## Thoughts about separation (Regularity and Complete Regularity)

So up until this point we’ve discussed spaces which are Kolomogorov, , or Hausdorff. We now discuss the next property in the progression.

**Regular (): **Let be a topological space. We call *regular* if it’s and given any closed subset of and any point there exists disjoint open neighborhoods of both.

*Remark:* It is quite common to call a space regular if it only has the second property mentioned and if is also . We, obviously, do not have that sentiment.

It clearly follows that every regular space is Hausdorff since its ness guarantees that points are closed sets.

We first discuss the most functional aspect of regular spaces.

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

**Proof:**

: Suppose that is regular and let be arbitrary. For any neighborhood of we have that is a closed set not containing . By assumption then there exists disjoint open neighborhoods of each. We claim that .

To see this, let then and since it follows by an earlier problem that . The conclusion follows.

: Let latex be closed and . Since they are disjoint we have that is an open set containing and by assumption there exists some neighborhood of such that . So, it follows quite readily that are the sets we seek.

From this we get the following almost corollary.

**Theorem:** Let be regular and be distinct. Then there exists neighborhoods of respectively such that .

**Proof: **Since is Hausdorff there exists disjoint open neighborhoods of respectively. By the previous problem these neighborhoods contain neighborhoods such that . These are clearly the neighborhoods we desired.

The following is the next step in the obvious progression which shows that in some sense compact sets act very similarly to points.

**Theorem:** Let be regular and be disjoint closed subsets such that is compact. There exists disjoint open sets such that .

**Proof:** Clearly for each we clearly have that and by the regularity of there exists disjoint open sets such that . Clearly the class is an open cover of and thus by ‘s compactness it has a finite subcover . Clearly if are the corresponding open sets containing that and and both are disjoint. The conclusion follows.

We note that regularity is hereditary.

**Theorem:** Let be regular and a subspace of then is regular.

**Proof:** Let be closed and . Since is closed we have that for some closed set in . Clearly otherwise . Thus, by ‘s regularity there exists open sets such that and . Clearly then are disjoint sets which contain and respectively. The conclusion follows.

Next is the obvious.

**Theorem:** Let be a non-empty collection of regular spaces, then is regular.

**Proof: **We actually need a little lemma first which we should have proved in the product topology section, namely:

**Lemma:** Let be a collection of topological spaces and . Then, given any class such that then .

**Proof: **

Let and let be any neighborhood of . By assumption for each and so there exists some point in . It follows that . Since was arbitrary it follows that is an adherent point of and so .

Conversely, let . Let be arbitrary, and let be an open set containing . Since is open in it contains a point . Then . The conclusion follows.

Let be arbitrary and let be any basic neighborhood of . Let be the finite number of indices such that . It is clear that is a neighborhood of and by ‘s regularity there exists some such that . Clearly then where . This is clearly a neighborhood of and since . The conclusion follows.

We now prove a partial converse.

**Theorem:** Let be a non-empty collection of topological spaces and be under the product topology. If is regular then so is for any

**Proof:** Let be closed and . Clearly then where is a closed subset of and where with where is a fixed but arbitrary set in . Thus, by ‘s regularity there exists disjoint open sets such that and . Clearly then and . It thus remains to show that they are disjoint.

Thus, assume that then where where where is as before. This is of course a contradiction. The conclusion follows.

**Theorem:** Let be a regular space and be perfect. Then, is regular.

**Proof:** Let be closed and . It follows from ‘s perfection that are disjoint closed subsets of such that is compact. By an earlier theorem there exists open sets such that . Clearly then are disjoint closed sets containing respectively. Once again, by ‘s perfection we see that are closed sets containing respectively. Thus, are open sets containing respectively. All that remains is to show that they are disjoint.

So, let then and so and so and thus and thus . A similar argument shows that . The conclusion follows.

Now, of course you might ask for an example a Hausdorff space which is not regular. Unfortunately, they are not as easy to come by. We *outline* such a topological space.

**Example:** Let be such that is open if where is open in the usual topology and is countable.

Our next example is a much more analysis motivated concept.

**Tychonoff (completely regular ):** Let be a topological space, then it is called Tychonoff (*completely regular/)* if it is and given any closed and any there exists some such that and

It is clear that every Tychonoff space is regular since given any closed , , and the corresponding the sets are disjoint open sets containing respectively.

**Theorem:** Let be Tychonoff and a subspace of . Then, is Tychonoff.

**Proof:** Let be closed and . Since is a subspace of we have that for some closed in . Clearly otherwise . Thus, and by ‘s complete regularity there exists some which is continuous and . Thus, is clearly continuous and . The conclusion follows.

**Theorem: **Let then is Tychonoff under the product topology if is Tychonoff for each .

**Proof: **Let be closed and . We may clearly choose a basic open set such that . Let be the finitely many indices such that . Since by assumption each of are Tychonoff we may find some function such that . Let then continuously and vanishes outside of . So, let be given by . Then this is a continuous mapping of into such that and . The conclusion follows.

: Let

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