# Abstract Nonsense

## Thoughts about separation (Tietze’s extension theorem)

In this post we shall discuss one of the most important theorems consequent of Urysohn’s lemma. It concerns the extension of continuous functions on closed subspaces of a normal space to the full space.

Theorem (Tietze’s Extension Theorem): Let $X$ be normal and $F\subseteq X$ a closed subspace. Then, given any $\varphi:F\mapsto [a,b]$ which is continuous there exists some $\varphi^*:X\mapsto [a,b]$ such that $\varphi^*\mid F=\varphi$. $\varphi^*$ is called an extension of $\varphi$.

Proof: if $a=b$ this is trivial (since the constant mapping extended to the full space is a trivial realization of $\varphi^*$). Thus, we may assume WLOG that $a. Also, we may assume that $[a,b]$ is the smallest closed interval containing $\varphi(X)$. Lastly, it is clear that we need only prove the case when $[a,b]=[-1,1]$.

We firstly define $\varphi_0=\varphi$. Clearly $\text{Dom }\varphi_0=F$, and we define $E_0,G_0\subseteq F$ to be $\varphi^{-1}\left(\left[-1,\frac{-1}{3}\right]\right)$ and $\varphi^{-1}\left(\left[\frac{1}{3},1\right]\right)$ respectively. Clearly $E_0$ and $G_0$ are disjoint closed subspaces of $F$ and since $F$ is itself  closed they are closed in $X$. Thus, by Urysohn’s lemma there exists some $\psi_0:X\mapsto \left[\frac{-1}{3},\frac{1}{3}\right]$ such that $\psi\left(E_0\right)=\frac{-1}{3},\psi(G_0)=\frac{1}{3}$.

We now define $\varphi_1$ on $F$ by $\varphi_1=\varphi_0-\psi_0$. Clearly we note that

$\displaystyle \varphi_1(x)=\begin{cases} \varphi_0(x)+\frac{1}{3} & \mbox{if} \quad x\in E_0 \\ \varphi_0(x)-\psi_0(x) & \mbox{if} \quad x\notin E_0\cup G_0 \\ \varphi_0(x)-\frac{1}{3} & \mbox{if} \quad x\in G_0 \end{cases}$.

From where it readily follows that $\left|\varphi_1(x)\right|\leqslant\frac{2}{3}$.

So, we know define $E_1=\varphi_1^{-1}\left(\left[-1,\frac{-1}{3}\cdot\frac{2}{3}\right]\right)$ and $G_1=\varphi_1^{-1}\left(\left[\frac{1}{3}\cdot \frac{2}{3},1\right]\right)$. Using the same logic above we may find some $\psi_1:X\mapsto \left[\left(\frac{-1}{3}\cdot\frac{2}{3},\frac{1}{3}\cdot\frac{2}{3}\right]\right)$ such that $\psi_1\left(E_1\right)=\frac{-1}{3}\cdot\frac{2}{3}$ and $\psi_1\left(G_1\right)=\frac{1}{3}\cdot\frac{2}{3}$. Now, define $\varphi_2$ on $F$ by $\varphi_2=\varphi_1-\psi_1=\varphi_0-\left(\psi_0+\psi_1\right)$. Using similar logic as before we see that $\left|\varphi_2(x)\right|\leqslant \frac{1}{3}\left(\frac{2}{3}\right)^2$.

Continuing in this way we get a sequence of functions $\left\{\varphi_n\right\}_{n\in\mathbb{N}\cup\{0\}}$ defined on $F$ such that $\left|\varphi_n(x)\right|\leqslant \left(\frac{2}{3}\right)^n$ and a sequence of functions $\left\{\psi_n\right\}_{n\in\mathbb{N}\cup\{0\}}$ defined on $X$ such that $\left|\psi_n(x)\right|\leqslant \frac{1}{3}\cdot\left(\frac{2}{3}\right)^n$, with the property that on $F$ we have that $\displaystyle \varphi_n=\varphi_0-\sum_{j=0}^{n}\psi_j(x)$.

We now define $\displaystyle \sigma_n(x)=\sum_{j=0}^{n}\psi_n(x)$. We therefore see that $\sigma_n$ is the partial sums of an infinite series of  functions in $\mathcal{C}\left[X,\mathbb{R}\right]$. We know that $\mathcal{C}\left[X,\mathbb{R}\right]$ is complete (this is a fundamental fact from basic analysis) and so by virtue of the facts that $\displaystyle \left|\psi_n(x)\right|\leqslant \frac{1}{3}\cdot\left(\frac{2}{3}\right)^n$ and $\displaystyle \sum_{n=0}^{\infty}\frac{1}{3}\cdot\left(\frac{2}{3}\right)^n=1$ we see that $\sigma_n\overset{\text{unif.}}{\longrightarrow}\varphi^*$ with the property that $\left|\varphi^*(x)\right|\leqslant 1$ and that it is continuous.  Also, note that since $\displaystyle \varphi_0-\sum_{j=0}^{n}\psi_j(x)=\varphi-\sum_{j=0}^{n}\psi_j(x)\leqslant \left(\frac{2}{3}\right)^n$ we note that $\sigma_n\overset{\text{unif.}}{\longrightarrow}\varphi$ on $F$ and so $\varphi^*\mid F=\varphi$. The conclusion follows. $\blacksquare$

Remark: A common exercise is to prove that the Tietze extension theorem implies Urysohn’s lemma. Merely note that if given a normal space $X$ and two closed disjoint subsets $G,E$ we see that $G\cup E$ is a closed subspace of $X$. Also, the function $\varphi:G\cup E\mapsto [0,1]$ given by $\displaystyle \varphi(x)=\begin{cases} 0 & \mbox{if} \quad x\in E \\ 1 & \mbox{if} \quad x\in G \end{cases}$ is readily verified to be continuous. Thus, there exists a continuous extension to the entirety of $X$. This satisfies the conditions of the function in Urysohn’s lemma.

In fact, we may prove a slight generalization to Tietze’s extension theorem (that will be generalized even further in subsequent posts).

Theorem (Semi-Extended Tietze’s Extension Theorem): Let $X$ be normal and $F\subseteq X$ a closed subspace. Then, given any continuous $\varphi:F\mapsto [a,b]^n$ there exists some $\varphi^*:X\mapsto\mathbb [a,b]^n$ which is also continuous and $\varphi^*\mid F=\varphi$.

Proof: It is clear taht we must only prove the case where $[a,b]^n=[0,1]^n$. Also, for the sake of notational convenience it is better to think of $[0,1]^n$ as the product space $X_1\times\cdots\times X_n$ where $X_k=[0,1]$ under the usual topology.

Now, as was proved earlier, since $\varphi:F\mapsto X_1\times\cdots\times X_n$ is continuous so is $\pi_k\varphi=\varphi_k:F\mapsto X_k$ for each $1\leqslant k\leqslant n$. It follows from the normal Tietze extension theorem that each $\varphi_k$ has an extension $\varphi_k^*:X\mapsto X_k$, such that $\varphi_k^*=\varphi_k$. We thus define

$\displaystyle \bigoplus_{j=1}^{n}\varphi_j^*=\varphi^*:X\mapsto X_1\times\cdots\times X_n$

by $x\mapsto\left(\varphi_1^*(x),\cdots,\varphi_n^*(x)\right)$. This is clearly continuous since (as was proven earlier) the sum of continuous maps is continuous. It remains to show that $\varphi^*\mid F=\varphi$.

To see this, let $x\in F$ then

$\varphi^*(x)=\left\{\pi_1\varphi(x)\right\}\times\cdots\times\left\{\pi_n\varphi(x)\right\}=\varphi(x)$.

This comes from remembering that on $F$ $\varphi^*_k=\varphi_k=\pi_k\varphi$. The conclusion follows. $\blacksquare$

Remark: It was pointed out to me that the notation $\varphi\times\cdots\times \psi$ is usually reserved for mappings where both the domain and codomain are product spaces. Thus, when just the codomain is a product space we shall use that $\oplus$ notation.

It should be noted that the closedness of $F$ is necessary in the above as can be illustrated by the following example.

Example: Let $\varphi:(0,1]\mapsto [-1,1]$ be given by $\varphi(x)=\sin\left(\frac{1}{x}\right)$. This clearly cannot be extended to $[0,1]$ since we would need to define $\varphi(0)=\lim_{x\to 0^+}\sin\left(\frac{1}{x}\right)$. But, of course this limit does not exist.