# 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 $h(B) = 1$.
2. Closedness is transitive: A closed subset (in the subspace topology) of a closed subset is closed in the whole space.

## Proof

This proof uses a tabular format for presentation. Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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 $f_0 = f$.

The key part of the proof is the inductive portions (6)-(10). Steps (1)-(5) can be thought of as a special case of these; however, these steps are provided separately so that the full details can be seen in the base case before the general inductive setup is used.

Step no. Construction/assertion Given data used Previous steps used Facts used Explanation
1 Let $\! C_1 = f^{-1}([-1,-1/3])$ and $\! D_1 = f^{-1}([1/3,1])$ Existence of $f:A \to [0,1]$ -- --
2 $C_1$ and $D_1$ are disjoint closed subsets of $A$ $f$ is continuous (1) inverse image of closed subset under continuous map is closed $C_1$ and $D_1$ are closed subsets of $A$ since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint since their images under $f$ are disjoint.
3 $C_1$ and $D_1$ are disjoint closed subsets of $X$ $A$ is closed in $X$ (2) (2) By step (2), $C_1$ and $D_1$ are closed in $A$, which is closed in $X$. Thus, $C_1$ and $D_1$ are closed in $X$.
4 Construct a continuous function $\! g_1: X \to [-1/3,1/3]$ such that $\! g_1(C_1) = -1/3$ and $\! g_1(D_1) = 1/3$ (3) (1) (Urysohn's lemma) Urysohn's lemma guarantees a continuous function $h:X \to [0,1]$ such that $h(C_1) = 0$ and $h(D_1) = 1$. Define $g_1(x) := (2h(x) - 1)/3$.
5 Define $\! f_1 = f_0 - g_1$ as a function on $A$. Note that $\! f_1$ is a continuous function on $\! A$ taking values in $\! [-2/3,2/3]$. ($f = f_0$ is continuous) (4) sum or difference of continuous functions on subsets of a topological space is continuous on their intersection. $f_1$ is continuous since both $f_0$ and $g_1$ are continuous. For the range, note the following. For $x \in A$, there are three possibilities for the interval in which $f_0(x)$ lies: $[-1,-1/3]$, $[-1/3,1/3]$, and $[1/3,1]$. In the first case, $g_1(x) = -1/3$, so $f_0(x) - g_1(x$ is in $[-1 - (-1/3), -1/3 - (-1/3)] = [-2/3,0]$. In the second case, both $f_0(x)$ and $g_1(x)$ are in $[-1/3,1/3]$, so by the triangle inequality, the difference is in $[-2/3,2/3]$. In the third case, $g_1(x) = 1/3$ so $f_0(x) - g_1(x)$ is in $[1/3 - 1/3,1 - 1/3] = [0,2/3]$.
6 We proceed iteratively as follows: from the previous stage, we have a continuous function $f_i$ on the closed subset $A$ taking values in $\! [-(2/3)^i,(2/3)^i]$.
7 Let $\! C_{i+1} = f_i^{-1}[-(2/3)^i,-(1/3)(2/3)^i]$ and $\! D_{i+1} = f_i^{-1}[(1/3)(2/3)^i,(2/3)^i]$.
8 $C_{i+1}$ and $D_{i+1}$ are disjoint closed subsets of $A$. (6): $f_i$ is continuous inverse image of closed subset under continuous map is closed $C_1$ and $D_1$ are closed subsets of $A$ since they are both the inverse image of a closed set under a continuous map. Moreover, they are disjoint since their images under $f$ are disjoint.
9 $C_{i+1}$ and $D_{i+1}$ are disjoint closed subsets of $X$ $A$ is closed in $X$ (8) (2) (closed subsets of closed subsets are closed) By step (8), $C_{i+1}$ and $D_{i+1}$ are closed in $A$, which is closed in $X$. Thus, $C_{i+1}$ and $D_{i+1}$ are closed in $X$.
10 Find a continuous 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 $\! g_{i+1}(D_{i+1}) = (1/3)(2/3)^i$. (9) (1) (Urysohn's lemma) We use fact (1) to get a function to $[0,1]$, then multiply by 2 and subtract 1 to get a function to $[-1,1]$, then scale it by the factor of $(1/3)(2/3)^i$.
11 Define $\! f_{i+1} = f_i - g_{i+1}$. $f_{i+1}$ is a continuous function on $A$ taking values in $\! [-(2/3)^{i+1},(2/3)^{i+1}]$. (6), (10) (Same logic as for (5))
12 Define the following function $g:X \to [-1,1]$: $g = \sum_{i=1}^\infty g_i$. This function is well defined. (10) 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 $[-1,1]$.
13 The function $g$ is continuous (10) 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
14 $\! g|_A = f$ 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 $x \in A$.