# A useful embedding for Cp(X)

Let $X$ be a Tychonoff space (also called completely regular space). By $C_p(X)$ we mean the space of all continuous real-valued functions defined on $X$ endowed with the pointwise convergence topology. In this post we discuss a scenario in which a function space can be embedded into another function space. We prove the following theorem. An example follows the proof.

Theorem 1
Suppose that the space $Y$ is a continuous image of the space $X$. Then $C_p(Y)$ can be embedded into $C_p(X)$.

Proof of Theorem 1
Let $t:X \rightarrow Y$ be a continuous surjection, i.e., $t$ is a continuous function from $X$ onto $Y$. Define the map $\psi: C_p(Y) \rightarrow C_p(X)$ by $\psi(f)=f \circ t$ for all $f \in C_p(Y)$. We show that $\psi$ is a homeomorphism from $C_p(Y)$ into $C_p(X)$.

First we show $\psi$ is a one-to-one map. Let $f,g \in C_p(Y)$ with $f \ne g$. There exists some $y \in Y$ such that $f(y) \ne g(y)$. Choose some $x \in X$ such that $t(x)=y$. Then $f \circ t \ne g \circ t$ since $(f \circ t)(x)=f(t(x))=f(y)$ and $(g \circ t)(x)=g(t(x))=g(y)$.

Next we show that $\psi$ is continuous. Let $f \in C_p(Y)$. Let $U$ be open in $C_p(X)$ with $\psi(f) \in U$ such that

$U=\left\{q \in C_p(X): \forall \ i=1,\cdots,n, \ q(x_i) \in U_i \right\}$

where $x_1,\cdots,x_n$ are arbitrary points of $X$ and each $U_i$ is an open interval of the real line $\mathbb{R}$. Note that for each $i$, $f(t(x_i)) \in U_i$. Now consider the open set $V$ defined by:

$V=\left\{r \in C_p(Y): \forall \ i=1,\cdots,n, \ r(t(x_i)) \in U_i \right\}$

Clearly $f \in V$. It follows that $\psi(V) \subset U$ since for each $r \in V$, it is clear that $\psi(r)=r \circ t \in U$.

Now we show that $\psi^{-1}: \psi(C_p(Y)) \rightarrow C_p(Y)$ is continuous. Let $\psi(f)=f \circ t \in \psi(C_p(Y))$ where $f \in C_p(Y)$. Let $G$ be open with $\psi^{-1}(f \circ t)=f \in G$ such that

$G=\left\{r \in C_p(Y): \forall \ i=1,\cdots,m, \ r(y_i) \in G_i \right\}$

where $y_1,\cdots,y_m$ are arbitrary points of $Y$ and each $G_i$ is an open interval of $\mathbb{R}$. Choose $x_1,\cdots,x_m \in X$ such that $t(x_i)=y_i$ for each $i$. We have $f(t(x_i)) \in G_i$ for each $i$. Define the open set $H$ by:

$H=\left\{q \in \psi(C_p(Y)) \subset C_p(X): \forall \ i=1,\cdots,m, \ q(x_i) \in G_i \right\}$

Clearly $f \circ t \in H$. Note that $\psi^{-1}(H) \subset G$. To see this, let $r \circ t \in H$ where $r \in C_p(Y)$. Now $r(t(x_i))=r(y_i) \in G_i$ for each $i$. Thus $\psi^{-1}(r \circ t)=r \in G$. It follows that $\psi^{-1}$ is continuous. The proof of the theorem is now complete. $\blacksquare$

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Example

The proof of Theorem 1 is not difficult. It is a matter of notating carefully the open sets in both function spaces. However, the embedding makes it easy in some cases to understand certain function spaces and in some cases to relate certain function spaces.

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. Furthermore consider these ordinals as topological spaces endowed with the order topology. As an application of Theorem 1, we show that $C_p(\omega_1+1)$ can be embedded as a subspace of $C_p(\omega_1)$. Define a continuous surjection $g:\omega_1 \rightarrow \omega_1+1$ as follows:

$g(\gamma) = \begin{cases} \omega_1 & \mbox{if } \ \gamma =0 \\ \gamma-1 & \mbox{if } \ 1 \le \gamma < \omega \\ \gamma & \mbox{if } \ \omega \le \gamma < \omega_1 \end{cases}$

The map $g$ is continuous from $\omega_1$ onto $\omega_1+1$. By Theorem 1, $C_p(\omega_1+1)$ can be embedded as a subspace of $C_p(\omega_1)$. On the other hand, $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. The function space $C_p(\omega_1+1)$ is a Frechet-Urysohn space, which is a property that is carried over to any subspace. The function $C_p(\omega_1)$ is not Frechet-Urysohn. Thus $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. A further comparison of these two function spaces is found in this subsequent post.

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

# Cp(X) is countably tight when X is compact

Let $X$ be a completely regular space (also called Tychonoff space). If $X$ is a compact space, what can we say about the function space $C_p(X)$, the space of all continuous real-valued functions with the pointwise convergence topology? When $X$ is an uncountable space, $C_p(X)$ is not first countable at every point. This follows from the fact that $C_p(X)$ is a dense subspace of the product space $\mathbb{R}^X$ and that no dense subspace of $\mathbb{R}^X$ can be first countable when $X$ is uncountable. However, when $X$ is compact, $C_p(X)$ does have a convergence property, namely $C_p(X)$ is countably tight.

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Tightness

Let $X$ be a completely regular space. The tightness of $X$, denoted by $t(X)$, is the least infinite cardinal $\kappa$ such that for any $A \subset X$ and for any $x \in X$ with $x \in \overline{A}$, there exists $B \subset A$ for which $\lvert B \lvert \le \kappa$ and $x \in \overline{B}$. When $t(X)=\omega$, we say that $Y$ has countable tightness or is countably tight. When $t(X)>\omega$, we say that $X$ has uncountable tightness or is uncountably tight. Clearly any first countable space is countably tight. There are other convergence properties in between first countability and countable tightness, e.g., the Frechet-Urysohn property. The notion of countable tightness and tightness in general is discussed in further details here.

The fact that $C_p(X)$ is countably tight for any compact $X$ follows from the following theorem.

Theorem 1
Let $X$ be a completely regular space. Then the function space $C_p(X)$ is countably tight if and only if $X^n$ is Lindelof for each $n=1,2,3,\cdots$.

Theorem 1 is the countable case of Theorem I.4.1 on page 33 of [1]. We prove one direction of Theorem 1, the direction that will give us the desired result for $C_p(X)$ where $X$ is compact.

Proof of Theorem 1
The direction $\Longleftarrow$
Suppose that $X^n$ is Lindelof for each positive integer. Let $f \in C_p(X)$ and $f \in \overline{H}$ where $H \subset C_p(X)$. For each positive integer $n$, we define an open cover $\mathcal{U}_n$ of $X^n$.

Let $n$ be a positive integer. Let $t=(x_1,\cdots,x_n) \in X^n$. Since $f \in \overline{H}$, there is an $h_t \in H$ such that $\lvert h_t(x_j)-f(x_j) \lvert <\frac{1}{n}$ for all $j=1,\cdots,n$. Because both $h_t$ and $f$ are continuous, for each $j=1,\cdots,n$, there is an open set $W(x_j) \subset X$ with $x_j \in W(x_j)$ such that $\lvert h_t(y)-f(y) \lvert < \frac{1}{n}$ for all $y \in W(x_j)$. Let the open set $U_t$ be defined by $U_t=W(x_1) \times W(x_2) \times \cdots \times W(x_n)$. Let $\mathcal{U}_n=\left\{U_t: t=(x_1,\cdots,x_n) \in X^n \right\}$.

For each $n$, choose $\mathcal{V}_n \subset \mathcal{U}_n$ be countable such that $\mathcal{V}_n$ is a cover of $X^n$. Let $K_n=\left\{h_t: t \in X^n \text{ such that } U_t \in \mathcal{V}_n \right\}$. Let $K=\bigcup_{n=1}^\infty K_n$. Note that $K$ is countable and $K \subset H$.

We now show that $f \in \overline{K}$. Choose an arbitrary positive integer $n$. Choose arbitrary points $y_1,y_2,\cdots,y_n \in X$. Consider the open set $U$ defined by

$U=\left\{g \in C_p(X): \forall \ j=1,\cdots,n, \lvert g(y_j)-f(y_j) \lvert <\frac{1}{n} \right\}$.

We wish to show that $U \cap K \ne \varnothing$. Choose $U_t \in \mathcal{V}_n$ such that $(y_1,\cdots,y_n) \in U_t$ where $t=(x_1,\cdots,x_n) \in X^n$. Consider the function $h_t$ that goes with $t$. It is clear from the way $h_t$ is chosen that $\lvert h_t(y_j)-f(x_j) \lvert<\frac{1}{n}$ for all $j=1,\cdots,n$. Thus $h_t \in K_n \cap U$, leading to the conclusion that $f \in \overline{K}$. The proof that $C_p(X)$ is countably tight is completed.

The direction $\Longrightarrow$
See Theorem I.4.1 of [1].

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Remarks

As shown above, countably tightness is one convergence property of $C_p(X)$ that is guaranteed when $X$ is compact. In general, it is difficult for $C_p(X)$ to have stronger convergence properties such as the Frechet-Urysohn property. It is well known $C_p(\omega_1+1)$ is Frechet-Urysohn. According to Theorem II.1.2 in [1], for any compact space $X$, $C_p(X)$ is a Frechet-Urysohn space if and only if the compact space $X$ is a scattered space.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2014 - 2015 \text{ by Dan Ma}$

# Working with the function space Cp(X)

This post provides basic information about the space of real-valued continuous functions with the pointwise convergence topology. The goal is to discuss the setting and to define the standard basic open sets in the function space, providing background information for subsequent posts.

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Completely Regular Spaces

The starting point is a completely regular space. 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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Defining the Function Space $C_p(X)$

Let $X$ be a completely regular 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 $\prod_{x \in X} \mathbb{R}=\mathbb{R}^X$. Thus $C(X)$ can be endowed with the subspace topology inherited from the product space $\mathbb{R}^X$. When this is the case, the function space $C(X)$ is denoted by $C_p(X)$. The topology on $C_p(X)$ is called the pointwise convergence topology.

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 $\prod_{x \in X} U_x$ where each $U_x$ is an open subset of $\mathbb{R}$ such that $U_x = \mathbb{R}$ 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:

$C(X) \cap \biggl(\prod_{x \in X} U_x \biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

where each $U_x$ is an open subset of $\mathbb{R}$ 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 simplify notation, the basic open sets as described in (1) can also be notated by:

$\prod_{x \in X} U_x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1a)$

Thus when working with open sets in $C_p(X)$, we take $\prod_{x \in X} U_x$ to mean the set of all $f \in C(X)$ such that $f(x) \in U_x$ for each $x \in X$.

To make the basic open sets of $C_p(X)$ more explicit, (1) or (1a) is translated as follows:

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

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 $f(x) \in O_x$.

There is another description of basic open sets that is useful. Let $f \in C_p(X)$. Let $F \subset X$ be finite. Let $\epsilon>0$. Let $B(f,F,\epsilon)$ be defined as follows:

$B(f,F,\epsilon)=\left\{g \in C(X): \forall \ x \in F, \lvert f(x)-g(x) \lvert< \epsilon \right\} \ \ \ \ \ \ \ (3)$

In proving results about $C_p(X)$, we can use basic open sets that are described in any one of the three forms (1), (2) and (3). If $U$ is a basic open subset of $C_p(X)$, as described in (1) or (1a), we use $supp(U)$ to denote the finite set of $x \in X$ such that $U_x \ne \mathbb{R}$. The set $supp(U)$ is called the support of $U$. The support for the basic open sets as described in (2) and (3) is already explicitly stated.

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Basic Discussion

The theory of $C_p(X)$ is a vast subject area. For a systematic introduction, see [1]. One fundamental theme in function space theory is the study of how properties of $X$ and $C_p(X)$ are related. The domain space $X$ and the function space $C(X)$ are not on the same footing. The domain $X$ only has a topological structure. The function space $C_p(X)$ carries a topology and two natural algebraic operations of addition and multiplication, making it a topological ring. In addition, $C_p(X)$ can be regarded as a topological group, or a linear topological space. In this post and in many subsequent posts, we narrow the focus to the topological properties of $X$ and $C_p(X)$, paying attention to the how the topological properties of $X$ and $C_p(X)$ are related.

In addition to the pointwise convergence topology, there are other topologies that can be defined on $C(X)$, e.g., the compact-open topology, the topology of uniform convergence and others. Both the pointwise convergence topology and the compact-open topology are examples of set-open topologies. In this post and in many of the subsequent posts, the focus is on the pointwise convergence topology, i.e., the subspace topology on $C(X)$ inherited from the product space.

The space $C_p(X)$ automatically inherits certain properties of the product space $\mathbb{R}^X$. Note that $C(X)$ is dense in $\mathbb{R}^X$. The product $\mathbb{R}^X$ has the countable chain condition (CCC) since it is a product of separable spaces. Hence $C_p(X)$ always has the CCC, i.e., there are no uncountably many pairwise disjoint open subsets of $C_p(X)$, regardless what the domain space $X$ is. One consequence of the CCC is that $C_p(X)$ is paracompact if and only if $C_p(X)$ is Lindelof.

It is well known that $\mathbb{R}^X$ is separable if and only if the cardinality of $X$ $\le$ continuum. Since $C(X)$ is dense in $\mathbb{R}^X$, $C_p(X)$ is not separable if the cardinality of $X$ $>$ continuum. Thus $C_p(X)$ is one way to get a CCC space that is not separable. There are non-separable $C_p(X)$ where the cardinality of $X$ $\le$ continuum. Obtaining such $C_p(X)$ would require more than the properties of the product space $\mathbb{R}^X$; using properties of $X$ would be necessary.

The properties of $C_p(X)$ discussed so far are inherited from the product space. Refer to chapter one of [1] for other elementary properties of $C_p(X)$. See this post for a discussion of $C_p(X)$ where $X$ is a separable metric space. See this post about a consequence of normality of $C_p(X)$.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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

# Cartesian Products of Two Paracompact Spaces

In some previous posts we discuss examples surrounding the Michael line showing that the product of a paracompact space and a complete metric space needs not be normal (see “Michael Line Basics”) and that the product of a Lindelof space and a separable metric space need not be normal (see “Bernstein Sets and the Michael Line”). These examples are classic counterexamples demonstrating that both paracompactness and Lindelofness are not preserved by taking two-factor cartesian products even when one of the factors is nice (complete metric space in the first example and separable metric space in the second example). We now show some positive results. Of course, these results require additional conditions on one or both of the factors. We prove the following results.

Result 1

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

Result 2

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Result 3

If $X$ is paracompact and perfectly normal and $Y$ is metrizable, then $X \times Y$ is paracompact and perfectly normal.

Result 4

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

With Results 1 and 2, compact spaces and $\sigma$-compact spaces can be called productively paracompact since the product of each of these spaces with any paracompact space is paracompact. We prove Result 1 and Result 2 below.

Result 3 and Result 4 are proved in another post Cartesian Products of Two Paracompact Spaces – Continued.

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Paracompact Spaces

First, recall some definitions. All spaces are at least regular (to us regular implies Hausdorff). Let $X$ be a space. A collection $\mathcal{A}$ of subsets of $X$ is said to be a cover of $X$ if $X=\bigcup \mathcal{A}$ (in words every point of the space belongs to one set in the collection). Furthermore, $\mathcal{A}$ is an open cover of $X$ is it is a cover of $X$ consisting of open subsets of $X$.

Let $\mathcal{A}$ and $\mathcal{B}$ be covers of the space $X$. The cover $\mathcal{B}$ is said to be a refinement of $\mathcal{A}$ ($\mathcal{B}$ is said to refine $\mathcal{A}$) if for every $B \in \mathcal{B}$, there is some $A \in \mathcal{A}$ such that $B \subset A$. The cover $\mathcal{B}$ is said to be an open refinement of $\mathcal{A}$ if $\mathcal{B}$ refines $\mathcal{A}$ and $\mathcal{B}$ is an open cover.

A collection $\mathcal{A}$ of subsets of $X$ is said to be a locally finite collection if for each point $x \in X$, there is a non-empty open subset $V$ of $X$ such that $x \in V$ and $V$ has non-empty intersection with at most finitely many sets in $\mathcal{A}$. An open cover $\mathcal{A}$ of $X$ is said to have a locally finite open refinement if there exists an open cover $\mathcal{C}$ of $X$ such that $\mathcal{C}$ refines $\mathcal{A}$ and $\mathcal{C}$ is a locally finite collection. We have the following definition.

Definition

The space $X$ is said to be paracompact if every open cover of $X$ has a locally finite open refinement.

A collection $\mathcal{U}$ of subsets of the space $X$ is said to be a $\sigma$-locally finite collection if $\mathcal{U}=\bigcup \limits_{i=1}^\infty \mathcal{U}_i$ such that each $\mathcal{U}_i$ is a locally finite collection of subsets of $X$. Consider the property that every open cover of $X$ has a $\sigma$-locally finite open refinement. This on the surface is a stronger property than paracompactness. However, Theorem 1 below shows that it is actually equivalent to paracompactness. The proof of Theorem 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).

Theorem 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover $\mathcal{U}$ of $X$ has a $\sigma$-locally finite open refinement.

Theorem 2 below is another characterization of paracompactness that is useful. For a proof of Theorem 2, see “Finite and Countable Products of the Michael Line”.

Theorem 2
Let $X$ be a regular space. Then $X$ is paracompact if and only if the following holds:

For each open cover $\left\{U_t: t \in T \right\}$ of $X$, there exists a locally finite open cover $\left\{V_t: t \in T \right\}$ such that $\overline{V_t} \subset U_t$ for each $t \in T$.

Theorem 3 below shows that paracompactness is hereditary with respect to $F_\sigma$-subsets.

Theorem 3
Every $F_\sigma$-subset of a paracompact space is paracompact.

Proof of Theorem 3
Let $X$ be paracompact. Let $Y \subset X$ such that $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is a closed subset of $X$. Let $\mathcal{U}$ be an open cover of $Y$. For each $U \in \mathcal{U}$, let $U^*$ be open in $X$ such that $U^* \cap Y=U$.

For each $i$, let $\mathcal{U}_i^*$ be the set of all $U^*$ such that $U \cap Y_i \ne \varnothing$. Let $\mathcal{V}_i^*$ be a locally finite refinement of $\mathcal{U}_i^* \cup \left\{X-Y_i \right\}$. Let $\mathcal{V}_i$ be the following:

$\mathcal{V}_i=\left\{V \cap Y: V \in \mathcal{V}_i^* \text{ and } V \cap Y_i \ne \varnothing \right\}$

It is clear that each $\mathcal{V}_i$ is a locally finite collection of open set in $Y$ covering $Y_i$. All the $\mathcal{V}_i$ together form a refinement of $\mathcal{U}$. Thus $\mathcal{V}=\bigcup \limits_{i=1}^\infty \mathcal{V}_i$ is a $\sigma$-locally finite open refinement of $\mathcal{U}$. By Theorem 1, the $F_\sigma$-set $Y$ is paracompact. $\blacksquare$
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Result 1

Result 1 is the statement that:

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

To prove Result 1, we use the Tube lemma (for a proof, see “The Tube Lemma”).

The Tube Lemma
Let $X$ be any space and $Y$ be compact. For each $x \in X$ and for each open set $U \subset X \times Y$ such that $\left\{x \right\} \times Y \subset U$, there is an open set $O \subset X$ such that $\left\{x \right\} \times Y \subset O \times Y \subset U$.

Proof of Result 1
Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, choose a finite $\mathcal{U}_x \subset \mathcal{U}$ such that $\mathcal{U}_x$ is a cover of $\left\{x \right\} \times Y$. By the Tube Lemma, for each $x \in X$, there is an open set $O_x \subset X$ such that $\left\{x \right\} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is paracompact, by Theorem 2, let $\left\{W_x: x \in X \right\}$ be a locally finite open refinement of $\left\{O_x: x \in X \right\}$ such that $W_x \subset O_x$ for each $x \in X$.

Let $\mathcal{W}=\left\{(W_x \times Y) \cap U: x \in X, U \in \mathcal{U}_x \right\}$. We claim that $\mathcal{W}$ is a locally finite open refinement of $\mathcal{U}$. First, this is an open cover of $X \times Y$. To see this, let $(a,b) \in X \times Y$. Then $a \in W_x$ for some $x \in X$. Furthermore, $a \in O_x$ and $(a,b) \in \cup \mathcal{U}_x$. Thus, $(a,b) \in (W_x \times Y) \cap U$ for some $U \in \mathcal{U}_x$. Secondly, it is clear that $\mathcal{W}$ is a refinement of the original cover $\mathcal{U}$.

It remains to show that $\mathcal{W}$ is locally finite. To see this, let $(a,b) \in X \times Y$. Then there is an open $V$ in $X$ such that $x \in V$ and $V$ can meets only finitely many $W_x$. Then $V \times Y$ can meet only finitely many sets in $\mathcal{W}$. $\blacksquare$

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Result 2

Result 2 is the statement that:

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Proof of Result 2
Note that the $\sigma$-compact space $Y$ is Lindelof. Since regular Lindelof are normal, $Y$ is normal and is thus completely regular. So we can embed $Y$ into a compact space $K$. For example, we can let $K=\beta Y$, which is the Stone-Cech compactification of $Y$ (see “Embedding Completely Regular Spaces into a Cube”). For our purpose here, any compact space containing $Y$ will do. By Result 1, $X \times K$ is paracompact. Note that $X \times Y$ can be regarded as a subspace of $X \times K$.

Let $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is compact in $Y$. Note that $X \times Y=\bigcup \limits_{i=1}^\infty X \times Y_i$ and each $X \times Y_i$ is a closed subset of $X \times K$. Thus the product $X \times Y$ is an $F_\sigma$-subset of $X \times K$. According to Theorem 3, $F_\sigma$-subsets of any paracompact space is paracompact space. Thus $X \times Y$ is paracompact. $\blacksquare$

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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.

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$\copyright \ \ 2012$

# Stone-Cech Compactifications – Another Two Characterizations

Let $X$ be a completely regular space. Let $\beta X$ be the Stone-Cech compactification of $X$. We present two characterizations of $\beta X$ in addition to three others that are discussed previously. In all, these five characterizations can help us derive many of the basic properties of $\beta X$. We prove the following theorems.

Theorem C4
Let $X$ be a completely regular space. Every two completely separated subsets of $X$ have disjoint closures in $\beta X$.

Theorem U4
The property described in Theorem C4 is unique to $\beta X$. That is, if $\alpha X$ is a compactification of $X$ satisfying the condition that every two completely separated subsets of $X$ have disjoint closures in $\alpha X$, then $\alpha X$ must be $\beta X$.

Theorem C5
Let $X$ be a normal space. Then every two disjoint closed subsets of $X$ have disjoint closures in $\beta X$.

Theorem U5
If $\alpha X$ is a compactification of $X$ satisfying the property that every two disjoint closed subsets of $X$ have disjoint closures in $\alpha X$, then $X$ is normal and $\alpha X$ must be $\beta X$.

The C theorem and U theorem with the same number work as a pair. The C theorem asserts that $\beta X$ has a certain property. The corresponding U theorem asserts that of all the compactifications of $X$, $\beta X$ is the only one with the property in question. Whenever we can show a given compactification does not possess the property described in the C-U theorem pair, we know that that compactification is not $\beta X$ (consequence of the C theorem). Whenever we can show that a given compactification has the property described in the C-U theorem pair, we know that that compactification must be $\beta X$ (a consequence of the U theorem).

Three other sets of characterizations (Theorems C1, U1, C2, U2, C3 and U3) have been established previously. See the links found below.
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Completely Separated Sets

Let $Y$ be a completely regular space. Let $H \subset Y$ and $K \subset Y$. The sets $H$ and $K$ are said to be completely separated in $Y$ if there is a continuous function $f:Y \rightarrow [0,1]$ such that for each $y \in H$, $f(y)=0$ and for each $y \in K$, $f(y)=1$ (this can also be expressed as $f(H) \subset \left\{0 \right\}$ and $f(K) \subset \left\{1 \right\}$). If $H$ and $K$ are completely separated, $\overline{H}$ and $\overline{K}$ are necessarily disjoint closed sets, since $\overline{H} \subset f^{-1}(0)$ and $\overline{K} \subset f^{-1}(1)$.

The Urysohn’s lemma can be stated as: a space is a normal space if and only if every two disjoint closed sets are completely separated. Thus disjoint closed sets are not necessarily completely separated (such sets can be found in non-normal spaces).

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To prove Theorem U4, we need a lemma and a theorem. Most of the work in proving Theorem U4 is carried out in Theorem 2 below.

Lemma 1
Let $Y$ be a compact space. Let $U$ be an open subset of $Y$. Let $\mathcal{C}$ be a collection of compact subsets of $Y$ such that $\cap \mathcal{C} \subset U$. Then there exists a finite collection $\left\{C_1,C_2,\cdots,C_n \right\} \subset \mathcal{C}$ such that $\bigcap \limits_{i=1}^n C_i \subset U$.

Proof of Lemma 1
Let $D=Y-U$, which is compact. Let $\mathcal{O}$ be the collection of all $Y-C$ where $C \in \mathcal{C}$. Note that $\cap \mathcal{C} \subset U$ implies that $D \subset \cup \mathcal{O}$. Thus $\mathcal{O}$ is a collection of open sets covering the compact set $D$. We have $\left\{O_1,O_2,\cdots,O_n \right\} \subset \mathcal{O}$ such that $D \subset \bigcup \limits_{i=1}^n O_i$. Each $O_i=Y-C_i$ for some $C_i \in \mathcal{C}$. Now $\left\{C_1,C_2,\cdots,C_n \right\}$ is the desired finite collection. $\blacksquare$

Theorem 2
Let $T$ be a completely regular space. Let $S$ be a dense subspace of $T$. Let $f:S \rightarrow K$ be a continuous function from $S$ into a compact space $K$. Suppose that every two completely separated subsets of $S$ have disjoint closures in $T$. Then $f$ can be extended to a continuous $F:T \rightarrow K$.

Proof
For each $t \in T$, let $\mathcal{O}(t)$ be the set of all open subsets of $T$ containing $t$. For each $t \in T$, let $\mathcal{W}(t)$ be the set of all $\overline{f(S \cap O)}$ where $O \in \mathcal{O}(t)$. Note that each $\mathcal{W}(t)$ consists of compact subsets of $K$. The theorem is established by proving the following claims.

Claim 1
For each $t \in T$, the collection $\mathcal{W}(t)$ has non-empty intersection.

For any $O_1, O_2, \cdots, O_n \in \mathcal{O}(t)$, we have the following:

$\overline{f(S \cap O_1 \cap O_2 \cap \cdots \cap O_n)} \subset \overline{f(S \cap O_1)} \cap \overline{f(S \cap O_2)} \cap \cdots \cap \overline{f(S \cap O_n)}$

The above shows that $\mathcal{W}(t)$ has the finite intersection property (f. i. p.). It is a well known fact that in a compact space, any collection of sets with f. i. p. has non-empty intersection (see [1] or [2] or see The Finite Intersection Property in Compact Spaces and Countably Compact Spaces in this blog).

Claim 2
For each $t \in T$, $\cap \mathcal{W}(t)$ has only one point.

Let $t \in T$. Suppose that

$\left\{k_1,k_2 \right\} \subset \cap \mathcal{W}(t)$ where $k_1 \ne k_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Then there exist open subsets $U_1$ and $U_2$ of $K$ such that $k_1 \in U_1$, $k_2 \in U_2$ and $\overline{U_1} \cap \overline{U_2} = \varnothing$. Since $K$ is compact, it is a normal space. By the Urysohn’s lemma, there exists a continuous $g:K \rightarrow [0,1]$ such that for each $k \in \overline{U_1}$, $g(k)=0$ and for each $k \in \overline{U_2}$, $g(k)=1$. Then because of the function $g \circ f:S \rightarrow [0,1]$, the sets $f^{-1}(\overline{U_1})$ and $f^{-1}(\overline{U_2})$ are completely separated sets in $S$. By assumption, these two sets have disjoint closures in $T$, i.e.,

$\text{ }$
$\overline{f^{-1}(\overline{U_1})} \cap \overline{f^{-1}(\overline{U_2})} = \varnothing \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
$\text{ }$

The point $t$ cannot be in both of the sets in $(2)$. Assume the following:

$\text{ }$
$t \notin \overline{f^{-1}(\overline{U_1})} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$
$\text{ }$

Then $H=T- \overline{f^{-1}(\overline{U_1})} \in \mathcal{O}(t)$. Note that $S \cap H=S-\overline{f^{-1}(\overline{U_1})}$. Furthermore, $\overline{f(S-\overline{f^{-1}(\overline{U_1})})} \in \mathcal{W}(t)$. Thus we have:

$\text{ }$
$k_1 \in \cap \mathcal{W}(t) \subset \overline{f(S-\overline{f^{-1}(\overline{U_1})})}=W$
$\text{ }$

Since $k_1 \in W$ and $U_1$ is an open set containing $k_1$, $U_1$ contains at least one point of $f(S-\overline{f^{-1}(\overline{U_1})})$. Choose $z \in U_1$ such that $z \in f(S-\overline{f^{-1}(\overline{U_1})})$. Now choose $a \in S-\overline{f^{-1}(\overline{U_1})}$ such that $f(a)=z$. First we have $a \notin \overline{f^{-1}(\overline{U_1})}$ and thus $a \notin f^{-1}(\overline{U_1})$. Secondly since $f(a)=z \in U_1$, we have $a \in f^{-1}(U_1) \subset f^{-1}(\overline{U_1})$. We now have $a \notin f^{-1}(\overline{U_1})$ and $a \in f^{-1}(\overline{U_1})$, a contradiction. If we assume $t \notin \overline{f^{-1}(\overline{U_2})}$, we can also derive a contradiction in a similar derivation. Thus the assumption in $(1)$ above is faulty. The intersection $\cap \mathcal{W}(t)$ can only have one point.

Claim 3
For each $t \in S$, $\cap \mathcal{W}(t) =\left\{f(t) \right\}$.

Let $t \in S$. Suppose that $\cap \mathcal{W}(t) =\left\{p \right\}$ where $p \ne f(t)$. the rest of the proof for Claim 3 is similar to that of Claim 2. For the sake of completeness, we give a sketch.

There exist open subsets $U_1$ and $U_2$ of $K$ such that $p \in U_1$, $f(t) \in U_2$ and $\overline{U_1} \cap \overline{U_2} = \varnothing$. By the same argument as in Claim 2, we have the condition $(2)$, i.e., $\overline{f^{-1}(\overline{U_1})} \cap \overline{f^{-1}(\overline{U_2})} = \varnothing$. Since $t \in f^{-1}(U_2)$, $t \notin \overline{f^{-1}(\overline{U_1})}$. The remainder of the proof of Claim 3 is the same as above starting with condition $(3)$ with $p=k_1$. A contradiction will be obtained. We can conclude that the assumption that $\cap \mathcal{W}(t) =\left\{p \right\}$ where $p \ne f(t)$ must be faulty. Thus Claim 3 is established.

Claim 4
For each $t \in T$, define $F:T \rightarrow K$ by letting $F(t)$ be the point in $\cap \mathcal{W}(t)$. Note that this function extends $f$. Furthermore, the map $F:T \rightarrow K$ is continuous.

To show $F$ is continuous, let $t \in T$ and let $F(t) \in E$ where $E$ is open in $K$. The collection $\mathcal{W}(t)$ is a collection of compact subsets of $K$ such that $\left\{F(t) \right\} =\cap \mathcal{W}(t) \subset E$. By Lemma 1, there exists $\left\{C_1,\cdots,C_n \right\} \subset \mathcal{W}(t)$ such that $\bigcap \limits_{i=1}^n C_i \subset E$. By the definition of $\mathcal{W}(t)$, there exists $\left\{O_1,O_2,\cdots,O_n \right\} \subset \mathcal{O}(t)$ such that each $C_i=\overline{f(S \cap O_i)}$. Let $O=O_1 \cap O_2 \cap \cdots \cap O_n$. We have:

$\text{ }$
$\overline{f(S \cap O)} \subset \bigcap \limits_{i=1}^n \overline{f(S \cap O_i)} \subset E \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$
$\text{ }$

Note that $O$ is an open subset of $T$ and $t \in O$. We show that $F(O) \subset E$. Pick $a \in O$. According to the definition of $\mathcal{W}(a)$, we have $\left\{F(a) \right\}=\bigcap \limits_{U \in \mathcal{O}(a)} \overline{f(S \cap U)}$. Since $O \in \mathcal{O}(a)$, we have $F(a) \in \overline{f(S \cap O)}$. Thus by $(4)$, we have $F(a) \in E$. Thus Claim 4 is established.

With all the above claims established, we completed the proof of Theorem 2. $\blacksquare$

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Theorem C4 and Theorem U4

Proof of Theorem C4
In proving C4, we use Theorem C3, which is found in C*-Embedding Property and Stone-Cech Compactification.

Let $E$ and $F$ be two completely separated sets in $X$. Then there exists some continuous $g:X \rightarrow [0,1]$ such that for each $x \in E$, $g(x)=0$ and for each $x \in F$, $g(x)=1$. By Theorem C3, $g$ is extended by some continuous $G:\beta X \rightarrow [0,1]$. The sets $G^{-1}(0)$ and $G^{-1}(1)$ are disjoint closed sets in $\beta X$. Furthermore, $E \subset G^{-1}(0)$ and $F \subset G^{-1}(1)$. Thus $E$ and $F$ have disjoint closures in $\beta X$. $\blacksquare$

Proof of Theorem U4
In proving U4, we use Theorem U1, which is stated and proved in Two Characterizations of Stone-Cech Compactification.

Suppose that $\alpha X$ is a compactification of $X$ satisfying the condition that every two completely separated subsets of $X$ have disjoint closures in $\alpha X$. Let $g:X \rightarrow Y$ be a continuous function from $X$ into a compact space $Y$. By Theorem 2, $g$ can be extended by a continuous $G:\alpha X \rightarrow Y$. By Theorem U1, $\alpha X$ must be $\beta X$. $\blacksquare$

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Theorem C5 and Theorem U5

Proof of Theorem C5
Let $X$ be a normal space. According to the Urysohn’s lemma, every two disjoint closed sets are completely separated. Thus by Theorem C4, every two disjoint closed subsets of $X$ have disjoint closures in $\beta X$. $\blacksquare$

Proof of Theorem U5
Suppose that $\alpha X$ is a compactification of $X$ satisfying the property that every two disjoint closed subsets of $X$ have disjoint closures in $\alpha X$. To show that $X$ is normal, let $H$ and $K$ be disjoint closed subsets of $X$. By assumption about $\alpha X$, $\overline{H}$ and $\overline{K}$ (closures in $\alpha X$) are disjoint. Since $\alpha X$ are compact and Hausdorff, $\alpha X$ is normal. Then $\overline{H}$ and $\overline{K}$ can be separated by disjoint open subsets $U$ and $V$ of $\alpha X$. Thus $U \cap X$ and $V \cap X$ are disjoint open subsets of $X$ separating $H$ and $K$.

We use Theorem U4 to prove Theorem U5. We show that $\alpha X$ satisfies Theorem U4. To this end, let $E$ and $F$ be two completely separated sets in $X$. We show that $E$ and $F$ have disjoint closures in $\alpha X$. There exists some continuous $f:X \rightarrow [0,1]$ such that for each $x \in E$, $f(x)=0$ and for each $x \in F$, $f(x)=1$. Then $f^{-1}(0)$ and $f^{-1}(1)$ are disjoint closed sets in $X$ such that $E \subset f^{-1}(0)$ and $F \subset f^{-1}(1)$. By assumption about $\alpha X$, $f^{-1}(0)$ and $f^{-1}(1)$ have disjoint closures in $\alpha X$. This implies that $E$ and $F$ have disjoint closures in $\alpha X$. Then by Theorem U4, $\alpha X$ must be $\beta X$. $\blacksquare$

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Blog Posts on Stone-Cech Compactification

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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.

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$\copyright \ \ 2012$

# Stone-Cech Compactification of the Integers – Basic Facts

This is another post Stone-Cech compactification. The links for other posts on Stone-Cech compactification can be found below. In this post, we prove a few basic facts about $\beta \omega$, the Stone-Cech compactification of the discrete space of the non-negative integers, $\omega=\left\{0,1,2,3,\cdots \right\}$. We use several characterizations of Stone-Cech compactification to find out what $\beta \omega$ is like. These characterizations are proved in the blog posts listed below. Let $c$ denote the cardinality of the real line $\mathbb{R}$. We prove the following facts.

1. The cardinality of $\beta \omega$ is $2^c$.
2. The weight of $\beta \omega$ is $c$.
3. The space $\beta \omega$ is zero-dimensional.
4. Every infinite closed subset of $\beta \omega$ contains a topological copy of $\beta \omega$.
5. The space $\beta \omega$ contains no non-trivial convergent sequence.
6. No point of $\beta \omega-\omega$ is an isolated point.
7. The space $\beta \omega$ fails to have many properties involving the existence of non-trivial convergent sequence. For example:
$\text{ }$

1. The space $\beta \omega$ is not first countable at each point of the remainder $\beta \omega-\omega$.
2. The space $\beta \omega$ is not a Frechet space.
3. The space $\beta \omega$ is not a sequential space.
4. The space $\beta \omega$ is not sequentially compact.

$\text{ }$

8. No point of the remainder $\beta \omega-\omega$ is a $G_\delta$-point.
9. The remainder $\beta \omega-\omega$ does not have the countable chain condition. In fact, it has a disjoint open collection of cardinality $c$.

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Characterization Theorems

For any completely regular space $X$, let $C(X,I)$ be the set of all continuous functions from $X$ into $I=[0,1]$. The Stone-Cech compactification $\beta X$ is the subspace of the product space $[0,1]^{C(X,I)}$ which is the closure of the image of $X$ under the evaluation map $\beta:X \rightarrow [0,1]^{C(X,I)}$ (for the details, see Embedding Completely Regular Spaces into a Cube).

The brief sketch of $\beta \omega$ we present here is not based on the definition using the evaluation map. Instead we reply on some characterization theorems that are stated here (especially Theorem U3.1). These theorems uniquely describe the Stone-Cech compactification $\beta X$ of a given completely regular space $X$. For example, $\beta X$ satisfies the function extension property in Theorem C3 below. Furthermore any compactification $\alpha X$ of $X$ that satisfies the same property must be $\beta X$ (Theorem U3.1). So a “C” theorem tells us a property possessed by $\beta X$. The corresponding “U” theorem tells us that there is only one compactification (up to equivalence) that has this property.

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Theorem C1
Let $X$ be a completely regular space. Let $f:X \rightarrow Y$ be a continuous function from $X$ into a compact Hausdorff space $Y$. Then there is a continuous $F: \beta X \rightarrow Y$ such that $F \circ \beta=f$.

$\text{ }$

Theorem U1
If $K$ is any compactification of $X$ that satisfies condition in Theorem C1, then $K$ must be equivalent to $\beta X$.

See Two Characterizations of Stone-Cech Compactification.
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$\text{ }$

Theorem C2
Let $X$ be a completely regular space. Among all compactifications of the space $X$, the Stone-Cech compactification $\beta X$ of the space $X$ is maximal with respect to the partial order $\le$.

$\text{ }$

Theorem U2
The property in Theorem C2 is unique to $\beta X$. That is, if, among all compactifications of the space $X$, $\alpha X$ is maximal with respect to the partial order $\le$, then $\alpha X \approx \beta X$.

See Two Characterizations of Stone-Cech Compactification.
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$\text{ }$

Theorem C3
Let $X$ be a completely regular space. The space $X$ is $C^*$-embedded in its Stone-Cech compactification $\beta X$.

$\text{ }$

Theorem U3.1
Let $X$ be a completely regular space. Let $I=[0,1]$. Let $\alpha X$ be a compactification of $X$ such that each continuous $f:X \rightarrow I$ can be extended to a continuous $\hat{f}:\alpha X \rightarrow I$. Then $\alpha X$ must be $\beta X$.

$\text{ }$

Theorem U3.2
If $\alpha X$ is any compactification of $X$ that satisfies the property in Theorem C3 (i.e., $X$ is $C^*$-embedded in $\alpha X$), then $\alpha X$ must be $\beta X$.

See C*-Embedding Property and Stone-Cech Compactification.
_______________________________________________________________________________________
$\text{ }$

The following discussion illustrates how we can use some of these characterizations theorem to obtain information about $\beta X$ and $\beta \omega$ in particular.

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Result 1 and Result 2

According to the previous post (Stone-Cech Compactification is Maximal), we have for any completely regular space $X$, $\lvert \beta X \lvert \le 2^{2^{d(X)}}$ where $d(X)$ is the density (the smallest cardinality of a dense set in $X$). With $\omega$ being a countable space, $\lvert \beta \omega \lvert \le 2^{2^{\omega}}=2^c$.

Result 1 is established if we have $2^c \le \lvert \beta \omega \lvert$. Consider the cube $I^I$ where $I$ is the unit interval $I=[0,1]$. Since the product space of $c$ many separable space is separable (see Product of Separable Spaces), $I^I$ is separable. Let $S \subset I^I$ be a countable dense set. Let $f:\omega \rightarrow S$ be a bijection. Clearly $f$ is a continuous function from the discrete space $\omega$ into $I^I$. By Theorem C1, $f$ is extended by a continuous $F:\beta \omega \rightarrow I^I$. Note that the image $F(\beta \omega)$ is dense in $I^I$ since $F(\beta \omega)$ contains the dense set $S$. On the other hand, $F(\beta \omega)$ is compact. So $F(\beta \omega)=I^I$. Thus $F$ is a surjection. The cardinality of $I^I$ is $2^c$. Thus we have $2^c \le \lvert \beta \omega \lvert$.

From the same previous post (Stone-Cech Compactification is Maximal), it is shown that $w(\beta X) \le 2^{d(X)}$. Thus $w(\beta \omega) \le 2^{\omega}=c$. The same function $F:\beta \omega \rightarrow I^I$ in the above paragraph shows that $c \le w(\beta \omega)$ (see Lemma 2 in Stone-Cech Compactification is Maximal). Thus we have $w(\beta \omega)=c$ $\blacksquare$

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Result 3

A space is said to be zero-dimensional whenever it has a base consisting of open and closed sets. The proof that $\beta X$ is zero-dimensional comes after the following lemmas and theorems.

Theorem 1
Let $X$ be a normal space. If $H$ and $K$ are disjoint closed subsets of $X$, then $H$ and $K$ have disjoint closures in $\beta X$.

Proof of Theorem 1
Let $H$ and $K$ be disjoint closed subsets of $X$. By the normality of $X$ and by the Urysohn’s lemma, there is a continuous function $g:X \rightarrow [0,1]$ such that $g(H) \subset \left\{0 \right\}$ and $g(K) \subset \left\{1 \right\}$. By Theorem C3.1, $g$ can be extended by $G:\beta X \rightarrow [0,1]$. Note that $\overline{H} \subset G^{-1}(0)$ and $\overline{K} \subset G^{-1}(1)$. Thus $\overline{H} \cap \overline{K} = \varnothing$. $\blacksquare$

Theorem 2
Let $X$ be a completely regular space. Let $H$ be a closed and open subset of $X$. Then $\overline{H}$ (the closure of $H$ in $\beta X$) is also a closed and open set in $\beta X$.

Proof of Theorem 2
Let $H$ be a closed and open subset of $X$. Let $K=X-H$. Define $\gamma:X \rightarrow [0,1]$ by letting $\gamma(x)=0$ for all $x \in H$ and $\gamma(x)=1$ for all $x \in K$. Since both $H$ and $K$ are closed and open, the map $\gamma$ is continuous. By Theorem C3, $\gamma$ is extended by some continuous $\Gamma:\beta X \rightarrow [0,1]$. Note that $\overline{H} \subset \Gamma^{-1}(0)$ and $\overline{K} \subset \Gamma^{-1}(1)$. Thus $H$ and $K$ have disjoint closures in $\beta X$, i.e. $\overline{H} \cap \overline{K} = \varnothing$. Both $H$ and $K$ are closed and open in $\beta X$ since $\beta X=\overline{H} \cup \overline{K}$. $\blacksquare$

Lemma 3
For every $A \subset \omega$, $\overline{A}$ (the closure of $A$ in $\beta \omega$) is both closed and open in $\beta \omega$.

Note that Lemma 3 is a corollary of Theorem 2.

Lemma 4
Let $O \subset \beta \omega$ be a set that is both closed and open in $\beta \omega$. Then $O=\overline{A}$ where $A= O \cap \omega$.

Proof of Lemma 4
Let $A=O \cap \omega$. Either $O \subset \omega$ or $O \cap (\beta \omega-\omega) \ne \varnothing$. Thus $A \ne \varnothing$. We claim that $O=\overline{A}$. Since $A \subset O$, it follows that $\overline{A} \subset \overline{O}=O$. To show $O \subset \overline{A}$, pick $x \in O$. If $x \in \omega$, then $x \in A$. So focus on the case that $x \notin \omega$. It is clear that $x \notin \overline{B}$ where $B=\omega -A$. But every open set containing $x$ must contain some points of $\omega$. These points of $\omega$ must be points of $A$. Thus we have $x \in \overline{A}$. $\blacksquare$

Proof of Result 3
Let $\mathcal{A}$ be the set of all closed and open sets in $\beta \omega$. Let $\mathcal{B}=\left\{\overline{A}: A \subset \omega \right\}$. Lemma 3 shows that $\mathcal{B} \subset \mathcal{A}$. Lemma 4 shows that $\mathcal{A} \subset \mathcal{B}$. Thus $\mathcal{A}= \mathcal{B}$. We claim that $\mathcal{B}$ is a base for $\beta \omega$. To this end, we show that for each open $O \subset \beta \omega$ and for each $x \in O$, we can find $\overline{A} \in \mathcal{B}$ with $x \in \overline{A} \subset O$. Let $O$ be open and let $x \in O$. Since $\beta \omega$ is a regular space, we can find open set $V \subset \beta \omega$ with $x \in V \subset \overline{V} \subset O$. Let $A=V \cap \omega$.

We claim that $x \in \overline{A}$. Suppose $x \notin \overline{A}$. There exists open $U \subset V$ such that $x \in U$ and $U$ misses $\overline{A}$. But $U$ must meets some points of $\omega$, say, $y \in U \cap \omega$. Then $y \in V \cap \omega=A$, which is a contradiction. So we have $x \in \overline{A}$.

It is now clear that $x \in \overline{A} \subset \overline{V} \subset O$. Thus $\beta \omega$ is zero-dimensional since $\mathcal{B}$ is a base consisting of closed and open sets. $\blacksquare$

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Result 4 and Result 5

Result 5 is a corollary of Result 4. We first prove two lemmas before proving Result 4.

Lemma 5
For each infinite $A \subset \omega$, $\overline{A}$ (the closure of $A$ in $\beta \omega$) is a homeomorphic copy of $\beta \omega$ and thus has cardinality $2^c$.

Proof of Lemma 5
Let $A \subset \omega$. Let $g:A \rightarrow [0,1]$ be any function (necessarily continuous). Let $f:\omega \rightarrow [0,1]$ be defined by $f(x)=g(x)$ for all $x \in A$ and $f(x)=0$ for all $x \in \omega-A$. By Theorem C3, $f$ can be extended by $F:\beta \omega \rightarrow [0,1]$. Let $G=F \upharpoonright \overline{A}$.

Note that the function $G: \overline{A} \rightarrow [0,1]$ extends $g:A \rightarrow [0,1]$. Thus by Theorem U3.1, $\overline{A}$ must be $\beta A$. Since $A$ is a countably infinite discrete space, $\beta A$ must be equivalent to $\beta \omega$. $\blacksquare$

Lemma 6
For each countably infinite $A \subset \beta \omega-\omega$ such that $A$ is relatively discrete, $\overline{A}$ (the closure of $A$ in $\beta \omega$) is a homeomorphic copy of $\beta \omega$ and thus has cardinality $2^c$.

Proof of Lemma 6
Let $A=\left\{t_1,t_2,t_3,\cdots \right\} \subset \beta \omega -\omega$ such that $A$ is discrete in the relative topology inherited from $\beta \omega$. There exist disjoint open sets $G_1,G_2,G_3,\cdots$ (open in $\beta \omega$) such that for each $j$, $t_j \in G_j$. Since $\beta \omega$ is zero-dimensional (Result 3), $G_1,G_2,G_3,\cdots$ can be made closed and open.

Let $f:A \rightarrow [0,1]$ be a continuous function. We show that $f$ can be extended by $F:\overline{A} \rightarrow [0,1]$. Once this is shown, by Theorem U3.1, $\overline{A}$ must be $\beta A$. Since $A$ is a countable discrete space, $\beta A$ must be equivalent to $\beta \omega$.

We first define $w:\omega \rightarrow [0,1]$ by:

$\displaystyle w(n)=\left\{\begin{matrix}f(t_j)& \exists \ j \text{ such that } n \in \omega \cap G_j\\{0}&\text{otherwise} \end{matrix}\right.$

The function $w$ is well defined since each $n \in \omega$ is in at most one $G_j$. By Theorem C3, the function $w$ is extended by some continuous $W:\beta \omega \rightarrow [0,1]$. By Lemma 4, for each $j$, $G_j=\overline{\omega \cap G_j}$. Thus, for each $j$, $t_j \in \overline{\omega \cap G_j}$. Note that $W$ is a constant function on the set $\omega \cap G_j$ (mapping to the constant value of $f(t_j)$). Thus $W(t_j)=f(t_j)$ for each $j$. So let $F=W \upharpoonright \overline{A}$. Thus $F$ is the desired function that extends $f$. $\blacksquare$

Proof of Result 4
Let $C \subset \beta \omega$ be an infinite closed set. Either $C \cap \omega$ is infinite or $C \cap (\beta \omega-\omega)$ is infinite. If $C \cap \omega$ is infinite, then by Lemma 5, $\overline{C \cap \omega}$ is a homeomorphic copy of $\beta \omega$. Now focus on the case that $C_0=C \cap (\beta \omega-\omega)$ is infinite. We can choose inductively a countably infinite set $A \subset C_0$ such that $A$ is relatively discrete. Then by Lemma 6 $\overline{A}$ is a copy of $\beta \omega$ that is a subset of $C$. $\blacksquare$

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Result 6

We prove that no point in the remainder $\beta \omega-\omega$ is an isolated point. To see this, pick $x \in \beta \omega-\omega$ and pick an arbitrary closed and open set $O \subset \beta \omega$ with $x \in O$. Let $V=O \cap (\beta \omega-\omega)$ (thus an arbitrary open set in the remainder containing $x$). By Lemma 4, $O=\overline{A}$ where $A=O \cap \omega$. According to Lemma 5, $O=\overline{A}$ is a copy of $\beta \omega$ and thus has cardinality $2^c$. The set $V$ is $O$ minus a subset of $\omega$. Thus $V$ must contains $2^c$ many points. This means that $\left\{ x \right\}$ can never be open in the remainder $\beta \omega-\omega$. In fact, we just prove that any open and closed subset of $\beta \omega-\omega$ (thus any open subset) must have cardinality at least $2^c$. $\blacksquare$

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Result 7

The results under Result 7 are corollary of Result 5 (there is no non-trivial convergent sequence in $\beta \omega$). To see Result 7.1, note that every point $x$ in the remainder is not an isolated point and hence cannot have a countable local base (otherwise there would be a non-trivial convergent sequence converging to $x$).

A space $Y$ is said to be a Frechet space if $A \subset Y$ and for each $x \in \overline{A}$, there is a sequence $\left\{ x_n \right\}$ of points of $A$ such that $x_n \rightarrow x$. A set $A \subset Y$ is said to be sequentially closed in $Y$ if for any sequence $\left\{ x_n \right\}$ of points of $A$, $x_n \rightarrow x$ implies $x \in A$. A space $Y$ is said to be a sequential space if $A \subset Y$ is a closed set if and only if $A$ is a sequentially closed set. If a space is Frechet, then it is sequential. It is clear that $\beta \omega$ is not a sequential space.

A space is said to be sequentially compact if every sequence of points in this space has a convergent subsequence. Even though $\beta \omega$ is compact, it cannot be sequentially compact.

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Result 8

Result 7.1 indicates that no point of remainder $\beta \omega-\omega$ can have a countable local base. In fact, no point of the remainder can be a $G_\delta$-point (a point that is the intersection of countably many open sets). The remainder $\beta \omega-\omega$ is a compact space (being a closed subset of $\beta \omega$). In a compact space, if a point is a $G_\delta$-point, then there is a countable local base at that point (see 3.1.F (a) on page 135 of [1] or 17F.7 on page 125 of [2]). $\blacksquare$

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Result 9

The space $\beta \omega$ is a separable space since $\omega$ is a dense set. Thus $\beta \omega$ has the countable chain condition. However, the remainder $\beta \omega-\omega$ does not have the countable chain condition. We show that there is a disjoint collection of $c$ many open sets in $\beta \omega-\omega$.

There is a family $\mathcal{A}$ of infinite subsets of $\omega$ such that for every $A,B \in \mathcal{A}$ with $A \ne B$, $A \cap B$ is finite. Such a collection of sets is said to be an almost disjoint family. There is even an almost disjoint family of cardinality $c$ (see A Space with G-delta Diagonal that is not Submetrizable). Let $\mathcal{A}$ be such a almost disjoint family.

For each $A \in \mathcal{A}$, let $U_A=\overline{A}$ and $V_A=\overline{A} \cap (\beta \omega -\omega)$. By Lemma 3, each $U_A$ is a closed and open set in $\beta \omega$. Thus each $V_A$ is a closed and open set in the remainder $\beta \omega-\omega$. Note that $\left\{V_A: A \in \mathcal{A} \right\}$ is a disjoint collection of open sets in $\beta \omega-\omega$. $\blacksquare$

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Blog Posts on Stone-Cech Compactification

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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.

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