# 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^{} $f\mathrm{:}A\mathrm{\to}\mathrm{R}$ such that $\mathrm{|}f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{|}\mathrm{\le}\mathrm{1}$, there is a continuous function $g\mathrm{:}X\mathrm{\to}\mathrm{R}$ such that $\mathrm{\left|}g\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\right|}\mathrm{\le}\frac{\mathrm{1}}{\mathrm{3}}$ for $x\mathrm{\in}X$, and $\mathrm{\left|}f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{-}g\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\right|}\mathrm{\le}\frac{\mathrm{2}}{\mathrm{3}}$ for $x\mathrm{\in}A$.

###### Proof.

The sets ${f}^{-1}\left((-\mathrm{\infty},-\frac{1}{3}]\right)$ and ${f}^{-1}\left([\frac{1}{3},\mathrm{\infty})\right)$ 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 $g:X\to [-\frac{1}{3},\frac{1}{3}]$ such that $g\left({f}^{-1}\left((-\mathrm{\infty},-\frac{1}{3}c]\right)\right)=-\frac{1}{3}$ and $g\left({f}^{-1}\left([\frac{1}{3},\mathrm{\infty})\right)\right)=\frac{1}{3}$. Thus $\left|g(x)\right|\le \frac{1}{3}$ for $x\in X$. Now if $-\le f(x)\le -\frac{1}{3}$, then $g(x)=-\frac{1}{3}$ and thus $\left|f(x)-g(x)\right|\le \frac{2}{3}$. Similarly if $\frac{1}{3}\le f(x)\le 1$, then $g(x)=\frac{1}{3}$ and thus $|f(x)-g(x)|\le \frac{2}{3}$. Finally, for $\left|f(x)\right|\le \frac{1}{3}$ we have that $\left|g(x)\right|\le \frac{1}{3}$, and so $\left|f(x)-g(x)\right|\le \frac{2}{3}$. Hence $\left|f(x)-g(x)\right|\le \frac{2}{3}$ 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 $f:C\cup D\to \mathbb{R}$ 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 $F:X\to \mathbb{R}$. 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 ${g}_{0}:X\to \mathbb{R}$ such that $\left|{g}_{0}(x)\right|\le \frac{1}{3}$ for $x\in X$ and $\left|f(x)-{g}_{0}(x)\right|\le \frac{2}{3}$ for $x\in A$. Since $(f-{g}_{0}):A\to \mathbb{R}$ is continuous, the lemma tells us there is a continuous function ${g}_{1}:X\to \mathbb{R}$ such that $\left|{g}_{1}(x)\right|\le \frac{1}{3}(\frac{2}{3})$ for $x\in X$ and $\left|f(x)-{g}_{0}(x)-{g}_{1}(x)\right|\le \frac{2}{3}(\frac{2}{3})$ for $x\in A$. By repeated application of the lemma we can construct a sequence^{} of continuous functions ${g}_{0},{g}_{1},{g}_{2},\mathrm{\dots}$ such that $\left|{g}_{n}(x)\right|\le \frac{1}{3}{(\frac{2}{3})}^{n}$ for all $x\in X$, and $\left|f(x)-{g}_{0}(x)-{g}_{1}(x)-{g}_{2}(x)-\mathrm{\cdots}\right|\le {(\frac{2}{3})}^{n}$ for $x\in A$.

Define $F(x)={\sum}_{n=0}^{\mathrm{\infty}}{g}_{n}(x)$. Since $\left|{g}_{n}(x)\right|\le \frac{1}{3}{(\frac{2}{3})}^{n}$ and ${\sum}_{n=0}^{\mathrm{\infty}}\frac{1}{3}{(\frac{2}{3})}^{n}$ converges^{} as a geometric series^{}, then ${\sum}_{n=0}^{\mathrm{\infty}}{g}_{n}(x)$ converges absolutely and uniformly, so $F$ is a continuous function defined everywhere. Moreover ${\sum}_{n=0}^{\mathrm{\infty}}\frac{1}{3}{(\frac{2}{3})}^{n}=1$ implies that $\left|F(x)\right|\le 1$.

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

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

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

Title | proof of Tietze extension theorem |
---|---|

Canonical name | ProofOfTietzeExtensionTheorem |

Date of creation | 2013-03-22 14:08:58 |

Last modified on | 2013-03-22 14:08:58 |

Owner | bbukh (348) |

Last modified by | bbukh (348) |

Numerical id | 10 |

Author | bbukh (348) |

Entry type | Proof |

Classification | msc 54C20 |