## Thoughts about separation (Tietze’s extension theorem)

In this post we shall discuss one of the most important theorems consequent of Urysohn’s lemma. It concerns the extension of continuous functions on closed subspaces of a normal space to the full space.

**Theorem (Tietze’s Extension Theorem): **Let be normal and a closed subspace. Then, given any which is continuous there exists some such that . is called an *extension* of .

**Proof:** if this is trivial (since the constant mapping extended to the full space is a trivial realization of ). Thus, we may assume WLOG that . Also, we may assume that is the smallest closed interval containing . Lastly, it is clear that we need only prove the case when .

We firstly define . Clearly , and we define to be and respectively. Clearly and are disjoint closed subspaces of and since is itself closed they are closed in . Thus, by Urysohn’s lemma there exists some such that .

We now define on by . Clearly we note that

.

From where it readily follows that .

So, we know define and . Using the same logic above we may find some such that and . Now, define on by . Using similar logic as before we see that .

Continuing in this way we get a sequence of functions defined on such that and a sequence of functions defined on such that , with the property that on we have that .

We now define . We therefore see that is the partial sums of an infinite series of functions in . We know that is complete (this is a fundamental fact from basic analysis) and so by virtue of the facts that and we see that with the property that and that it is continuous. Also, note that since we note that on and so . The conclusion follows.

*Remark:* A common exercise is to prove that the Tietze extension theorem implies Urysohn’s lemma. Merely note that if given a normal space and two closed disjoint subsets we see that is a closed subspace of . Also, the function given by is readily verified to be continuous. Thus, there exists a continuous extension to the entirety of . This satisfies the conditions of the function in Urysohn’s lemma.

In fact, we may prove a slight generalization to Tietze’s extension theorem (that will be generalized even further in subsequent posts).

**Theorem (Semi-Extended Tietze’s Extension Theorem):** Let be normal and a closed subspace. Then, given any continuous there exists some which is also continuous and .

**Proof: **It is clear taht we must only prove the case where . Also, for the sake of notational convenience it is better to think of as the product space where under the usual topology.

Now, as was proved earlier, since is continuous so is for each . It follows from the normal Tietze extension theorem that each has an extension , such that . We thus define

by . This is clearly continuous since (as was proven earlier) the sum of continuous maps is continuous. It remains to show that .

To see this, let then

.

This comes from remembering that on . The conclusion follows.

*Remark:* It was pointed out to me that the notation is usually reserved for mappings where both the domain and codomain are product spaces. Thus, when just the codomain is a product space we shall use that notation.

It should be noted that the closedness of is necessary in the above as can be illustrated by the following example.

**Example:** Let be given by . This clearly cannot be extended to since we would need to define . But, of course this limit does not exist.

[…] this is true (although, the following is related, though more parallel than a generalization viz. Tietze’s Extension Theorem) I can show that we can’t forgo either completely. In particular (consider all domain spaces […]

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