Tietze extension theorem

This article gives the statement, and possibly proof, of a basic fact in topology.

Statement

Suppose $X$ is a normal space (i.e., a topological space that is T1 and where any two disjoint closed subsets can be separated by disjoint open subsets). Suppose $A$ is a closed subset of $X$, and $f:A\to [0,1]$ is a continuous map. Then, there exists a continuous map $g:X\to [0,1]$ such that the restriction of $g$ to $A$ is $f$.

Facts used

1. Urysohn's lemma: This states that for, given two closed subsets $A,B$ of a normal space $X$, there is a continuous function $h:X\to [0,1]$ such that $h(A)=0$ and $h(B)=1$.

Proof

Note that since $[0,1]$ is homeomorphic to $[-1,1]$, it suffices to prove the result replacing $[0,1]$ with $[-1,1]$. We will also freely use that any closed interval is homeomorphic to $[0,1]$, so Urysohn's lemma can be stated replacing $[0,1]$ by any closed interval.

Given: A normal space $X$. A closed subset $A$ of $X$. A continuous function $f:A\to [-1,1]$.

To prove: There exists a continuous function $g:X\to [-1,1]$ such that the restriction of $g$ to $A$ is $f$.

Proof: We write $f_{0}=f$.

1. Let $\!C_{1}=f^{-1}([-1,-1/3])$ and $\!D_{1}=f^{-1}([1/3,1])$. These are both closed subsets of $A$ since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint by definition. Further, since $A$ is closed in $X$, they are both closed in $X$. So, by fact (1), there exists a continuous function $\!g_{1}:X\to [-1/3,1/3]$ such that $\!g_{1}(C_{1})=-1/3$ and $\!g_{1}(D_{1})=1/3$.
2. Define $\!f_{1}=f_{0}-g_{1}$ as a function on $A$. Note that $\!f_{1}$ is a function on $\!A$ taking values in $\![-2/3,2/3]$. Iteratively, we proceed as follows:
   1. From the previous stage, we have a continuous function $f_{i}$ on the closed subset $A$ taking values in $\![-(2/3)^{i},(2/3)^{i}]$.
   2. Let $\!C_{i+1}=f_{i}^{-1}[-(2/3)^{i},-(1/3)(2/3)^{i}]$ and $\!D_{i+1}=f_{i}^{-1}[(1/3)(2/3)^{i},(2/3)^{i}]$. Note that both $C_{i+1}$ and $D_{i+1}$ are closed subsets of $A$ (since $f_{i}$ is continuous) and hence in $X$, since $A$ is closed in $X$.
   3. By fact (1), find a function $g_{i+1}:X\to [-(1/3)(2/3)^{i},(1(3/(2/3)^{i}]$ such that $\!g_{i+1}(C_{i+1})=(-1/3)(2/3)^{i}$ and $\!g_{i+1}(D_{i+1})=(1/3)(2/3)^{i}$.
   4. Define $\!f_{i+1}=f_{i}-g_{i+1}$. We see that $f_{i+1}$ is a function on $A$ taking values in $\![-(2/3)^{i+1},(2/3)^{i+1}]$.
3. Define $g=\sum _{i=1}^{\infty }g_{i}$. This sum is well-defined at each point and takes values in $[-1,1]$: Note that the absolute value of $g_{i}$ is bounded by the geometric progression $(1/3)+(1/3)(2/3)+(1/3)(2/3)^{2}+\dots =1$. Similarly, the lower bound is $-1$. Further, since $g_{n}$ are bounded by the geometric progression in absolute value, the series $g_{n}$ converges. So the sum is well-defined and takes values in $[-1,1]$.
4. The function $g$ is continuous: This follows from the fact that each $g_{i}$ is continuous and the co-domain of the $g_{i}$s approaches zero. Fill this in later
5. $\!g|_{A}=f$: Let $x\in A$. Then, $\!f_{1}(x)=f(x)-g_{1}(x)$. Inductively, $f_{n}(x)=f(x)-\sum _{i=1}^{n}g_{i}(x)$. Since the upper and lower bound on $f_{n}$ tend to zero as $n\to \infty$, $f_{n}(x)\to 0$ as $n\to \infty$. Thus, $f(x)=\sum _{i=1}^{\infty }g_{i}(x)=g(x)$ for $x\in A$.