# Embedding Completely Regular Spaces into a Cube

This is a continuation of a discussion on completely regular spaces (continuing from these two posts: Completely Regular Spaces and Pseudocompact Spaces, Completely Regular Spaces and Function Spaces). This post gives another reason (one of the most important ones) why the class of completely regular spaces occupies a central place in general topology, which is that completely regular spaces are precisely the spaces that can be embedded in a cube (the product of copies of the unit interval). This theorem was proved by Tychonoff in 1930 . The tool that makes this theorem possible also allowed Stone and Cech in 1937 to construct for any completely regular space $X$, a compact Hausdorff space $\beta X$ that contains $X$ as a dense set ($\beta X$ is called the Stone-Cech compactification of $X$). In this post we discuss the role played by complete regularity in this construction. We discuss the following theorem.

Theorem 1
Let $X$ be a space. Then $X$ is completely regular if and only if $X$ is homeomorphic to a subspace of a cube.

Original articles are [1], [4] and [5]. The Stone-Cech compactification and the related concepts are classic topics that are covered in standard texts (see [2] and [6]).

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Completely Regular Spaces

A space $X$ is said to be completely regular if $X$ is a $T_0$ space and for each $x \in X$ and for each closed subset $A$ of $X$ with $x \notin A$, there is a continuous function $f:X \rightarrow [0,1]$ such that $f(A) \subset \left\{0 \right\}$ and $f(x)=1$. Note that the $T_0$ axiom and the existence of the continuous function imply the $T_1$ axiom, which is equivalent to the property that single points are closed sets. Completely regular spaces are also called Tychonoff spaces.

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The Evaluation Map

The evaluation map can be defined in a more general setting. We define it here just to deal with the task at hand, namely to discuss Theorem 1.

Let $X$ be a completely regular space. Let $I=[0,1]$ be the unit interval in the real line $\mathbb{R}$. A cube is of the form $I^\mathcal{K}$, i.e., the product of $\mathcal{K}$ many copies of $I$ where $\mathcal{K}$ is some cardinal. Let $C(X,I)$ be the set of all continuous real-valued functions defined on the space $X$. Consider the product space $\prod \limits_{f \in C(X,I)} I_f$ where each $I_f=I$. This is the cube where $X$ is embedded as a subspace. We can represent each point in the cube $\prod \limits_{f \in C(X,I)} I_f$ as a function $H:C(X,I) \rightarrow I$ or as a sequence $< H_f >_{f \in C(X,I)}$ such that each term (or coordinate) $H_f \in I=[0,1]$.

The key to embed $X$ into a cube is through the evaluation map, which is a map $E$ from $X$ into the product space $\prod \limits_{f \in C(X,I)} I_f$. Thus we have:

$\displaystyle (1) \ \ \ \ \ \ E:X \longrightarrow \prod \limits_{f \in C(X,I)} I_f$

We now define the map $E$. For each $x \in X$, $E(x)=< H_f >_{f \in C(X,I)}$ such that $H_f=f(x)$ for each $f \in C(X,I)$. In other words, $E(x)$ is the point in the product space $\prod \limits_{f \in C(X,I)} I_f$ whose $f^{th}$ coordinate is $f(x)$.

We show that because $X$ is completely regular, the evaluation map $E$ is a homeomorphism. We show the following:

• The map $E$ is continuous.
• The map $E$ is one-to-one.
• The map $E^{-1}$ is continuous.

The continuity of the map $E$ follows from the fact that each $f \in C(X,I)$ is continuous. The map $E$ being one-to-one follows from the fact that for each pair $x,y \in X$ with $x \ne y$, there is an $f \in C(X,I)$ such that $f(x) \ne f(y)$.

We now show $E^{-1}$, the inverse of $E$, is continuous. This is where $X$ must be completely regular. Let $U$ be open in $X$ such that $x =E^{-1}(E(x)) \in U$. Since $X$ is completely regular, there is a continuous $g:X \rightarrow I$ such that $g(X-U) \subset \left\{0 \right\}$ and $g(x)=1$. Let $V=\prod \limits_{f \in C(X,I)} I_f$ where $I_f=I$ for all $f \ne g$ and $I_g=(0,1]$. Then $V_0=V \cap E(X)$ is open in $E(X)$ and $E(x) \in V_0$. We claim that $E^{-1}(V_0) \subset U$. To see this, pick $T=E(y) \in V_0$. Note that $T_g \in (0,1]$ since $T \in V_0$. If $y \in X-U$, then $T_g =0$. Thus we must have $y =E^{-1}(E(y)) \in U$.

The above discussion shows that any completely regular space $X$ is homeomorphic to a subspace of the product space $[0,1]^\mathcal{K}$ where $\mathcal{K}$ is the cardinality of $C(X,I)$. This establishes one direction of Theorem 1. The other direction is clear. Note that any cube is a compact Hausdorff space and any subspace of a compact Hausdorff space is completely regular.

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The Stone-Cech Compactification

Let $X$ be a completely regular space. Let $E$ be the evaluation as defined in $(1)$ above. According to the above discussion $E$ is a homeomorphism. Hence $E(X)$ is a topological copy of $X$ as a subspace of the product space $\prod \limits_{f \in C(X,I)} I_f$. Consider $\overline{E(X)}$ where the closure is taken in the product space. The Stone-Cech compactification of $X$ is denoted by $\beta X$ and is defined to be this closure $\beta X=\overline{E(X)}$.

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Reference

1. Cech, E., On bicompact spaces, Ann. Math. (2) 38, 823-844, 1937.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
4. Stone, M. H., Applications of the Theorey of Boolean Rings to General Topology, Trans. Amer. Math. Soc., 41, 375-481, 1937.
5. Tychonoff, A., Uber die topologische von Raumen, Math., Ann., 102, 544-561, 1930.
6. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# Completely Regular Spaces and Function Spaces

This is a continuation of a discussion on completely regular spaces (continuing from this previous post). These spaces are an integral part of many discussions involving topological spaces and/or properties. Some notions are contingent on the existence of certain real-valued continuous functions. Discussion of such notions can be greatly facilitated by working in the class of completely regular spaces. One example given in a previous post is on a discussion of pseudocompact spaces. Another example for requiring complete regularity is when working with function spaces. When the object being studied is the space of real-valued continuous functions defined on a topological space $X$, it is desirable to have enough continuous functions in the function space being studied. In this post, we illustrate this point by giving the proofs of two simple results in $C_p(X)$, the space of real-valued continuous functions endowed with the pointwise convergence topology.

Basic references are [2] and [4]. Refer to [3] for a discussion of where complete regularity is placed among the separation axioms. An in-depth treatment for $C_p$ theory is [1].
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Completely Regular Spaces

A space $X$ is said to be completely regular if $X$ is a $T_0$ space and for each $x \in X$ and for each closed subset $A$ of $X$ with $x \notin A$, there is a continuous function $f:X \rightarrow [0,1]$ such that $f(A) \subset \left\{0 \right\}$ and $f(x)=1$. Note that the $T_0$ axiom and the existence of the continuous function imply the $T_1$ axiom, which is equivalent to the property that single points are closed sets. Completely regular spaces are also called Tychonoff spaces.

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Function Spaces

Let $X$ be a space. Let $C(X)$ be the set of all real-valued continuous functions defined on the space $X$. The set $C(X)$ is naturally a subspace of the product space $\prod \limits_{x \in X} Y_x$ where each $Y_x = \mathbb{R}$. We can also write $\prod \limits_{x \in X} Y_x = \mathbb{R}^X$. Thus $C(X)$ can be endowed with the subspace topology inherited from the product space $\mathbb{R}^X$. When this is the case, the resulting function space is denoted by $C_p(X)$.

Now we need a good handle on the open sets in the function space $C_p(X)$. A basic open set in the product space $\mathbb{R}^X$ is of the form $\prod \limits_{x \in X} U_x$ where each $U_x$ is open in $\mathbb{R}$ and $U_x = \mathbb{R}$ for all but finitely many $x \in X$ (equivalently $U_x \ne \mathbb{R}$ for only finitely many $x \in X$). Thus a basic open set in $C_p(X)$ is of the form:

$(1) \ \ \ \ \ \ \ \ C(X) \cap \prod \limits_{x \in X} U_x$

where each $U_x$ is open in $\mathbb{R}$ and $U_x = \mathbb{R}$ for all but finitely many $x \in X$. In addition, when $U_x \ne \mathbb{R}$, we can take $U_x$ to be an open interval of the form $(a,b)$. To make the basic open sets of $C_p(X)$ more explicit, $(1)$ is translated as follows:

$(2) \ \ \ \ \ \ \ \ \bigcap \limits_{x \in F} [x, O_x]$

where $F \subset X$ is a finite set, for each $x \in F$, $O_x$ is an open interval of $\mathbb{R}$, and $[x,O_x]$ is the set of all $f \in C(X)$ such that $f(x) \in O_x$. In proving results about $C_p(X)$, we can use basic open sets that are described in $(2)$.

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Completely Regular Spaces and Function Spaces

We now give two simple results in $C_p(X)$ that are good illustrations that complete regularity is the ideal setting for the domain space of $C_p$ function spaces. In Theorem 1, complete regularity is used to generate a non-separable subspace of the function space given a non-Lindelof subspace of the domain space. In Theorem 2, complete regularity is used to generate a non-Lindelof subspace of the function space given a non-separable subspace of the domain space.

Theorem 1
Let $X$ be a completely regular space. Then if $C_p(X)$ is hereditarily separable, then $X$ is hereditarily Lindelof.

Proof
We show that if $X$ has a subspace that is not Lindelof, then there is a subspace of $C_p(X)$ that is not separable. We use complete regularity to generate the continuous functions that form a non-separable subspace.

Let $Y \subset X$ be a non-Lindelof subspace. There exists an open cover $\mathcal{U}$ of $Y$ such that $\mathcal{U}$ has no countable subcover. Open sets in $\mathcal{U}$ are open sets in $Y$. We wish to expand these to open sets in $X$. Let $\mathcal{U}^*$ be the collection of open subsets $U$ of $X$ such that $U \cap Y \in \mathcal{U}$. It is clear that no countable subcollection of $\mathcal{U}^*$ can cover $Y$.

For each $y \in Y$, choose $U(y) \in \mathcal{U}^*$ such that $y \in U(y)$. Here’s where we use complete regularity. For each $y \in Y$, there is a continuous function $f_y:X \rightarrow [0,1]$ such that $f_y(X-U(y)) \subset \left\{0 \right\}$ and $f_y(y) =1$. Let $T=\left\{f_y:y \in Y \right\}$, which is a subspace of $C_p(X)$.

We claim that $T$ is not separable. To see this, let $A=\left\{g_1,g_2,g_3,\cdots \right\}$ be a countable subset of $T$ such that for each $i$, $g_i$ is obtained from the point $y(i) \in Y$ and the open set $U_i=U(y(i))$, i.e., $g_i=f_{y(i)}$. Note that $\left\{U_1,U_2,U_3,\cdots \right\}$ cannot be a cover of $Y$. Let $a \in Y$ be a point that is not in all $U_i$.

Consider the function $f_a$ chosen above using complete regularity. Note that $f_a(a)=1$ and $f_a(X-U(a)) \subset \left\{0 \right\}$. On the other hand, $g_i(a)=f_{y(i)}(a)=0$ for all $i$ since $a \notin U_i$ for all $i$. This means that $f_a$ is not in the closure of $A$. For example, $[a,(0.9,1.1)]$ is a basic open set containing $f_a$ such that $g_i \notin [a,(0.9,1.1)]$ for all $i$. Thus no countable subset of $T$ can be dense in $T$, completing the proof. $\blacksquare$

Theorem 2
Let $X$ be a completely regular space. Then if $C_p(X)$ is hereditarily Lindelof, then $X$ is hereditarily separable.

Proof
Let $Y$ be a non-separable subspace of $X$. For each countable $A \subset Y$, there must be some point $y(A) \in Y$ such that $y(A)$ is not a member of the closure of $A$ (relative to $Y$). For each countable $A \subset Y$, let $\overline{A}$ be the closure of $A$ in the entire space $X$. Clearly $y(A) \notin \overline{A}$.

Now apply the complete regularity of the space $X$. For each countable $A \subset Y$, let $f_A: X \rightarrow [0,1]$ be continuous such that $f_A(\overline{A}) \subset \left\{0 \right\}$ and $f_A(y(A))=1$. Let $W$ be the set of all $f_A$ where $A \subset Y$ is countable.

We now show that $W$ is a non-Lindelof subspace of $C_p(X)$. For each $f_A \in W$, let $U_A=[y(A),(0.9, 1.1)] \cap W$, which is open in $C_p(X)$ and contains $f_A$. Let $\mathcal{U}$ be the collection of all such open sets $U_A$.

Then $\mathcal{U}$ is an open cover of $W$ that has no countable subcover. To see this, suppose we have $\left\{U_1,U_2,U_3,\cdots \right\}$ such that for each $i$, $U_i=U_{A(i)}$ where $A(i) \subset Y$ is countable. Then let $B$ be the set of all $A(i)$ and all points $y(A(i))$. Note that the set $B$ is still a countable subset of $Y$. Consider the point $y(B)$ and the continuous function $f_B$, which is a member of $W$. We have $f_B(y(B))=1$ and $f_B(t)=0$ for all $t \in \overline{B}$. Note that each $y(A(i)) \in B$ and thus $f_B(y(A(i))=0$ for all $i$. This shows that $f_B \notin U_i=U_{A(i)}$ for all $i$. Then $\mathcal{U}$ is an open cover of $W$ that has no countable subcover, leading to the conclusion that $W$ is a non-Lindelof subspace of $C_p(X)$. $\blacksquare$

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Remark

The above proofs of Theorem 1 and Theorem 2 are only meant as demonstration of the role played by complete regularity in working with function spaces. They are much weaker versions of a deeper result. These two results can serve to motivate a deeper result that explores the relationship between the hereditary separability (respectively hereditary Lindelof property) of the domain space and the hereditary Lindelof property (respectively the hereditary separability) of the function space. The following theorem is the countable version of a theorem found in [5].

Theorem 3
Let $X$ be a completely regular space. The following conditions are equivalent.

1. $C_p(X)$ is hereditarily separable (respectively hereditarily Lindelof).
2. $X^\omega$ is hereditarily Lindelof (respectively hereditarily separable).
3. For each positive integer $n$, $X^n$ is hereditarily Lindelof (respectively hereditarily separable).

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Reference

1. Arhangel’skii, A., Topological Function Spaces, Kluwer Academic Publishers, Boston, 1992.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.
5. Zenor, P., Hereditarily m-separability and the hereditarily m-lindelof property in product spaces and function spaces, Fund. Math. 106, 175-180, 1980.

# Completely Regular Spaces and Pseudocompact Spaces

In proving theorems about properties in abstract topological spaces, it makes sense that the spaces in questions satisfy some axioms in addition to the ones stipulated in the definition of topological spaces. For example, authors typically assume certain separation axioms. Doing so will help authors know in advance what basic properties the spaces will have. For example, it is desirable to know that singleton sets (and finite sets) are closed (assuming the $T_1$ axiom or a $T_1$ space). In some circumstances, it may be desirable to be able to separate a single point from a closed set not containing it (assuming $T_3$ axiom or regularity). In some situation, it may be advantageous (and even necessary) to know in advance that there is a sufficient quantity in continuous real-valued functions that can be defined on the spaces in question. We give several reasons of needing enough continuous functions (this list is not meant to be exhaustive).

1. Certain notions involve continuous real-valued function defined on the space. Pseudocompactness is one such notion (this view point is discussed in this post).
2. Spaces of continuous functions are the objects being studied (this view point is discussed in this post).
3. Completely regular spaces are precisely the spaces that can be embedded in a cube (the product space of copies of the unit interval). Discussed in this post.

In this post, we discuss the importance of complete regularity from the first view point and use pseudocompactness as an illustration. See [1] and [2] and [3] for any notions not defined here. Steen and Seebach (section 2 of [2] starting on p.11) has an excellent discussion of separation axioms.

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Completely Regular Spaces and Pseudocompact Spaces

A space $X$ is said to be completely regular if $X$ is a $T_0$ space and for each $x \in X$ and for each closed subset $A$ of $X$ with $x \notin A$, there is a continuous function $f:X \rightarrow [0,1]$ such that $f(A) \subset \left\{0 \right\}$ and $f(x)=1$. Note that the $T_0$ axiom and the existence of the continuous function imply the $T_1$ axiom, which is equivalent to the property that single points are closed sets. Completely regular spaces are also called Tychonoff spaces.

As defined above, in a completely regular space, for any closed set and a point not in the closed set, we can always find a continuous function mapping the closed set to 0 and the point to 1. Essentially, to be a completely regular space, it suffices to provide a continuous function that maps a given closed set and a point (not in the closed set) to two different real numbers $a$ and $b$. So in a space that is not completely regular, there exist a closed set $H$ and a point $x \notin H$ such that every real-valued continuous function $g$ that can be defined on the space maps $H$ and the point $x$ to the same real number. Thus outside of completely regular spaces, notions that are based on continuous real-valued functions may be difficult to work with.

A space $X$ is said to be pseudocompact if every real-valued continuous function $f$ defined on $X$ is a bounded function (i.e. $f(X)$ is a bounded set in the real line $\mathbb{R}$). Even though the definition does not include complete regularity, an effective discussion of pseudocompactness typically make use of complete regularity. We illustrate this point using a proof of a theorem that gives a characterization of pseudocompactness.

Theorem 1
Let $X$ be a space. The following conditions are equivalent:

1. The space $X$ is pseudocompact.
2. If $\mathcal{V}$ is a locally finite family of non-empty open subsets of $X$, then $\mathcal{V}$ is finite.
3. If $\mathcal{V}$ is a locally finite open cover of $X$, then $\mathcal{V}$ is finite.
4. If $\mathcal{V}$ is a locally finite open cover of $X$, then $\mathcal{V}$ has a finite subcover.

Proof
This theorem is discussed in this discussion of pseudocompactness. We repeat the proof of $1 \Rightarrow 2$ to illustrate an application of complete regularity.

$1 \Rightarrow 2$
Suppose that condition $2$ does not hold. Then there is an infinite locally finite family of non-empty open sets $\mathcal{V}$ such that $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$. We wish to define an unbounded continuous function using $\mathcal{V}$.

This is where we need to invoke the assumption of complete regularity. For each $n$ choose a point $x_n \in V_n$. Then for each $n$, there is a continuous function $f_n:X \rightarrow [0,n]$ such that $f_n(x_n)=n$ and $f_n(X-V_n) \subset \left\{ 0 \right\}$. Define $f:X \rightarrow [0,\infty)$ by $f(x)=f_1(x)+f_2(x)+f_3(x)+\cdots$.

Because $\mathcal{V}$ is locally finite, the function $f$ is essentially pointwise the sum of finitely many $f_n$. In other words, for each $x \in X$, for some positive integer $N$, $f_j(x)=0$ for all $j \ge N$. Thus the function $f$ is well defined and is continuous at each $x \in X$. Note that for each $x_n$, $f(x_n) \ge n$, showing that it is unbounded. $\blacksquare$

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Remark

Complete regularity is an integral part of the above proof. It simplifies the proof and clarifies the argument. As a direct corollary to the Theorem 1, any pseudocompact paracompact space is compact. A less direct corollary is that any pseudocompact metacompact space is compact (see this discussion of pseudocompactness). These results are made possible by assuming that the pseudocompact spaces are also completely regular. They are restated below.

Corollary 2
Let $X$ be a completely regular space. If $X$ is pseudocompact and paracompact, then $X$ is compact.

Corollary 3
Let $X$ be a completely regular space. If $X$ is pseudocompact and metacompact, then $X$ is compact.

It could be a valid math question to ask whether the above two results are valid outside of the class of completely regular space. We do not know the answer. We also feel that it is also a valid approach to just assume complete regularity and focus our attention on exploring the main concepts involved (in this case pseudocompactness, paracompactness and metacompactness).

We would like to remark that in working with pseudocompactness, we also need to take care that we do not assume too much. For example, we do not want to assume normality since any normal pseudocompact space is countably compact (see Theorem 2 in this post). Then working with pseudocompactness is turned into the problem of working with the stronger concept of countably compactness.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# Pseudocompact + Metacompact implies Compact

It is a well known result that any countably compact and metacompact space is compact (see Theorem 5.3.2 in [1]). A discussion of this result is also found in this blog (countably compact + metacompact). Since countably compactness implies pseudocompactness, a natural question arises: can this result be generalized to “any pseudocompact and metacompact space is compact?” The answer is yes and was established in [2] and [3]. In this post, we put together a proof of this result by using building blocks already worked out in this blog.

All spaces considered here are Tychonoff (completely regular). Refer to [1] and [4] for any terms and notions not defined here (or refer to elsewhere in this blog).

A space $X$ is said to be almost compact if for every open cover $\mathcal{U}$ of $X$, there is a finite $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{V}$ is dense in $X$. It can be shown that for any regular space, almost compactness implies compactness. We have the following lemma.

Lemma 1
Let $X$ be a regular space. Then $X$ is compact if and only if $X$ is almost compact.

Theorem 2, Theorem 3 and Theorem 4 are building blocks proved in previous posts. Theorem 5 below is the main theorem. Corollary 6, the intended result, is obtained from applying Theorem 5 and Lemma 1.

Theorem 2 (see Theorem 4B in this post)
Every regular pseudocompact is a Baire space.

Theorem 3 (see Main Theorem in this post)
Let $X$ be a space. The following conditions are equivalent.

1. $X$ is a Baire space.
2. For any point-finite open cover $\mathcal{V}$ of $X$, the set $D=\left\{x \in X: \mathcal{V} \text{ is locally finite at } x \right\}$ is a dense set in $X$.

Theorem 4 (see Theorem 1 in this post)
Let $X$ be a space. The following conditions are equivalent.

1. $X$ is a pseudocompact space.
2. If $\mathcal{W}$ is a locally finite family of non-empty open subsets of $X$, then $\mathcal{W}$ is finite.

Theorem 5
Let $X$ be a pseudocompact and metacompact space. Then $X$ is almost compact.

Proof
Let $\mathcal{U}$ be an open cover of $X$. By metacompactness, there is a $\mathcal{V}$ which is a point-finite open refinement of $\mathcal{U}$. It suffices to find a finite $\mathcal{W} \subset \mathcal{V}$ such that $\mathcal{W}$ covers a dense set.

By Theorem 2, $X$ is a Baire space. By Theorem 3, the set $D$ is dense in $X$ where $D=\left\{x \in X: \mathcal{V} \text{ is locally finite at } x \right\}$. Let $\mathcal{W}$ be the collection of all $V \in \mathcal{V}$ such that $V \cap D \ne \varnothing$. Note that $\bigcup \mathcal{W}$ is open and dense in $X$. Furthermore, it is straightforward to show that $\mathcal{W}$ is locally finite at each point $x \in D$. By Theorem 4, $\mathcal{W}$ is finite. $\blacksquare$

Corollary 6
Let $X$ be a pseudocompact and metacompact space. Then $X$ is compact.

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Scott, B., M., Pseudocompact Metacompact Spaces are Compact, Topology Proc., 4, 577-587, 1979.
3. Watson, W. S., Pseudocompact Metacompact Spaces are Compact, Proc. Amer. Math. Soc., 81, 151-152, 1981.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# A Characterization of Baire Spaces

We present a useful characterization of Baire spaces. A Baire space is a topological space $X$ in which the conclusion of Baire category theorem holds, that is, for each countable family $\left\{U_1,U_2,U_3,\cdots \right\}$ of open and dense subsets of $X$, the intersection $\bigcap \limits_{n=1}^\infty U_n$ is dense in $X$. This definition is equivalent to the statement that every non-empty open subset of $X$ is of second category in $X$. An elementary discussion of Baire spaces is found in this blog post. Baire spaces can also be characterized in terms of the Banach-Mazur game (see Theorem 1 in this post). We add one more characterization in terms of point-finite open cover and locally finite open family (from [2] and [3]). We prove the following theorem.

A collection $\mathcal{S}$ of subsets of a space $X$ is said to be point-finite if every point in the space $X$ belongs to at most finitely many members of $\mathcal{S}$. A collection $\mathcal{S}$ of subsets of $X$ is said to be locally finite at the point $x \in X$ if there is an open set $V \subset X$ such that $x \in V$ and $V$ meets at most finitely many members of $\mathcal{S}$. The collection $\mathcal{S}$ is said to be locally finite in the space $X$ if it is locally finite at every $x \in X$. For any terms and concepts not explicitly defined here, refer to [1] (Engelking) or [4]) (Willard).

Theorem
Let $X$ be a space. The following conditions are equivalent.

1. $X$ is a Baire space.
2. For any point-finite open cover $\mathcal{U}$ of $X$, the set $D=\left\{x \in X: \mathcal{U} \text{ is locally finite at } x \right\}$ is a dense set in $X$.
3. For any countable point-finite open cover $\mathcal{U}$ of $X$, the set $D=\left\{x \in X: \mathcal{U} \text{ is locally finite at } x \right\}$ is a dense set in $X$.

Proof
$1 \Rightarrow 2$
Let $\mathcal{U}$ be a point-finite open cover of $X$. Let $O$ be a non-empty open subset of $X$. We wish to show that $O \cap D \ne \varnothing$ where $D$ is the set defined in condition 2. For each $n$, define

$\displaystyle . \ \ \ \ \ F_n=\left\{x \in O: x \text{ belongs to exactly n members of } \mathcal{U} \right\}$.

Note that $O=\bigcup \limits_{n=1}^\infty F_n$. Since $X$ is a Baire space, $O$ must be of second category in $X$. None of the sets $F_n$ can be a nowhere dense set. Thus for some $n$, $F_n$ has non-empty interior. Choose some non-empty open set $W$ such that $W \subset F_n$.

Pick $y \in W$. Since $y \in F_n$, let $U_1,U_2,\cdots,U_n$ be the $n$ members of $\mathcal{U}$ that contain $y$. Let $V=W \cap \bigcap \limits_{j=1}^n U_n$. Note that $V \subset W \subset F_n \subset O$. Observe that $V$ is a non-empty open set that meets exactly $n$ members of $\mathcal{U}$. Therefore $\mathcal{U}$ is locally finite at points of $V$, leading to the conclusion that $V \subset D$ and $O \cap D \ne \varnothing$.

The direction $2 \Rightarrow 3$ is immediate.

$3 \Rightarrow 1$
Suppose condition 3 holds. We claim that $X$ is a Baire space. Suppose not. Let $U$ be a non-empty open subset of $X$ such that $U=\bigcup \limits_{n=1}^\infty K_n$ where each $K_n$ is nowhere dense in $X$. Let $\mathcal{U}$ be defined as the following:

$\displaystyle . \ \ \ \ \ \mathcal{U}=\left\{X \right\} \cup \left\{U_n: n=1,2,3,\cdots\right\}$,

where $U_n=U - (\overline{K_1} \cup \cdots \cup \overline{K_n})$. Clearly, $\mathcal{U}$ is a point-finite open cover of $X$. By condition 3, $D$ is dense in $X$ ($D$ is defined in condition 3). In particular, $U \cap D \ne \varnothing$. Choose $y \in U \cap D$. Since $\mathcal{U}$ is locally finite at $y$, we can choose some open set $V \subset U$ such that $y \in V$ and such that $V$ meets only finitely many $U_j$, say only up to $U_1,\cdots, U_m$ (so $V \cap U_j = \varnothing$ for all $j > m$).

On the other hand, all sets $K_j$ are nowhere dense. So we can choose some open set $V_0 \subset V$ such that $V_0$ misses the nowhere dense set $\overline{K_1} \cup \cdots \cup \overline{K_m} \cup \overline{K_{m+1}}$. In particular, this means that $V_0 \cap U_{m+1} \ne \varnothing$, contradicting that $V \cap U_j = \varnothing$ for all $j > m$. So $X$ must be a Baire space if condition 3 holds. $\blacksquare$

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Fletcher, P., Lindgren, W. F., A note on spaces of second category, Arch, Math., 24, 186-187, 1973.
3. McCoy, R. A., A Baire space extension, Proc. Amer. Math. Soc., 33, 199-202, 1972.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# Baire Category Theorem and the Finite Intersection Property

A Baire space is a topological space in which the intersection of any countable family of open and dense sets is dense (equivalently every non-empty open subset is of second category). One version of the Baire category theorem states that every complete metric space is a Baire space. Another common version states that every compact Hausdorff space is a Baire space. Another version states that every locally compact Hausdorff space is a Baire space. The commonality among these versions is the finite intersection property (whenever a collection of a certain type of sets satisfies the property that any finite subcollection has non-empty intersection, the whole collection has non-empty intersection). For each of these classes of spaces, in addition to countably compact spaces and pseudocompact spaces, Baire category theorem is derived from having one specific form of the finite intersection property. In this post, we explore this relationship.

In each of the following theorem pairs, the B Theorem follows from the A theorem. The A theorem is a form of the finite intersection property and the B theorem is a version of Baire category theorem.

Another interesting observation is that the finite intersection properties discussed here can give a stronger property than being a Baire space. This stronger property is defined by the Banach-Mazur game.

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Theorem 1A
Let $(X, \rho)$ be a metric space. Then the following conditions are equivalent.

1. $(X, \rho)$ is a complete metric space.
2. For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$ such that the diameters of the sets $C_n$ converge to zero, we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$.

Theorem 1B
Every complete metric space is a Baire space.

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Theorem 2A
Let $X$ be a Hausdorff space. Then the following conditions are equivalent.

1. $X$ is a compact space.
2. For every family $\mathcal{F}$ consisting of non-empty closed subsets of $X$, if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.

Theorem 2B

• Every compact Hausdorff space is a Baire space.
• Every locally compact Hausdorff space is a Baire space.

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Theorem 3A
Let $X$ be a Hausdorff space. Then the following conditions are equivalent.

1. $X$ is a countably compact space.
2. For every countable family $\mathcal{F}$ consisting of non-empty closed subsets of $X$, if $\mathcal{F}$ has the finite intersection property, then $\mathcal{F}$ has non-empty intersection.
3. For each decreasing sequence $C_1 \supset C_2 \supset C_3 \supset \cdots$ of non-empty closed subsets of $X$, we have $\bigcap \limits_{n=1}^\infty C_n \ne \varnothing$.

Theorem 3B
Every countably compact Hausdorff space is a Baire space.

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Theorem 4A
Let $X$ be a regular space. The following conditions are equivalent:

1. The space $X$ is pseudocompact.
2. If $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $O_n \supset O_{n+1}$ for each $n$, then $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$.
3. If $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $\mathcal{V}$ has the finite intersection property, then $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$.

Theorem 4B
Every regular pseudocompact space is a Baire space.

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Remark
Theorem 1A (the Cantor Theorem) can be found in Engelking (page 269 in [1]). Theorem 2A and Theorem 3A can also be found in Engelking (they are also proved in this post). Theorem 4B is also found in Engelking (Theorem 3.10.23 in page 207 of [1]) and is proved this post.

We would like to explicitly point out that between Thoerem 1A and Theorem 2A, none of the two theorems implies the other. For example, even though both complete metric spaces and compact Hausdorff spaces are Baire spaces, complete metric spaces are not necessarily compact and there are compact spaces that are not even metrizable. However, the finite intersection property of Theorem 2A implies that of Theorem 3A, which in turn implies the finite intersection property of Theorem 4A.
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Baire Category Theorem

The proofs of all four B theorems are amazingly similar. It is a matter of exploiting the fact that whenever a decreasing sequence of open sets satisfying the condition that each closure is a subset of the previous open set (and satisfying some other condition), the sequence of open sets has non-empty intersection. For example, for complete metric space, make sure that the closures of the open sets have diameters going to zero. For any reader who is new to this material, it will be very instructive to walk through the arguments of these Baire category theorems. The proof of Theorem 1A can be found this post. We prove Theorem 4B.

Recall that $X$ is a Baire space if $\left\{U_1,U_2,U_3,\cdots \right\}$ is a countable family of open and dense sets in $X$, $\bigcap \limits_{i=1}^\infty U_i$ is dense in $X$, or equivalently every non-empty open subset of $X$ is of second category in $X$. For more background about the concepts of Baire space and category (see [1] or this post).

Proof of Theorem 4B
Let $X$ be a regular pseudocompact space. Let $\left\{U_1,U_2,U_3,\cdots \right\}$ be a countable family of open and dense sets in $X$. Let $O$ be a non-empty open subset of $X$. We show that $O$ has to contain points of $\bigcap \limits_{n=1}^\infty U_n$. We let $O_1=O \cap U_1$. We find open $O_2$ such that $O_2 \subset U_2$ and $\overline{O_2} \subset O_1$ (using regularity). Continue this inductive process, we have for each $n$, an open $O_n$ such that $O_n \subset U_n$ and $\overline{O_n} \subset O_{n-1}$. Then we have a decreasing sequence of open sets $O_n$ as in condition 2 of Theorem 4A. Then we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$. Since $\overline{O_{n+1}} \subset O_n$ for each $n$, we also have $\bigcap \limits_{n=1}^\infty O_n \ne \varnothing$. It is clear that $\bigcap \limits_{n=1}^\infty O_n \subset \bigcap \limits_{n=1}^\infty U_n$. $\blacksquare$

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Banach-Mazur Game

In proving the above versions of Baire category theorem, we can exploit the appropriate version of the finite intersection property – the situation that any nested decreasing sequence of open sets (under some specified conditions) has non-empty intersection. In fact, the finite intersection property offers more than just Baire category theorem; it can endow the space in question a type of completeness property stronger than Baire space. This completeness property is defined using the Banach-Mazur game.

The Banach-Mazur game is a two-person game played on a topological space. Let $X$ be a space. There are two players, $\alpha$ and $\beta$. They take turn choosing nested decreasing nonempty open subsets of $X$ as follows. The player $\beta$ goes first by choosing a nonempty open subset $U_0$ of $X$. The player $\alpha$ then chooses a nonempty open subset $V_0 \subset U_0$. At the nth play where $n \ge 1$, $\beta$ chooses an open set $U_n \subset V_{n-1}$ and $\alpha$ chooses an open set $V_n \subset U_n$. The player $\alpha$ wins if $\bigcap \limits_{n=0}^\infty V_n \ne \varnothing$. Otherwise the player $\beta$ wins. For more detailed discussion of the game, see this post.

One interesting point that we like to make about the finite intersection property ranging from Theorem 1A to Theorem 4A is that the player $\alpha$ can always win the Banach-Mazur game as long as he/she plays the game according to each specific version of the finite intersection. For example, playing the game in a complete metric space, player $\alpha$ always wins as long as he/she makes the diameters of the closures of the open sets going to zero. In a regular pseudocompact space, player $\alpha$ can always win by making the closure of each of his/her open sets a subset of the previous move of other player.

A topological space in which the player $\alpha$ has a winning strategy is said to be a weakly $\alpha$-favorable space. Thus complete metric spaces, compact Hausdorff spaces, locally compact Hausdorff spaces, countably compact Hausdorff spaces, regular pseudocompact spaces are all weakly $\alpha$-favorable.

There is characterization of Baire spaces in terms of the Banach-Mazur game. A space $X$ is a Baire space if and only if the player $\beta$ has no winning strategy in the Banach-Mazur game played on the space $X$ (see theorem 1 in this post). If the player $\alpha$ can always win, then player $\beta$ can never win. In terms of game terminology, if player $\alpha$ has a winning strategy, then the other player (player $\beta$) has no winning strategy. Thus a space is weakly $\alpha$-favorable implies that it is a Baire space. But the implication is not reversible (see example in this post).

So all the spaces discussed from Theorem 1A to Theorem 4A are all weakly $\alpha$-favorable, a property stronger than Baire spaces. These observations are summarized in the following theorems.

Theorem 1C
Every complete metric space is a weakly $\alpha$-favorable space.

Theorem 2C

• Every compact Hausdorff space is a weakly $\alpha$-favorable space.
• Every locally compact Hausdorff space is a weakly $\alpha$-favorable space.

Theorem 3C
Every countably compact Hausdorff space is a weakly $\alpha$-favorable space.

Theorem 4C
Every regular pseudocompact space is a weakly $\alpha$-favorable space.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# A Space with G-delta Diagonal that is not Submetrizable

The property of being submetrizable implies having a $G_\delta$-diagonal. There are several other properties lying between these two properties (see [1]). Before diving into these other properties, it may be helpful to investigate a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The diagonal of a space $X$ is the set $\Delta=\left\{(x,x): x \in X \right\}$, a subset of the square $X \times X$. An interesting property is when the diagonal of a space is a $G_\delta$-set in $X \times X$ (the space is said to have a $G_\delta$-diagonal). Any compact space or a countably compact space with this property must be metrizable (see compact and countably compact space). A space $(X,\tau)$ is said to be submetrizable if there is a topology $\tau^*$ that can be defined on $X$ such that $(X,\tau^*)$ is a metrizable space and $\tau^* \subset \tau$. In other words, a submetrizable space is a space that has a coarser (weaker) metrizable topology. Every submetrizable space has a $G_\delta$-diagonal. Note that when $X$ has a weaker metric topology, the diagonal $\Delta$ is always a $G_\delta$-set in the metric square $X \times X$ (hence in the square in the original topology). The property of having a $G_\delta$-diagonal is strictly weaker than the property of having a weaker metric topology. In this post, we discuss the Mrowka space, which is a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The Mrowka space (also called Psi space) was discussed previously in this blog (see this post). For the sake of completeness, the example is defined here.

First, we define some basic notions. Let $\omega$ be the first infinite ordinal (or more conveniently the set of all nonnegative integers). 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. An 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 $A \in \mathcal{A}$. In other words, if you put one more set into a maximal almost disjoint family, it ceases to be almost disjoint.

A natural question is whether there is an uncountable almost disjoint family of subsets of $\omega$. In fact, there is one whose cardinality is continuum (the cardinality of the real line). To see this, identify $\omega$ with $\mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace$ (the set of all rational numbers). Let $\mathbb{P}=\mathbb{R}-\mathbb{Q}$ be the set of all irrational numbers. For each $x \in \mathbb{P}$, choose a subsequence of $\mathbb{Q}$ consisting of distinct elements that converges to $x$ (in the Euclidean topology). 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 this post).

Let $\mathcal{A}$ be an infinite almost disjoint family of subsets of $\omega$. We now define a Mrowka space (or $\Psi$-space), denoted by $\Psi(\mathcal{A})$. 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 is 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 would like to point out that the definition of a Mrowka space $\Psi(\mathcal{A})$ only requires that the family $\mathcal{A}$ is an almost disjoint family and does not necessarily have to be maximal. For the example discribed in the title, $\mathcal{A}$ needs to be a maximal almost disjoint family of subsets of $\omega$.

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Example
Let $\mathcal{A}$ be a maximal almost disjoint family of subsets of $\omega$. Then $\Psi(\mathcal{A})$ as defined above is a space in which there is a $G_\delta$-diagonal that is not submetrizable.

Note that $\Psi(\mathcal{A})$ is pseudocompact (proved in this post). Because there is no countable maximal almost disjoint family of subsets of $\omega$, $\mathcal{A}$ must be an uncountable in addition to being a closed and discrete subspace of $\Psi(\mathcal{A})$ (thus the space is not Lindelof). Since $\Psi(\mathcal{A})$ is separable and is not Lindelof, $\Psi(\mathcal{A})$ is not metrizable. Any psuedocompact submetrizable space is metrizable (see Theorem 4 in this post). Thus $\Psi(\mathcal{A})$ must not be submetrizable.

On the other hand, any $\Psi$-space $\Psi(\mathcal{A})$ (even if $\mathcal{A}$ is not maximal) is a Moore space. It is well known that any Moore space has a $G_\delta$-diagonal. The remainder of this post has a brief discussion of Moore space.

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Moore Space

A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$ and each open set $U \subset X$ with $x \in U$, there is some $n$ such that any open set in $\mathcal{D}_n$ containing the point $x$ is contained in $U$. A developable space is one that has a development. A Moore space is a regular developable space.

Suppose that $X$ is a Moore space. We show that $X$ has a $G_\delta$-diagonal. That is, we wish to show that $\Delta=\left\{(x,x): x \in X \right\}$ is a $G_\delta$-set in $X \times X$.

Let $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ be a development. For each $n$, let $U_n=\bigcup \lbrace{V \times V:V \in \mathcal{D}_n}\rbrace$. Clearly $\Delta \subset \bigcap_{n<\omega} U_n$. Let $(x,y) \in \bigcap_{n<\omega} U_n$. For each $n$, $(x,y) \in V_n \times V_n$ for some $V_n \in \mathcal{D}_n$. We claim that $x=y$. Suppose that $x \ne y$. By the definition of development, there exists some $m$ such that every open set in $\mathcal{D}_m$ containing the point $x$ has to be a subset of $X-\left\{y \right\}$. Then $V_m \subset X-\left\{y \right\}$, which contradicts $y \in V_m$. Thus we have $\Delta = \bigcap_{n<\omega} U_n$.

The remaining thing to show is that $\Psi(\mathcal{A})$ is a Moore space. For each positive integer $n$, let $F_n=\left\{0,1,\cdots,n-1 \right\}$ and let $F_0=\varnothing$. The development is defined by $\lbrace{\mathcal{E}_n}\rbrace_{n<\omega}$, where for each $n$, $\mathcal{E}_n$ consists of open sets of the form $\lbrace{A}\rbrace \cup (A-F_n)$ where $A \in \mathcal{A}$ plus any singleton $\left\{j \right\}$ ($j \in \omega$) that has not been covered by the sets $\lbrace{A}\rbrace \cup (A-F_n)$.

Reference

1. Arhangel’skii, A. V., Buzyakova, R. Z., The rank of the diagonal and submetrizability, Commentationes Mathematicae Universitatis Carolinae, Vol. 47 (2006), No. 4, 585-597.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.