## Thoughts about separation (Normality)

We have discussed up until now Kolomogorov, , Hausdorff, regular, and Tychonoff. We now discuss what (maybe bar Hausdoffness) the most fruitful separation axiom.

**Normal (): Let be a topological space. Then, $late X$ is called normal () if it is and given any disjoint closed **there exists disjoint open which contain respectively.

We first prove a similar (and very functional) theorem that came up in regularity.

**Theorem:** Let be a topological space, then is normal if and only if given any closed and any neighborhood of there exists some neighborhood of such that .

**Proof:**

: Since is normal and are disjoint open sets there exists open sets which respectively contain them. We claim that . But, this follows from an earlier problem since if are disjoint open sets then . Thus, given any we have that . It follows that . The conclusion follows.

: Conversely, let be disjoint and closed. Clearly then is a neighborhood of and by assumption there exists some neighborhood of such that . Clearly then which are disjoint and open. The conclusion follows. .

The following theorem follows.

**Theorem:** Let be normal, then given any disjoint closed there exists neighborhoods of them respectively such that .

**Proof:** It follows from definition of normality that there exists disjoint open sets which contain each. But, by the previous theorem there exists neighborhoods of such that . Clearly then . The conclusion follows. .

Unfortunately normality doesn’t react as well with subspaces and products as the other separation axioms. Namely:

**Theorem:** Let be normal a closed subspace of . Then is normal.

**Proof:** Let be disjoint and closed. Clearly then for some closed . But, this clearly implies that are closed in and thus by its normality there exists disjoint open sets such that . Clearly then are disjoint open sets in that contain respectively. The conclusion follows.

This in fact cannot be strengthened to a more general case. To come up with your own example try proving the general case and see where you run into issues. Things likes disjointness and openess in will be your main hinderances.

In fact, the product of just **two** normal spaces need not be normal. The spaces that is commonly used to illustrate this is complicated and so I suggest than an interested reader Google Sorgenfrey Plane.

The next theorem gives a nice result which enables many of the commonly encountered spaces to be automatically labeled normal.

**Theorem:** Let be a compact Hausdorff space, then is normal.

**Proof:** Let be disjoint and closed. By an earlier problem we know that since is Hausdorff and Hausdorff for each there exists disjoint open sets such that . Doing this for each we obtain an open cover for . But, by ‘s compactness there exists a finite subcover of , say . Thus, if are the corresponding open sets containing but not we see that are open sets such that and . The conclusion follows.

**Corollary: **Since the product of Hausdorff spaces is Hausdorff and the product of compact spaces is compact we see that the previous theorem says that the product of compact Hausdorff spaces are normal.

The next theorem validates a strong suspicion one may have regarding normality and metric spaces.

**Theorem:** Any metric space (and thus any metrizable space) is normal.

**Proof:** Let be a metric space and be disjoint and closed. Since we have that . Similarly, define . We leave it to the reader to verify that and that these sets are disjoint. .

Our next theorem is fairly deep.

**Theorem: **Let be a second countable regular space, then is normal.

**Proof:** Let be disjoint and closed. For each there exists a neighborhood of such that . Using regularity we may find some neighborhood of such that . Clearly covers , but since every second countable space is Lindelof this open cover has a countable subcover . Form similary such that . The sets are clearly open sets containing respectively. But, there is no need for them to be disjoint.

Thus, given define and . Clearly each is open. Also, covers since for each , belongs to some and none of the sets . Similarly covers .

But, if we define then . To see this, suppose that , then for some . Clearly either or . But, if then it follows by definition that , and since it follows (once again by definition) that which is of course a contradiction. A similar contradiction occurs if we assume that . It follows that they are in fact disjoint. The conclusion follows.

We end this discussion on normality for now and begin next time with Urysohn’s lemma.

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