# Some subspaces of Bing’s Example G

In the previous post called Compact Subspaces of Bing’s Example G, we give a characterization of the compact subspaces of Bing’s Example G (the space is denoted by the letter $F$). In this post, we discuss how this characterization of compact subsets of Bing’s G can shed light on certain subspaces of Bing’s G. This post should be read (or studied) alongside the previous post on the characterization of compact sets of $F$. The only thing we repeat is the definition of Bing’s Example G.

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Defining Bing’s Example G

First we repeat the definition of Bing’s Example G. Let $P$ be any uncountable set. Let $Q$ be the power set of $P$, i.e., the set of all subsets of $P$. Let $F=2^Q$ be the set of all functions $f: Q \rightarrow 2=\left\{0,1 \right\}$. Obviously $2^Q$ is simply the Cartesian product of $\lvert Q \lvert$ many copies of the two-point discrete space $\left\{0,1 \right\}$, i.e., $\prod \limits_{q \in Q} \left\{0,1 \right\}$. For each $p \in P$, define the function $f_p: Q \rightarrow 2$ by the following:

$\forall q \in Q$, $f_p(q)=1$ if $p \in q$ and $f_p(q)=0$ if $p \notin q$

Let $F_P=\left\{f_p: p \in P \right\}$. Let $\tau$ be the set of all open subsets of $2^Q$ in the product topology. The following is another topology on $2^Q$:

$\tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}$

Bing’s Example G is the set $F=2^Q$ with the topology $\tau^*$. In other words, each $x \in F-F_P$ is made an isolated point and points in $F_P$ retain the usual product open sets.

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Compact Subspaces of Bing’s Example G

We do not repeat the results given in the previous post. For the purposes in this post, the following observation is crucial (see Theorem 2 in the previous post).

Any infinite compact subspace of $F$ is the union of a finite set and finitely many other compact sets each of which is a compact set with only one limit point.

Any compact subset of $F$ that has exactly one limit point is a member of the collection of sets $\mathcal{K}_p$ for some $p \in P$ (see Theorem 1 in the previous post). For $p \in P$, the set $\mathcal{K}_p$ is defined by:

$\mathcal{K}_p=\left\{K \in \mathcal{C}_p: \forall q \in Q, \left\{f \in K: f(q) \ne f_p(q) \right\} \text{ is finite} \right\}$

where $\mathcal{C}_p$ is the collection of all closed subsets of $F$ each of which has the point $f_p$ as the only limit point. For the results shown below, it suffices to work with a member of some $\mathcal{K}_p$ when we work with an infinite compact subset of $F$.

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Some Subspaces of Example G

For each $f \in F$, let $supp(f)$ be the support of $f$, i.e., $supp(f)=\left\{q \in Q:f(q) \ne 0 \right\}$. For any infinite cardinal number $\theta \le \lvert Q \lvert$, we consider the following subspace:

$M_{\theta}=F_P \cup \left\{f \in F: \lvert supp(f) \lvert <\theta \right\}$

The subspace $M_{\theta}$ consists of all points $f_p \in F_P$ and all other $f \in F$ such that $f(q)=1$ for less than $\theta$ many $q \in Q$. So the support of these functions is small (in relation to the size of the domain $Q$). Among these subspaces, of particular interest are the following two subspaces:

$M_{\lvert Q \lvert}=F_P \cup \left\{f \in F: \lvert supp(f) \lvert <\lvert Q \lvert \right\}$
$M_{\omega}=F_P \cup \left\{f \in F: supp(f) \text{ is finite} \right\}$

The subspace $M_{\omega}$ was discussed by Michael in [1] and is discussed in our blog in the post called A subspace of Bing’s example G. Michael in [1] used the letter $G$ to denote the space $M_{\omega}$. We choose another letter to distinguish it from Example G. The subspace $M_{\omega}$ consists of all points $f_p \in F_P$ and all other $f \in F$ such that $f(q)=1$ for only finitely many $q \in Q$. Just like Example G, the space $M$ is normal and not collectionwise Hausdorff (hence not collectionwise normal and not paracompact). By eliminating points $f \in F$ that have values of $1$ for infinitely many $q \in Q$, we obtain a subspace that is metacompact.

We show the following claim about the subspace $M_{\lvert Q \lvert}$:

Proposition 1
All compact subsets of the space $M_{\lvert Q \lvert}$ are finite.

Proof of Proposition 1
In light of the comment in the preceding section, we only need to show for any compact subset $K$ of $F$ such that $K \in \mathcal{K}_p$ for some $p \in P$, $K \cap M_{\lvert Q \lvert}$ is finite.

Suppose $K \cap M_{\lvert Q \lvert}$ is infinite for $K \in \mathcal{K}_p$. Choose $\left\{g_1,g_2,g_3,\cdots \right\} \subset K \cap M_{\lvert Q \lvert}$ such that $g_i \ne g_j$ for $i \ne j$. Note that $K_0=\left\{g_1,g_2,g_3,\cdots \right\} \cup \left\{f_p \right\}$ is a closed subset of $K$ and is thus compact.

For each $j$ let $Q_j=\left\{q \in Q: g_j(q)=1 \right\}$. Since each $g_j \in M_{\lvert Q \lvert}$, each $Q_j$ has cardinality less than $\lvert Q \lvert$. Thus $Q_1 \cup Q_2 \cup \cdots$ has cardinality less than $\lvert Q \lvert$ too. On the other hand, let $Q_\omega=\left\{q \in Q: p \in q \right\}$. Since $Q_\omega$ has cardinality equal to $\lvert Q \lvert$, we can pick $r \in Q_\omega-(Q_1 \cup Q_2 \cup \cdots)$.

Right away we know that $f_p(r)=1$ and $g_j(r)=0$ for all $j$. Let $V=\left\{f \in 2^Q: f(r)=1 \right\}$ which is an open set that contains $f_p$. But $g_j \notin V$ for all $j$. Thus the following collection

$\left\{V \right\} \cup \left\{ \left\{g_j \right\}:j=1,2,3,\cdots \right\}$

is an open cover of $K_0$ that has no finite subcover, contradicting the fact that $K_0$ is compact. Thus for any $p \in P$, for any compact set $K \in \mathcal{K}_p$, $K \cap M_{\lvert Q \lvert}$ is finite. In other words, for any compact subset $K$ of $F$ with only one limit point, $K \cap M_{\lvert Q \lvert}$ must be finite. It follows that in the space $M_{\lvert Q \lvert}$, all compact sets are finite. $\blacksquare$

Note that $M_\theta \subset M_{\lvert Q \lvert}$ for all infinite cardinal numbers $\theta < \lvert Q \lvert$. Thus the compact subsets of all such subspaces $M_\theta$ are finite. In particular, for the subspace $M_{\omega}$, there are no infinite compact subsets. Thus we have the following two easy propositions.

Proposition 2
For any infinite cardinal number $\theta < \lvert Q \lvert$, all compact subsets of the space $M_{\theta}$ are finite.
Proposition 3
In particular, all compact subsets of the space $M_{\omega}$ are finite.

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Another Proposition

The proof of Proposition 1 can be modified to show that every “small” subspace of the space $M_{\lvert Q \lvert}$ has no limit point. We have the following proposition.

Proposition 4
For any $Y \subset M_{\lvert Q \lvert}$ with $\lvert Y \lvert <\lvert Q \lvert$, the subspace $Y$ has no limit point (i.e. cluster point) in $M_{\lvert Q \lvert}$.

Once Proposition 4 is established, we have the following two propositions.

Proposition 5
Let $\theta$ be any infinite cardinal number $\theta < \lvert Q \lvert$. Then for any set $Y \subset M_{\theta}$ with $\lvert Y \lvert <\lvert Q \lvert$, the subspace $Y$ has no limit point (i.e. cluster point) in $M_{\theta}$.
Proposition 6
In particular, for any $Y \subset M_{\omega}$ with $\lvert Y \lvert <\lvert Q \lvert$, the subspace $Y$ has no limit point (i.e. cluster point) in $M_{\omega}$.

The limit points in Bing’s Example G have large character. In the subspaces of Bing’s Example G discussed here, the characters at the points of $F_P$ are still large. In these subspaces, the closure of any “small” subset cannot reach the limit points in the set $F_P$. So even by narrowing the focus on just the subspaces of points with “small” support, we still obtain subspaces that have large characters. For example, $M_{\omega}$ (the subspace with finite support on the isolated points) is not only not first countable; it cannot even have any convergent sequence. In fact, any long as a subset is small (cardinality less than the cardinality of $Q$), the closure cannot reach any limit points at all.

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Reference

1. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.

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

# Compact Subspaces of Bing’s Example G

In a previous post, we discuss basic properties of Bing’s Example G, a classic and influential example of a normal but not collectionwise normal space. In another post, we discuss a subspace of Bing’s Example G which is also normal but not collectionwise normal but is metacompact. In this post, we further discuss Bing’s Example G by characterizing its compact subspaces.

The ideas discussed here can be found in [1] with a lot of details omitted. In this post, we provide all the necessary details in understanding these ideas. In the next post called Some subspaces of Bing’s Example G, we discuss some subspaces of Bing’s Example G based on the characterization of compact sets given in this post.

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Defining Bing’s Example G

First we repeat the definition of Bing’s Example G. Let $P$ be any uncountable set. Let $Q$ be the power set of $P$, i.e., the set of all subsets of $P$. Let $F=2^Q$ be the set of all functions $f: Q \rightarrow 2=\left\{0,1 \right\}$. Obviously $2^Q$ is simply the Cartesian product of $\lvert Q \lvert$ many copies of the two-point discrete space $\left\{0,1 \right\}$, i.e., $\prod \limits_{q \in Q} \left\{0,1 \right\}$. For each $p \in P$, define the function $f_p: Q \rightarrow 2$ by the following:

$\forall q \in Q$, $f_p(q)=1$ if $p \in q$ and $f_p(q)=0$ if $p \notin q$

Let $F_P=\left\{f_p: p \in P \right\}$. Let $\tau$ be the set of all open subsets of $2^Q$ in the product topology. The following is another topology on $2^Q$:

$\tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}$

Bing’s Example G is the set $F=2^Q$ with the topology $\tau^*$. In other words, each $x \in F-F_P$ is made an isolated point and points in $F_P$ retain the usual product open sets.

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Basic Open Sets in Bing’s Example G

To facilitate the discussion below, we now fix now some notation for basic open sets of the points $f_p \in F_P$. For any finite $S \subset Q$, the following describes the basic open sets containing the point $f_p \in F_P$.

$U(f_p,S)=\left\{f \in 2^Q: \forall q \in S,f(q)=f_p(q) \right\}$

If $S=\left\{q_1,q_2,\cdots,q_n \right\}$, the open set $U(f_p,S)$ can be denoted by

$V(f_p,q_1,q_2,\cdots,q_n)=U(f_p,S)$

Recall that for any space $X$ and for any $A \subset X$ and for any point $p \in X$, the point $p$ is a limit point of $A$ if every open subset of $X$ containing $p$ contains a point of $A$ that is different from $p$.

Because points of $F_P$ retain the product open sets, points of $F_P$ are the only limit points in the space $F$. The set $F_P$ is a closed and discrete set in the space $F$. To see this, consider the open set $V(f_p,q)$ where $q=\left\{p \right\}$ and $p \in P$. It contains $f_p \in F_P$ and does not contain $f_t$ for any $t \ne p$.

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One Example of Compact Subsets of Bing’s Example G

Since the set $F_P$ is a closed and discrete set in the space $F$, any compact subset of $F$ can contain at most finitely many points of $F_P$. We first present an example of an infinite compact subset of $F$ whose intersection with $F_P$ has only one point (i.e. the compact set has only one limit point).

For each $p \in P$ and for each $q \in Q$, define a function $H(p,q):Q \rightarrow 2$ by the following:

$H(p,q)(r) = \begin{cases} f_p(r), & \mbox{if } r \in Q-\left\{q \right\} \\ \ne f_p(r), & \mbox{if } r=q \end{cases}$

Essentially, the function $H(p,q)$ agrees with the function $f_p$ except on the point $q$. For each $p \in P$, consider the following subset of the space $F$.

$D(p)=\left\{H(p,q): q \in Q \right\}$

We show that for a fixed $p$, the functions $H(p,q)$ are distinct for distinct $q$. So for $q \ne q'$, we show that $H(p,q) \ne H(p,q')$. By definition, it is clear that

$H(p,q)(q')=f_p(q')$ and $H(p,q')(q') \ne f_p(q')$.

Thus the set $D(p)$ consists of distinct elements. Since $Q$ is uncountable, $D(p)$ is uncountable. If $\lvert P \lvert$ is the cardinal number $\theta$, then $\lvert D(p) \lvert=\lvert Q \lvert=2^\theta$.

We make the following claims about the set $D(p)$.

• For each $p \in P$, the set $D(p)$ contains no point of $F_P$.
• Every open set containing $f_p$ contains all but finitely many points of $D(p)$.
• The point $f_p$ is the only limit point of $D(p)$.
• For each $p \in P$, $A(p)=D(p) \cup \left\{f_p \right\}$ is compact.
• Thus, $A(p)$ is like the one-point compactification of $D(p)$.

To see the first bullet point, clearly $f_p \ne H(p,q)$ for all $q \in Q$. The function $f_p$ and the function $H(p,q)$ differ at the point $q$. It is also the case that for $t \ne p$, $f_t \ne H(p,q)$ for all $q \in Q$. Suppose that $f_t=H(p,q)$ for some $q \in Q$. There are two cases to consider: $t \in q$ or $t \notin q$. Suppose $t \in q$. Then $f_t(q)=1$. So $H(p,q)(q)=f_t(q)=1$. But $H(p,q)(q) \ne f_p(q)$. Thus $f_p(q)=0$. This means that $p \notin q$. Now let $r=q \cup \left\{p \right\}-\left\{t \right\}$. First $H(p,q)(r)=f_p(r)=1$. On the other hand, $H(p,q)(r)=f_p(r)=0$, leading to a contradiction. It can be shown that assuming $t \notin q$ will also lead to a contradiction too. Thus for any $t \in P$ with $t \ne p$, $f_t \notin D(p)$.

To see the second bullet point, let $S \subset Q$ be a finite set and let $U(f_p,S)$ be an arbitrary basic open set containing $f_p$. For any $q \subset Q-S$, for all $r \in S$, $r \ne q$ and $H(p,q)(r)=f_p(r)$, implying that $H(p,q) \in U(f_p,S)$. Thus every open set containing $f_p$ contains all but finitely many points of $D(p)$.

The second bullet shows that $f_p$ is a limit point of $D(p)$. We now show that $f_p$ is the only limit point of $D(p)$. Let $t \in P-\left\{p \right\}=v$. Consider the basic open set $V(f_t,v)$, which contains $f_t$. For each $q \in Q$ with $q \ne v$, $H(p,q)(v)=f_p(v)=0$, showing that $H(p,q) \notin V(f_t,v)$. Thus the open set $V(f_t,v)$ can contain at most one point of $D(p)$, namely $H(p,v)$. So $f_t$ cannot be a limit point of $D(p)$ for all $t \in P-\left\{p \right\}$.

Thus the set $A(p)$ is a compact set in the space $F$. It has only one limit point, namely the point $f_p$. Viewing $A(p)$ as a space by itslef, the open set at the point $f_p$ is co-finite (second bullet point). Thus $A(p)$ is like the one-point compactification of $D(p)$.

We have demonstrated a specific example of infinite compact subsets of the space $F$. In the characterization of compact sets in the next section, we can always refer to $A(p)$ and know that the sets described by the characterization below exist.

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Characterizing Compact Subsets of Bing’s Example G

Let $\mathcal{C}$ be the collection of all infinite closed subsets of the space $F$. For each $p \in P$, let $\mathcal{C}_p$ be the following collection:

$\mathcal{C}_p=\left\{C \in \mathcal{C}: f_p \text{ is the only limit point of } C \text{ and if } t \ne p, f_t \notin C \right\}$

Note that the set $A(p)=D(p) \cup \left\{f_p \right\}$ as discussed above is a member of $\mathcal{C}_p$. Let $\mathcal{K}_p$ be the following set:

$\mathcal{K}_p=\left\{K \in \mathcal{C}_p: \forall q \in Q, \left\{f \in K: f(q) \ne f_p(q) \right\} \text{ is finite} \right\}$

Note that $\mathcal{K}_p \ne \varnothing$ since $A(p) \in \mathcal{K}_p$. It turns out that $\mathcal{K}_p$ characterizes the compact subsets with $f_p$ as the only limit point. We now prove the following theorem.

Theorem 1

Let $p \in P$. Then $K$ is an infinite compact subset of the space $F$ with $f_p$ being the only limit point of $K$ if and only if $K \in \mathcal{K}_p$.

Proof of Theorem 1
$\Longrightarrow$
Suppose that $K$ is an infinite compact subset of the space $F$ with $f_p$ being the only limit point of $K$. Suppose $K \notin \mathcal{K}_p$. Then for some $q \in Q$, the set

$T_q=\left\{f \in K: f(q) \ne f_p(q) \right\}$

is infinite. Now choose an infinite subset $\left\{f_1,f_2,f_3,\cdots \right\} \subset T_q$ where $f_i \ne f_j$ for $i \ne j$. Consider the open set $V(f_p,q)=\left\{f \in 2^Q: f(q)=f_p(q) \right\}$. Note that $f_p \in V(f_p,q)$ and $f_j \notin V(f_p,q)$ for all $j$. Thus $V(f_p,q)$ is an open containing $f_p$ that misses infinitely many points of $K$, a contradiction. Thus $K \in \mathcal{K}_p$.

$\Longleftarrow$
Suppose $K \in \mathcal{K}_p$. Let $q_1,q_2,\cdots,q_n \in Q$ be finitely many sets from $Q$. Consider the basic open set $V(f_p,q_1,\cdots,q_n)$ containing $f_p$. Recall that this is just the set of all $f \in 2^Q$ that agree with $f_p$ on the elements $q_1,q_2,\cdots,q_n \in Q$. Because $K \in \mathcal{K}_p$, for each $q_j$, the following set is finite.

$A_j=\left\{f \in K: f(q_j) \ne f_p(q_j) \right\}$

For each $f \in K-(A_1 \cup A_2 \cup \cdots \cup A_n)$, and for each $j$, $f(q_j)=f_p(q_j)$. Thus the open set $V(f_p,q_1,\cdots,q_n)$ contains all but finitely many points of $K$. So $f_p$ is a limit point of $K$.

We now show that $f_p$ is the only limit point of $K$. Let $t \in P-\left\{p \right\}$. We show that $f_t$ is not a limit point of $K$. Let $r=P-\left\{p \right\}$. Consider the basic open set $V(f_t,r)$. Note that $f_t \in V(f_t,r)$ and $f_p(r)=0$. Consider the set $A$:

$A=\left\{f \in K: f(r) \ne f_p(r) \right\}$

The set $A$ is finite. For each $f \in K-A$, $f(r)=f_p(r)=0 \ne 1=f_t(r)$ and thus $f \notin V(f_t,r)$. So the open set $V(f_t,r)$ can contain at most finitely many points of $K$. Thus $f_t$ is not a limit point of $K$. We have shown that $K$ is an infinite compact subset of $F$ with $f_p$ as the only limit point. $\blacksquare$

Theorem 1 characterizes the compact subsets of Bing’s Example G with only one limit point. We now characterize all compact subsets of the space $F$. We have the following theorem.

Theorem 2

Let $K \subset F$. Then $K$ is a compact subspace of $F$ if and only if $K$ is the union of finitely many sets in $\mathcal{K}$ where

$\mathcal{K}=\biggl( \bigcup \limits_{p \in P} \mathcal{K}_p \biggr) \cup \mathcal{H}$

with $\mathcal{H}$ being the collection of all finite subsets of $F$.

Proof of Theorem 2
The direction $\Longleftarrow$ is clear. For the direction $\Longrightarrow$, let $K$ be a compact subset of $F$. If $K$ is finite, then $K \in \mathcal{H}$. So assume that $K$ is infinite. Since the set $F_P$ is closed and discrete in $F$, $K$ can only contain finitely many points of $F_P$, say $f_{p_1},f_{p_2},\cdots,f_{p_n}$.

Choose open sets $U_1,U_2,\cdots,U_n$ such that for each $j$, $f_{p_j} \in U_j$ and such that $\overline{U_i} \cap \overline{U_j} = \varnothing$ for $i \ne j$. For each $j$, let $K_j=\overline{U_j} \cap K$. Let $H$ be the set of points of $K$ not in any of the $K_j$. Note that $H \in \mathcal{H}$ and $K_j \in \mathcal{K}_{p_j}$. So $K$ is the union of finitely many sets in the collection $\mathcal{K}$. $\blacksquare$

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Comment

Essentially an infinite compact subset of Bing’s Example G is the union of finitely many sets from finitely many $\mathcal{K}_p$. In some cases, when working with compact subsets of Example G, it is sufficient to work with sets from $\mathcal{K}_p$ for one arbitrary $p \in P$. See the next post for an example.

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Reference

1. Boone, J. R., Some characterizations of paracompactness in k-spaces, Fund. Math., 72, 145-155, 1971.

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

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