y$ is in fact compact, and hence this is in fact the statement of lemma .

Next to show that $(X,\tau)$ is indeed normal, we apply the idea of the proof of lemma once more:

Let $Y_1, Y_2 \subset X$ be two disjoint closed subspaces. By the previous statement then for every point $y_1 \in Y$ we find disjoint open neighbourhoods $U_{y_1} \subset \{y_1\}$ and $U_{Y_2,y_1} \supset Y_2$. The union of the $U_{y_1}$ is a cover of $Y_1$, and by compactness of $Y_1$ there is a finite subset $S \subset Y$ such that

$U_{Y_1} \coloneqq \underset{s \in S \subset Y_1}{\bigcup} U_{y_1}$

is an open neighbourhood of $Y_1$ and

$U_{Y_2} \coloneqq \underset{s \in S \subset Y}{\bigcap} U_{Y_2,s}$

is an open neighbourhood of $Y_2$, and both are disjoint.