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

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 $f_{0} = f$ .

Let $C_{1} = f^{- 1} ([- 1, - 1 / 3])$ and $D_{1} = f^{- 1} ([1 / 3, 1])$ . These are both closed subsets of $X$ since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint by definition. 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 $g_{1} (D_{1}) = 1 / 3$ .
Define f1=f0−g1 as a function on A. Note that h1 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 $C_{i} = f_{i}^{- 1} [- (2 / 3)^{i}, - (1 / 3) (2 / 3)^{i}]$ and $D_{i} = f_{i}^{- 1} [(1 / 3) (2 / 3)^{i}, (2 / 3)^{i}]$ . Note that both $C_{i}$ and $D_{i}$ are closed subsets of $A$ (since $f_{i}$ is continuous) and hence in $X$ , since $A$ is closed in <mah>X</math>.
3. By fact (1), find a function $g_{i + 1} : X \to [- (1 / 3) (2 / 3)^{i}, (1 (3 / (2 / 3)^{i}]$ such that $g_{i + 1} (C_{i}) = (- 1 / 3) (2 / 3)^{i}$ and $g_{i + 1} (D_{i}) = (1 / 3) (2 / 3)^{i}$ .
4. Define $f_{i + 1} = f_{i} - g_{i + 1}$ . We see that $f_{i + 1}$ is a function on $A$ taking values in $[- (2 / 3)^{i + 1}, (2 / 3)^{i + 1}$ .
Define $g$ as $\sum g_{i}$ . 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 $g_{n}$ coverges. So the sum is well-defined and takes values in $[- 1, 1]$ .
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
$g |_{A} = f$ : 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 $x \in A$ .