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

# Ψ-Spaces – spaces from almost disjoint families

As the title suggests the $\Psi-$spaces are defined using almost disjoint families, in our case, of subsets of $\omega$. This is a classic example of a pseudocompact space that is not countably compact. This example is due to Mrowka ([3]) and Isbell (credited in [2]), It is sometimes called the Mrowka space in the literature. This is another example that is a useful counterexample and set-theoretic construction. This being an introduction, I prove that the $\Psi-$space, when it is defined using a maximal almost disjoint family of subsets of $\omega$, is pseudocompact and not countably compact. On the other hand, I show that if a normal space is pseudocompact space, then it is countably compact. All spaces in this note is at least Hausdorff.

A space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. According to Theorem 3.10.3 in [1], a space $X$ is countably compact if and only if every infinite subset of $X$ has an accumulation point. A space $X$ is pseudocompact if every real-valued continuous function defined on $X$ is bounded. It is clear that if $X$ is a countably compact space, then it is pseudocompact. We show that the converse does not hold by using the example of $\Psi-$space. We also show that the converse does hold for normal spaces.

Let $\mathcal{A}$ be a family of infinite subsets of $\omega$. The family $\mathcal{A}$ is said to be an almost disjoint family if for each two distinct $A,B \in \mathcal{A}$, $A \cap B$ is finite. The almost disjoint family $\mathcal{A}$ is said to be a maximal almost disjoint family if $B$ is an infinite subset of $\omega$ such that $B \notin \mathcal{A}$, then $B \cap A$ is infinite for some $A \in \mathcal{A}$.

There is an almost disjoint family $\mathcal{A}$ of subsets of $\omega$ such that $\lvert \mathcal{A} \lvert=\text{continuum}$. To see this, identify $\omega$ (the set of all natural numbers) with $\mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace$ (the set of all rational numbers). For each real number $x$, choose a subsequence of $\mathbb{Q}$ consisting of distinct elements that converges to $x$. Then the family of all such sequences of rational numbers would be an almost disjoint family. By a Zorn’s Lemma argument, this almost disjoint family is contained within a maximal almost disjoint family. Thus we also have a maximal almost disjoint family of cardinality continuum. On the other hand, there is no countably infinite maximal almost disjoint family of subsets of $\omega$. See comment below Theorem 2.

Let $\mathcal{A}$ be an infinite almost disjoint family of subsets of $\omega$. Let’s define the space $\Psi(\mathcal{A})$. The underlying set is $\Psi(\mathcal{A})=\mathcal{A} \cup \omega$. Points in $\omega$ are isolated. For $A \in \mathcal{A}$, a basic open set of of the form $\lbrace{A}\rbrace \cup (A-F)$ where $F \subset \omega$ is finite. It is straightforward to verify that $\Psi(\mathcal{A})$ is Hausdorff, first countable and locally compact. It has a countable dense set of isolated points. Note that $\mathcal{A}$ is an infinite discrete and closed set in the space $\Psi(\mathcal{A})$. Thus $\Psi(\mathcal{A})$ is not countably compact. We have the following theorems.

Theorem 1. Let $\mathcal{A}$ be an infinite maximal almost disjoint family of subsets of $\omega$. Then $\Psi(\mathcal{A})$ is pseudocompact.

Proof. Suppose we have a continuous $f:\Psi(\mathcal{A}) \rightarrow \mathbb{R}$ that is unbounded. We can find an infinite $B \subset \omega$ such that $f$ is unbounded on $B$. If $B \in \mathcal{A}$, then we have a contradiction since $\lbrace{f(n):n \in B}\rbrace$ is a sequence that does not converge to $f(B)$. So we have $B \notin \mathcal{A}$. By the maximality of $\mathcal{A}$, $C=B \cap A$ is infinite for some $A \in \mathcal{A}$. Then $\lbrace{f(n):n \in C}\rbrace$ is a sequence that does not converge to $f(A)$, another contradiction. So $\Psi(\mathcal{A})$ is pseudocompact.

Theorem 2. For a normal space $X$, if $X$ is pseudocompact, then $X$ is countably compact.

Proof. Suppose $X$ is not countably compact. Then we have an infinite closed and discrete set $A=\lbrace{a_0,a_1,a_2,...}\rbrace$ in $X$. Define $f:A \rightarrow \mathbb{R}$ by $f(a_n)=n$. According to the Tietze-Urysohn Theorem, in a normal space, any continuous function defined on a closed subset of $X$ can be extended to a continuous function defined on all of $X$. Then $f:A \rightarrow \mathbb{R}$ can be extended to a continuous $f^*:X \rightarrow \mathbb{R}$, making $X$ not pseudocompact.

Comment
If there is a countably infinite maximal almost disjoint family $\mathcal{B}$ of subsets of $\omega$, then $\Psi(\mathcal{B})$ is a countable first countable space and is thus has a countable base (hence is normal). By Theorem 1, it is pseudocompact. By Theorem 2, it is countably compact. Yet $\mathcal{B}$ is an infinite closed and discrete subset of $\Psi(\mathcal{B})$, contradicting that it is countably compact. Thus there is no countably infinite maximal almost disjoint family $\mathcal{B}$ of subsets of $\omega$. In fact, we have the following corollary.

Corollary. If $\mathcal{A}$ is an infinite maximal almost disjoint family of subsets of $\omega$, then $\Psi(\mathcal{A})$ cannot be normal.

Reference

1. Engelking, R., General Topology, Revised and Completed Edition, 1988, Heldermann Verlag Berlin.
2. Gillman, L., and Jerison, M., Rings of Continuous Functions, 1960, Van Nostrand, Princeton, NJ.
3. Mrowka, S., On completely regular spaces, Fund. Math., 41, (1954) 105-106.