Tietze extension theorem: Difference between revisions

Revision as of 04:28, 17 July 2009

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:

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 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 $h_{1}=f-g_{1}$ $h_{1}=f-g_{1}$ as a function on $A$ $A$ . Note that $h_{1}$ $h_{1}$ is a function on $A$ $A$ taking values in [-2/3,2/3]</math>. Iteratively, we proceed as follows:
1. At any stage, we have $g_{i}:X\to [-(1/3)(2/3)^{i-1},(1/3)(2/3)^{i-1}]$ .
2. We define $h_{i}=h_{i-1}-g_{i}$ on the closed subset $A$ . $h_{i}$ is thus a function on $A$ taking values in $[-(2/3)^{i},(2/3)^{i}]$ .
3. Consider the subsets $C_{i}=h_{i}^{-1}[-(2/3)^{i},-(1/3)(2/3)^{i}]$ and $D_{i}=h_{i}^{-1}[(1/3)(2/3)^{i},(2/3)^{i}]$ . Note that both $C_{i}$ and $D_{i}$ are closed subsets of $A$ (since $h_{i}$ is continuous) and hence in $X$ , since $A$ is closed in <mah>X</math>.
4. 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 $g_{i+1}(D_{i})=(1/3)(2/3)^{i}$ .
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, $h_{1}(x)=f(x)-g_{1}(x)$ . Inductively, $h_{n}(x)=f(x)-\sum _{i=1}^{n}g_{i}(x)$ . Since the upper and lower bound on $h_{n}$ tend to zero as $n\to \infty$ , $h_{n}(x)\to 0$ as $n\to \infty$ . Thus, $f(x)=\sum _{i=1}^{\infty }g_{i}(x)=g(x)$ for $x\in A$ .

@@ Line 7: / Line 7: @@
 ==Facts used==
-* [[Urysohn's lemma]]
+# [[uses::Urysohn's lemma]]: This states that for, given two closed subsets <math>A,B</math> of a [[normal space]] <math>X</math>, there is a continuous function <math>h:X \to [0,1]</math> such that <math>h(A) = 0</math> and <math>h(B) = 1</math>.
+==Proof==
+Note that since <math>[0,1]</math> is homeomorphic to <math>[-1,1]</math>, it suffices to prove the result replacing <math>[0,1]</math> with <math>[-1,1]</math>. We will also freely use that any closed interval is homeomorphic to <math>[0,1]</math>, so Urysohn's lemma can be stated replacing <math>[0,1]</math> by any closed interval.
+'''Given''': A normal space <math>X</math>. A closed subset <math>A</math> of <math>X</math>. A continuous function <math>f:A \to [-1,1]</math>.
+'''To prove''': There exists a continuous function <math>g:X \to [-1,1]</math> such that the restriction of <math>g</math> to <math>A</math> is <math>f</math>.
+'''Proof''':
+# Let <math>C_1 = f^{-1}([-1,-1/3])</math> and <math>D_1 = f^{-1}([1/3,1])</math>. These are both closed subsets of <math>X</math> 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 function <math>g_1: X \to [-1/3,1/3]</math> such that <math>g_1(C_1) = -1/3</math> and <math>g_1(D_1) = 1/3</math>.
+# Define <math>h_1 = f - g_1</math> as a function on <math>A</math>. Note that <math>h_1</math> is a function on <math>A</math> taking values in [-2/3,2/3]</math>. Iteratively, we proceed as follows:
+## At any stage, we have <math>g_i:X \to [-(1/3)(2/3)^{i-1},(1/3)(2/3)^{i-1}]</math>.
+## We define <math>h_i = h_{i-1} - g_i</math> on the closed subset <math>A</math>. <math>h_i</math> is thus a function on <math>A</math> taking values in <math>[-(2/3)^i,(2/3)^i]</math>.
+## Consider the subsets <math>C_i = h_i^{-1}[-(2/3)^i,-(1/3)(2/3)^i]</math> and <math>D_i = h_i^{-1}[(1/3)(2/3)^i,(2/3)^i]</math>. Note that both <math>C_i</math> and <math>D_i</math> are closed subsets of <math>A</math> (since <math>h_i</math> is continuous) and hence in <math>X</math>, since <math>A</math> is closed in <mah>X</math>.
+## By fact (1), find a function <math>g_{i+1}:X \to [-(1/3)(2/3)^i,(1(3/(2/3)^i]</math> such that <math>g_{i+1}(C_i) = (-1/3)(2/3)^i</math> and <math>g_{i+1}(D_i) = (1/3)(2/3)^i</math>.
+# Define <math>g</math> as <math>\sum g_i</math>. This sum is well-defined at each point and takes values in <math>[-1,1]</math>: Note that the absolute value of <math>g_i</math> is bounded by the geometric progression <math>(1/3) + (1/3)(2/3) + (1/3)(2/3)^2 + \dots = 1</math>. Similarly, the lower bound is <math>-1</math>. Further, since <math>g_n</math> are bounded by the geometric progression in absolute value, the series <math>g_n</math> coverges. So the sum is well-defined and takes values in <math>[-1,1]</math>.
+# The function <math>g</math> is continuous: This follows from the fact that each <math>g_i</math> is continuous and the co-domain of the <math>g_i</math>s approaches zero. {{fillin}}
+# <math>g|_A = f</math>: Let <math>x \in A</math>. Then, <math>h_1(x) = f(x) - g_1(x)</math>. Inductively, <math>h_n(x) = f(x) - \sum_{i=1}^n g_i(x)</math>. Since the upper and lower bound on <math>h_n</math> tend to zero as <math>n \to \infty</math>, <math>h_n(x) \to 0</math> as <math>n \to \infty</math>. Thus, <math>f(x) = \sum_{i=1}^\infty g_i(x) = g(x)</math> for <math>x \in A</math>.