## Thoughts about separation (Urysohn’s lemma)

In this chapter we discuss one of the most important theorems in point-set topology. Please, forgive my in-eloquence is the description of the creation of the sets.

**Theorem (Urysohn’s Lemma):** Let be a normal space and be disjoint and closed. Then, there exists some which is continuous such that .

**Proof:** We will do this in steps.

Step 1: Clearly is a neighborhood of and thus by normality we may find some open such that . Appealing to normality again we may find open such that .

So now we now we use the principle of recursive definition to form a class of sets such that which is trivially dense in .

So, define and supposing that the first placement (the meaning of this term will be apparent in a second) define the th placement as follows. Let be the set of all indexes of ‘s such that was created at or before the th placement.

By normality we may find some open with such that . Similarly we may find some with such that . Also, for any we may find some with such that .

Doing this we clearly arrive at a set such that .

Step 2: We now define by .

It is clear that since for every that and since for every we see that . Thus, it remains to show that is continuous.

Since the set of all such that form an open subbase for we must merely show that their inverse images are open.

**Claim**: .

**Proof: **Let then . By ‘s density in we know there exists some such that . Now, suppose that , then which is clearly a contradiction. It follows that and consequently .

Conversely, let then for some . But, as stated earlier and so for every . It follows that and so . The conclusion follows. .

It follows that is the union of open sets, and thus open.

**Claim:** .

**Proof:** Let then . By the density of we know that there exists some such that . Now, assume that then clearly and so by earlier reasoning for every and so which is of course a contradiction. It follows that and so .

Conversely, let then and consequently for some . It follows that . Thus, . The conclusion follows. .

We thus see that is the union of open sets and thus open. By previous comments this finishes the proof.

**Corollary: **If is normal and are closed disjoint subsets of there exists a continuous mapping such that .

*Remark:* Having, for example with , and is preference. Taking gives a continuous mapping such that and

**Corollary:** If is normal then given any and any neighborhood of there exists a function such that and for every .

**Proof:** Let be arbitrary and any neighborhood of . By normality there exists some neighborhood of such that . Thus, and are disjoint closed sets and so there exists some such that and . The conclusion follows.

**Corollary: **Every normal space is completely regular.

There is in fact a nice function which works in place of the above for metric spaces.

**Theorem:** Let be a metric space and disjoint closed subspaces. Then, there exists a mapping such that .

**Proof: **Define by . It is clear that (just noting that the mapping is well-defined).

We first remember from basic analysis that if are continuous real valued functions then are continuous provided that is never zero on . So, is continuous (since as was proved the mapping is continuous) as long as for any . But, this is clear since and similarly and since to have we must have that and thus by previous comment which of course contradicts the disjointness of these two sets.

It follows that is a continuous function, and since if we have that ; and if we have that we may conclude that . The conclusion follows.

While there are many more interesting problems/theorems consequent of Urysohn’s lemma we discuss only a few in the subsequent posts.

For example there is a “strong” Urysohn lemma where but this requires the introduction of the idea of sets which I am weary to do…considering the workload.

No comments yet.

## Leave a Reply