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 $g=\sum _{i=1}^{\mathrm{\infty }}{g}_i$. 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 $n\to \mathrm{\infty }$, $f_n(x)\to 0$ as $n\to \mathrm{\infty }$. Thus, $f(x)=\sum _{i=1}^{\mathrm{\infty }}{g}_i(x)=g(x)$ for $x\in A$.