proof of Tietze extension theorem
To prove the Tietze Extension Theorem, we first need a lemma.
The sets and are disjoint and closed in . Since is closed, they are also closed in . Since is normal, then by Urysohn’s lemma and the fact that is homeomorphic to , there is a continuous function such that and . Thus for . Now if , then and thus . Similarly if , then and thus . Finally, for we have that , and so . Hence holds for all . ∎
This puts us in a position to prove the main theorem.
Proof of the Tietze extension theorem.
First suppose that for any continuous function on a closed subset there is a continuous extension. Let and be disjoint and closed in . Define by for and for . Now is continuous and we can extend it to a continuous function . By Urysohn’s lemma, is normal because is a continuous function such that for and for .
Conversely, let be normal and be closed in . By the lemma, there is a continuous function such that for and for . Since is continuous, the lemma tells us there is a continuous function such that for and for . By repeated application of the lemma we can construct a sequence of continuous functions such that for all , and for .
Now for , we have that and as goes to infinity, the right side goes to zero and so the sum goes to . Thus Therefore extends .∎
Remarks: If was a function satisfying , then the theorem can be strengthened as follows. Find an extension of as above. The set is closed and disjoint from because for . By Urysohn’s lemma there is a continuous function such that and . Hence is a continuous extension of , and has the property that .
If is unbounded, then Tietze extension theorem holds as well. To see that consider . The function has the property that for , and so it can be extended to a continuous function which has the property . Hence is a continuous extension of .
|Title||proof of Tietze extension theorem|
|Date of creation||2013-03-22 14:08:58|
|Last modified on||2013-03-22 14:08:58|
|Last modified by||bbukh (348)|