# Tietze-Urysohn-like theorems for completely regular spaces

Completely regular spaces (also called Tychonoff spaces) are topological spaces that come with a guarantee of having continuous real-valued functions in sufficient quantity. Thus the class of completely regular spaces is an ideal setting for many purposes that require the use of continuous real-valued functions (one example is working with function spaces). In a completely regular space $X$, for any closed set $B$ and for any point $x$ not in the closed set $B$, there always exists a continuous function $f:X \rightarrow [0,1]$ such that $f(x)=0$ and $f$ maps $B$ to $1$. It turns out that in such a space, we can replace the point $x$ with any compact set $A$ that is disjoint from the closed set $B$. This is a useful tool for proving theorems as well as for constructing objects. In this post, we discuss and prove this result (which resembles Urysohn’s lemma) and another useful fact about completely regular spaces that works very much like Tietze’s extension theorem. Specifically we prove the following results.

Theorem 1

Let $X$ be a completely regular space. For any compact set $A \subset X$ and for any closed set $B \subset X$ that is disjoint from $A$, there exists a continuous function $f:X \rightarrow [0,1]$ such that

• $f(x)=0$ for all $x \in A$,
• $f(x)=1$ for all $x \in B$.
Theorem 2

Let $X$ be a completely regular space. For any compact set $A \subset X$, any continuous function $f:A \rightarrow \mathbb{R}$ can be continuously extended over $X$, i.e., there exists a continuous function $\hat{f}:X \rightarrow \mathbb{R}$ such that $\hat{f}(x)=f(x)$ for all $x \in A$ (in symbol we write $\hat{f} \upharpoonright A=f$).

A space $X$ is normal if any two disjoint closed sets $A \subset X$ and $B \subset X$ can be separated by disjoint open sets, i.e., $A \subset U$ and $B \subset V$ for some disjoint open subsets $U$ and $V$ of $X$. Normal spaces are usually have the additional requirement that singleton sets are closed (i.e. $T_1$ spaces).

These two theorems remind us of two important tools for normal spaces, namely Urysohn’s lemma and Tietze’s extension theorem.

Urysohn’s lemma indicates that for any two disjoint closed sets in a normal space, the space can be mapped continuously to the closed unit interval $[0,1]$ such that one closed set is mapped to $0$ and the other closed set is mapped to $1$. Theorem 1 is like a weakened version of Urysohn’s lemma in that one of the two disjoint closed sets must be compact.

Tietze’s extension theorem indicates that in a normal space, any continuous real-valued function defined on a closed subspace can be extended to the entire space. Theorem 2 is like a weakened version of Tiezte’s extension theorem in that the continuous extension only works for continuous functions defined on a compact subspace.

So if one only works in a completely regular space, one can still apply these two theorems about normal spaces (the weakened versions of course). For the sake of completeness, we state these two theorems about normal spaces.

Urysohn’s Lemma

Let $X$ be a normal space. For any two disjoint closed sets $A \subset X$ and $B \subset X$, there exists a continuous function $f:X \rightarrow [0,1]$ such that

• $f(x)=0$ for all $x \in A$,
• $f(x)=1$ for all $x \in B$.
Tietze’s Extension Theorem

Let $X$ be a normal space. For any closed set $A \subset X$, any continuous function $f:A \rightarrow \mathbb{R}$ can be continuously extended over $X$, i.e., there exists a continuous function $\hat{f}:X \rightarrow \mathbb{R}$ such that $\hat{f}(x)=f(x)$ for all $x \in A$ (in symbol we write $\hat{f} \upharpoonright A=f$).

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Proof of Theorem 1

We now prove Theorem 1. Let $X$ be a completely regular space. Let $A \subset X$ and $B \subset X$ be two disjoint closed sets where $A$ is compact. For each $x \in A$, there exists a continuous function $f_x:X \rightarrow [0,1]$ such that $f_x(x)=0$ and $f_x(B) \subset \left\{1 \right\}$. The following collection is an open cover of the compact set $A$.

$\left\{f_x^{-1}([0,\frac{1}{10})): x \in A \right\}$

Finitely many sets in this collection would cover $A$ since $A$ is compact. Choose $x_1,x_2,\cdots,x_n \in A$ such that $A \subset \bigcup \limits_{j=1}^n f_{x_n}^{-1}([0,\frac{1}{10}))$. Define $h:X \rightarrow [0,1]$ by, for each $x \in X$, letting $h(x)$ be the minimum of $f_{x_1}(x),\cdots,f_{x_n}(x)$. It can be shown that the function $h$, being the minimum of finitely many continuous real-valued functions, is continuous. Furthermore, we have:

• $A \subset h^{-1}([0,\frac{1}{10}))$, and
• $h(B) \subset \left\{1 \right\}$

Now define $w:X \rightarrow [0,1]$ by, for each $x \in X$, letting $w(x)$ be as follows:

$\displaystyle w(x)=\frac{10}{9} \cdot \biggl[ \text{max} \left\{h(x)-\frac{1}{10},0\right\} \biggr]$

It can be shown that the function $w$ is continuous. It is clear that $w(x)=0$ for all $x \in A$ and $w(x)=1$ for all $x \in B$. $\blacksquare$

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Proof of Theorem 2

Interestingly, the proof of Theorem 2 given here uses Tietze’s extension theorem even Theorem 2 is described earlier as a weakened version of Tietze’s extension theorem. Beside using Tietze’s extension theorem, we also use the fact that any completely regular space can be embedded in a cube (see the previous post called Embedding Completely Regular Spaces into a Cube).

The proof is quite short once all the deep results that are used are understood. Let $X$ be a completely regular space. Then $X$ can be embedded in a cube, which is a product of the closed unit interval $[0,1]$. Thus $X$ is homeomorphic to a subspace of the following product space

$Y=\prod \limits_{a \in S} I_a$

for some index set $S$ where $I_a=[0,1]$ for all $a \in S$. We can now regard $X$ as a subspace of the compact space $Y$. Let $A \subset X$ be a compact subset of $X$. Let $f:A \rightarrow \mathbb{R}$ be a continuous function.

The set $A$ is a subset of $X$ and can also be regarded as a subspace of the compact space $Y$, which is normal. Hence Tietze’s extension theorem is applicable in $Y$. Let $\bar{f}:Y \rightarrow \mathbb{R}$ be a continuous extension of $f$. Let $\hat{f}=\bar{f} \upharpoonright X$. Then $\hat{f}$ is the required continuous extension. $\blacksquare$

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$\copyright \ 2014 \text{ by Dan Ma}$