# proof of Tietze extension theorem

To prove the Tietze Extension Theorem, we first need a lemma.

###### Lemma 1.

If $X$ is a normal topological space and $A$ is closed in $X$, then for any continuous function $f\colon A\to\mathbb{R}$ such that $|f(x)|\leq 1$, there is a continuous function $g\colon X\to\mathbb{R}$ such that $\left\lvert g(x)\right\rvert\leq\frac{1}{3}$ for $x\in X$, and $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$ for $x\in A$.

###### Proof.

The sets $f^{-1}\bigl{(}(-\infty,-\frac{1}{3}]\bigr{)}$ and $f^{-1}\bigl{(}[\frac{1}{3},\infty)\bigr{)}$ are disjoint and closed in $A$. Since $A$ is closed, they are also closed in $X$. Since $X$ is normal, then by Urysohn’s lemma and the fact that $[0,1]$ is homeomorphic to $[-\frac{1}{3},\frac{1}{3}]$, there is a continuous function $g\colon X\to[-\frac{1}{3},\frac{1}{3}]$ such that $g\Bigl{(}f^{-1}\bigl{(}(-\infty,-\frac{1}{3}c]\bigr{)}\Bigr{)}=-\frac{1}{3}$ and $g\Bigl{(}f^{-1}\bigl{(}[\frac{1}{3},\infty)\bigr{)}\Bigr{)}=\frac{1}{3}$. Thus $\left\lvert g(x)\right\rvert\leq\frac{1}{3}$ for $x\in X$. Now if $-\leq f(x)\leq-\frac{1}{3}$, then $g(x)=-\frac{1}{3}$ and thus $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$. Similarly if $\frac{1}{3}\leq f(x)\leq 1$, then $g(x)=\frac{1}{3}$ and thus $|f(x)-g(x)|\leq\frac{2}{3}$. Finally, for $\left\lvert f(x)\right\rvert\leq\frac{1}{3}$ we have that $\left\lvert g(x)\right\rvert\leq\frac{1}{3}$, and so $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$. Hence $\left\lvert f(x)-g(x)\right\rvert\leq\frac{2}{3}$ holds for all $x\in A$. ∎

This puts us in a position to prove the main theorem.

###### Proof of the Tietze extension theorem.

First suppose that for any continuous function on a closed subset there is a continuous extension. Let $C$ and $D$ be disjoint and closed in $X$. Define $f\colon C\cup D\to\mathbb{R}$ by $f(x)=0$ for $x\in C$ and $f(x)=1$ for $x\in D$. Now $f$ is continuous and we can extend it to a continuous function $F\colon X\to\mathbb{R}$. By Urysohn’s lemma, $X$ is normal because $F$ is a continuous function such that $F(x)=0$ for $x\in C$ and $F(x)=1$ for $x\in D$.

Conversely, let $X$ be normal and $A$ be closed in $X$. By the lemma, there is a continuous function $g_{0}\colon X\rightarrow\mathbb{R}$ such that $\left\lvert g_{0}(x)\right\rvert\leq\frac{1}{3}$ for $x\in X$ and $\left\lvert f(x)-g_{0}(x)\right\rvert\leq\frac{2}{3}$ for $x\in A$. Since $(f-g_{0})\colon A\rightarrow\mathbb{R}$ is continuous, the lemma tells us there is a continuous function $g_{1}\colon X\rightarrow\mathbb{R}$ such that $\left\lvert g_{1}(x)\right\rvert\leq\frac{1}{3}(\frac{2}{3})$ for $x\in X$ and $\left\lvert f(x)-g_{0}(x)-g_{1}(x)\right\rvert\leq\frac{2}{3}(\frac{2}{3})$ for $x\in A$. By repeated application of the lemma we can construct a sequence of continuous functions $g_{0},g_{1},g_{2},\ldots$ such that $\left\lvert g_{n}(x)\right\rvert\leq\frac{1}{3}(\frac{2}{3})^{n}$ for all $x\in X$, and $\left\lvert f(x)-g_{0}(x)-g_{1}(x)-g_{2}(x)-\cdots\right\rvert\leq(\frac{2}{3}% )^{n}$ for $x\in A$.

Define $F(x)=\sum_{n=0}^{\infty}g_{n}(x)$. Since $\left\lvert g_{n}(x)\right\rvert\leq\frac{1}{3}(\frac{2}{3})^{n}$ and $\sum_{n=0}^{\infty}\frac{1}{3}(\frac{2}{3})^{n}$ converges as a geometric series, then $\sum_{n=0}^{\infty}g_{n}(x)$ converges absolutely and uniformly, so $F$ is a continuous function defined everywhere. Moreover $\sum_{n=0}^{\infty}\frac{1}{3}(\frac{2}{3})^{n}=1$ implies that $\left\lvert F(x)\right\rvert\leq 1$.

Now for $x\in A$, we have that $\left\lvert f(x)-\sum_{n=0}^{k}g_{n}(x)\right\rvert\leq(\frac{2}{3})^{k+1}$ and as $k$ goes to infinity, the right side goes to zero and so the sum goes to $F(x)$. Thus $\left\lvert f(x)-F(x)\right\rvert=0$ Therefore $F$ extends $f$.∎

Remarks: If $f$ was a function satisfying $\left\lvert f(x)\right\rvert<1$, then the theorem can be strengthened as follows. Find an extension $F$ of $f$ as above. The set $B=F^{-1}(\{-1\}\cup\{1\})$ is closed and disjoint from $A$ because $\left\lvert F(x)\right\rvert=\left\lvert f(x)\right\rvert<1$ for $x\in A$. By Urysohn’s lemma there is a continuous function $\phi$ such that $\phi(A)=\{1\}$ and $\phi(B)=\{0\}$. Hence $F(x)\phi(x)$ is a continuous extension of $f(x)$, and has the property that $\left\lvert F(x)\phi(x)\right\rvert<1$.

If $f$ is unbounded, then Tietze extension theorem holds as well. To see that consider $t(x)=\tan^{-1}(x)/(\pi/2)$. The function $t\circ f$ has the property that $(t\circ f)(x)<1$ for $x\in A$, and so it can be extended to a continuous function $h\colon X\to\mathbb{R}$ which has the property $\left\lvert h(x)\right\rvert<1$. Hence $t^{-1}\circ h$ is a continuous extension of $f$.

Title proof of Tietze extension theorem ProofOfTietzeExtensionTheorem 2013-03-22 14:08:58 2013-03-22 14:08:58 bbukh (348) bbukh (348) 10 bbukh (348) Proof msc 54C20