proof of Tietze extension theorem

To prove the Tietze Extension Theorem, we first need a lemma.

Lemma 1.

If X is a normal topological space and A is closed in X, then for any continuous functionMathworldPlanetmathPlanetmath f:AR such that |f(x)|1, there is a continuous function g:XR such that |g(x)|13 for xX, and |f(x)-g(x)|23 for xA.


The sets f-1((-,-13]) and f-1([13,)) are disjoint and closed in A. Since A is closed, they are also closed in X. Since X is normal, then by Urysohn’s lemma and the fact that [0,1] is homeomorphicMathworldPlanetmath to [-13,13], there is a continuous function g:X[-13,13] such that g(f-1((-,-13c]))=-13 and g(f-1([13,)))=13. Thus |g(x)|13 for xX. Now if -f(x)-13, then g(x)=-13 and thus |f(x)-g(x)|23. Similarly if 13f(x)1, then g(x)=13 and thus |f(x)-g(x)|23. Finally, for |f(x)|13 we have that |g(x)|13, and so |f(x)-g(x)|23. Hence |f(x)-g(x)|23 holds for all xA. ∎

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 extensionPlanetmathPlanetmath. Let C and D be disjoint and closed in X. Define f:CD by f(x)=0 for xC and f(x)=1 for xD. Now f is continuous and we can extend it to a continuous function F:X. By Urysohn’s lemma, X is normal because F is a continuous function such that F(x)=0 for xC and F(x)=1 for xD.

Conversely, let X be normal and A be closed in X. By the lemma, there is a continuous function g0:X such that |g0(x)|13 for xX and |f(x)-g0(x)|23 for xA. Since (f-g0):A is continuous, the lemma tells us there is a continuous function g1:X such that |g1(x)|13(23) for xX and |f(x)-g0(x)-g1(x)|23(23) for xA. By repeated application of the lemma we can construct a sequencePlanetmathPlanetmath of continuous functions g0,g1,g2, such that |gn(x)|13(23)n for all xX, and |f(x)-g0(x)-g1(x)-g2(x)-|(23)n for xA.

Define F(x)=n=0gn(x). Since |gn(x)|13(23)n and n=013(23)n convergesPlanetmathPlanetmath as a geometric seriesMathworldPlanetmath, then n=0gn(x) converges absolutely and uniformly, so F is a continuous function defined everywhere. Moreover n=013(23)n=1 implies that |F(x)|1.

Now for xA, we have that |f(x)-n=0kgn(x)|(23)k+1 and as k goes to infinityMathworldPlanetmath, the right side goes to zero and so the sum goes to F(x). Thus |f(x)-F(x)|=0 Therefore F extends f.∎

Remarks: If f was a function satisfying |f(x)|<1, then the theorem can be strengthened as follows. Find an extension F of f as above. The set B=F-1({-1}{1}) is closed and disjoint from A because |F(x)|=|f(x)|<1 for xA. By Urysohn’s lemma there is a continuous function ϕ such that ϕ(A)={1} and ϕ(B)={0}. Hence F(x)ϕ(x) is a continuous extension of f(x), and has the property that |F(x)ϕ(x)|<1.

If f is unboundedPlanetmathPlanetmath, then Tietze extension theorem holds as well. To see that consider t(x)=tan-1(x)/(π/2). The function tf has the property that (tf)(x)<1 for xA, and so it can be extended to a continuous function h:X which has the property |h(x)|<1. Hence t-1h is a continuous extension of f.

Title proof of Tietze extension theorem
Canonical name ProofOfTietzeExtensionTheorem
Date of creation 2013-03-22 14:08:58
Last modified on 2013-03-22 14:08:58
Owner bbukh (348)
Last modified by bbukh (348)
Numerical id 10
Author bbukh (348)
Entry type Proof
Classification msc 54C20