Compact + Hausdorff = Normal

 
 

The notion of a topological space being Hausdorff or normal identifies the degree to which points or sets can be "separated." In a Hausdorff space, it's guaranteed that if you pick any two distinct points in the space -- say x and y -- then you can always find an open set containing x and an open set containing y such that those two sets don't overlap. In other words, you can separate the two points. A normal topological space is very similar - not only can we separate points, we can separate sets. (See below for the formal definition.)

While it is true that every normal space is a Hausdorff space, it is not true that every Hausdorff space is normal. That is, Hausdorff is a necessary condition for a space to be normal, but it is not sufficient. We need one extra condition, namely compactness

In this post, we prove that if a space is both Hausdorff and compact, then it is normal.

 
 

First we give the formal definition of a normal topological space:

Definition: A topological space $X$ is said to be normal if $X$ is Hausdorff and for all closed, disjoint subsets $A,B\subset X$, there exist open disjoing subsets $U,V\subset X$ such that $A\subset U$ and $B\subset V$.

In the proof of the theorem below, we will use the fact that a comapct subspace of a Hausdorff space is closed. (You can find the proof in [1] p. 32.)

Theorem: A compact Hausdorff space is normal. In fact, if $A,B$ are compact subsets of a Hausdorff space, and are disjoint, there exist disjoint open sets $U,V$, such that $A\subset U$ and $B\subset V$.

Proof: Let $X$ be a Hausdorff space and let $A,B\subset X$ be disjoint compact subsets of $X$. We wish to show the existence of disjoint open sets $U,V\subset X$ such that $A\subset U$ and $B\subset V$.

Since $X$ is Hausdorff, for any points $a\in A$ and $b\in B$, there exist disjoint open subsets $U_{a,b}$ and $V_{a,b}$ so that $a\in U_{a,b}$ and $b\in V_{a,b}$. For now, fix the point $a$ and consider the collection $\mathcal{V}_a=\{V_{a,b}\: : \:b\in B\}$. This forms an open cover for $B$ and, since $B$ is compact, $\mathcal{V}_a$ has a finite subcover, $$V_a'=\bigcup_{V_{a,b}\in \mathcal{V}_a}V_{a,b}$$ (where the union is finite).

Recall that each $V_{a,b}\in V_a'$ has a corresponding $U_{a,b}$ (this is because $a$ is fixed, and as we range throughout all the $b\in B$, we get a different open set $U_{a,b}$ which contains $a$). So consider the intersection of the collection of these $U_{a,b}$s: $$U_a'=\cap\{U_{a,b}\: : \: V_{a,b}\in V_a'\}.$$ Observe that both $V_a'$ and $U_a'$ are open and disjoint and $$B\subset V_a'.$$

Notice that the collection of all such $U_a'$ forms a cover for $A$ (now letting $a$ range throughout $A$), and we'll denote this by $\mathcal{U}=\{U_a'\: : \: a\in A\}$. Since $A$ is compact, $\mathcal{U}$ has a finite subcover $U'$. So denote the union of elements in $U'$ by $U$, that is $$U=\bigcup_{U_a'\in U'}U_a'$$ (where the union is finite). Finally, recall that for each $U_a'$ there is a corresponding $V_a'$. Let $V$ denote the intersection of these, that is $$V=\cap\{V_a'\: : \: U_a'\in U'\}.$$ Then $A\subset U$ and $B\subset V$ and by construction $U$ and $V$ are open disjoint subsets of $X$. This shows that $X$ is normal. $\square$

Honestly, the first time I went through this I got lost with all the notation. But in topology picture proofs are always helpful! So I created the GIF below to try and help illustrate what's really being done (which is actually very simple!). Just note, the (unlabeled!) little black ovals are the $U_{a,b}$ and $V_{a,b}$ mentioned above.

And yes! I realize this looks like graffiti at the end....

Abstract art anyone?

 
 

Reference: 

[1] Lang, Serge. Real and Functional Analysis (3ed.)Proposition 3.5 p. 33.