# Stone-Cech Compactification is Maximal

Let $X$ be a completely regular space. Let $\beta X$ be the Stone-Cech compactification of $X$. In a previous post, we show that among all compactifcations of $X$, the Stone-Cech compactification $\beta X$ is maximal with respect to a partial order $\le$ (see Theorem C2 in Two Characterizations of Stone-Cech Compactification). As a result of the maximality, $\beta X$ is the largest among all compactifications of $X$ both in terms of cardinality and weight. We also establish an upper bound for the cardinality of $\beta X$ and an upper bound for the weight of $\beta X$. As a result, we have upper bounds for cardinalities and weights for all compactifications of $X$. We prove the following points.

Upper Bounds for Stone-Cech Compactification

1. $\lvert \beta X \lvert \le 2^{2^{d(X)}}$.
2. $w(\beta X) \le 2^{d(X)}$.
3. Stone-Cech Compactification is Maximal

4. For every compactification $\alpha X$ of the space $X$, $\lvert \alpha X \lvert \le \lvert \beta X \lvert$.
5. For every compactification $\alpha X$ of the space $X$, $w(\alpha X) \le w(\beta X)$.
6. Upper Bounds for all Compactifications

7. For every compactification $\alpha X$ of the space $X$, $w(\alpha X) \le 2^{d(X)}$.
8. For every compactification $\alpha X$ of the space $X$, $\lvert \alpha X \lvert \le 2^{2^{d(X)}}$.

It is clear that Results 5 and 6 follow from the preceding results. The links for other posts on Stone-Cech compactification can be found toward the end of this post

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Some Cardinal Functions

Let $X$ be a space. The density of $X$ is denoted by $d(X)$ and is defined to be the smallest cardinality of a dense set in $X$. For example, if $X$ is separable, then $d(X)=\omega$. The weight of the space $X$ is denoted by $w(X)$ and is defined to be the smallest cardinality of a base of the space $X$. For example, if $X$ is second countable (i.e. having a countable space), then $w(X)=\omega$. Both $d(X)$ and $w(X)$ are cardinal functions that are commonly used in topological discussion. Most authors require that cardinal functions only take on infinite cardinals. We also adopt this convention here. We use $c$ to denote the cardinality of the continuum (the cardinality of the real line $\mathbb{R}$).

If $\mathcal{K}$ is a cardinal number, then $2^{\mathcal{K}}$ refers to the cardinal number that is the cardinallity of the set of all functions from $\mathcal{K}$ to $2=\left\{0,1 \right\}$. Equivalently, $2^{\mathcal{K}}$ is also the cardinality of the power set of $\mathcal{K}$ (i.e. the set of all subsets of $\mathcal{K}$). If $\mathcal{K}=\omega$ (the first infinite ordinal), then $2^\omega=c$ is the cardinality of the continuum.

If $X$ is separable, then $d(X)=\omega$ (as noted above) and we have $2^{d(X)}=c$ and $2^{2^{d(X)}}=2^c$. Result 5 and Result 6 imply that $2^c$ is an upper bound for the cardinality of all compactifications of any separable space $X$ and $c$ is an upper bound of the weight of all compactifications of any separable space $X$.

In general, Result 5 and Result 6 indicate that the density of $X$ bounds the cardinality of any compactification of $X$ by two exponents and the density of $X$ bounds the weight of any compactification of $X$ by one exponent.

Another cardinal function related to weight is that of the network weight. A collection $\mathcal{N}$ of subsets of the space $X$ is said to be a network for $X$ if for each point $x \in X$ and for each open subset $U$ of $X$ with $x \in U$, there is some set $A \in \mathcal{N}$ with $x \in A \subset U$. Note that sets in a network do not have to be open. However, any base for a topology is a network. The network weight of the space $X$ is denoted by $nw(X)$ and is defined to be the least cardinality of a network for $X$. Since any base is a network, we have $nw(X) \le w(X)$. It is also clear that $nw(X) \le \lvert X \lvert$ for any space $X$. Our interest in network and network weight is to facilitate the discussion of Lemma 2 below. It is a well known fact that in a compact space, the weight and the network weight are the same (see Result 5 in Spaces With Countable Network).
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Some Basic Facts

We need the following two basic results.

Lemma 1
Let $X$ be a space. Let $C(X)$ be the set of all continuous functions $f:X \rightarrow \mathbb{R}$. Then $\lvert C(X) \lvert \le 2^{d(X)}$.

Lemma 2
Let $S$ be a space and let $T$ be a compact space. Suppose that $T$ is the continuous image of $S$. Then $w(T) \le w(S)$.

Proof of Lemma 1
Let $A \subset X$ be a dense set with $\lvert A \lvert=2^{d(X)}$. Let $\mathbb{R}^A$ be the set of all functions from $A$ to $\mathbb{R}$. Consider the map $W:C(X) \rightarrow \mathbb{R}^A$ by $W(f)= f \upharpoonright A$. This is a one-to-one map since $f=g$ whenever $f$ and $g$ agree on a dense set. Thus we have $\lvert C(X) \lvert \le \lvert \mathbb{R}^A \lvert$. Upon doing some cardinal arithmetic, we have $\lvert \mathbb{R}^A \lvert=2^{d(X)}$. Thus Lemma 1 is established. $\blacksquare$

Proof of Lemma 2
Let $g:S \rightarrow T$ be a continuous function from $S$ onto $T$. Let $\mathcal{B}$ be a base for $S$ such that $\lvert \mathcal{B} \lvert=w(S)$. Let $\mathcal{N}$ be the set of all $g(B)$ where $B \in \mathcal{B}$. Note that $\mathcal{N}$ is a network for $T$ (since $g$ is a continuous function). So we have $nw(T) \le \lvert \mathcal{N} \lvert \le \lvert \mathcal{B} \lvert = w(S)$. Since $T$ is compact, $w(T)=nw(T)$ (see Result 5 in Spaces With Countable Network). Thus we have $nw(T)=w(T) \lvert \le w(S)$. $\blacksquare$

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Results 1 and 2

Let $X$ be a completely regular space. Let $I$ be the unit interval $[0,1]$. We show that the Stone-Cech compactification $\beta X$ can be regarded as a subspace of the product space $I^{\mathcal{K}}$ where $\mathcal{K}= 2^{d(X)}$ (the product of $2^{d(X)}$ many copies of $I$). The cardinality of $I^{\mathcal{K}}$ is $2^{2^{d(X)}}$, thus leading to Result 1.

Let $C(X,I)$ be the set of all continuous functions $f:X \rightarrow I$. The Stone-Cech compactification $\beta X$ is constructed by embedding $X$ into the product space $\prod \limits_{f \in C(X,I)} I_f$ where each $I_f=I$ (see Embedding Completely Regular Spaces into a Cube or A Beginning Look at Stone-Cech Compactification). Thus $\beta X$ is a subspace of $I^{\mathcal{K}_1}$ where $\mathcal{K}_1=\lvert C(X,I) \lvert$.

Note that $C(X,I) \subset C(X)$. Thus $\beta X$ can be regarded as a subspace of $I^{\mathcal{K}_2}$ where $\mathcal{K}_2=\lvert C(X) \lvert$. By Lemma 1, $\beta X$ can be regarded as a subspace of the product space $I^{\mathcal{K}}$ where $\mathcal{K}= 2^{d(X)}$.

To see Result 2, note that the weight of $I^{\mathcal{K}}$ where $\mathcal{K}= 2^{d(X)}$ is $2^{d(X)}$. Then $\beta X$, as a subspace of the product space, must have weight $\le 2^{d(X)}$. $\blacksquare$

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Results 3 and 4

What drives Result 3 and Result 4 is the following theorem (established in Two Characterizations of Stone-Cech Compactification).

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 $\le$.

$\text{ }$

To define the partial order, for $\alpha_1 X$ and $\alpha_2 X$, both compactifications of $X$, we say that $\alpha_2 X \le \alpha_1 X$ if there is a continuous function $f:\alpha_1 X \rightarrow \alpha_2 X$ such that $f \circ \alpha_1=\alpha_2$. See the following figure.

Figure 1

In this post, we use $\le$ to denote this partial order as well as the order for cardinal numbers. Thus we need to rely on context to distinguish this partial order from the order for cardinal numbers.

Let $\alpha X$ be a compactification of $X$. Theorem C2 indicates that $\alpha X \le \beta X$ (partial order), which means that there is a continuous $f:\beta X \rightarrow \alpha X$ such that $f \circ \beta=\alpha$ (the same point in $X$ is mapped to itself by $f$). Note that $\alpha X$ is the image of $\beta X$ under the function $f:\beta X \rightarrow \alpha X$. Thus we have $\lvert \alpha X \lvert \le \lvert \beta X \lvert$ (cardinal number order). Thus Result 3 is established.

By Lemma 2, the existence of the continuous function $f:\beta X \rightarrow \alpha X$ implies that $w(\alpha X) \le w(\beta X)$ (cardinal number order). Thus Result 4 is established.

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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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$\copyright \ \ 2012$

# C*-Embedding Property and Stone-Cech Compactification

This is a continuation of an introduction of Stone-Cech compactification started in two previous posts (first post: A Beginning Look at Stone-Cech Compactification; second post: Two Characterizations of Stone-Cech Compactification). In this post, we present another characterization of the Stone-Cech compactification, that is, for any completely regular space $X$, $X$ is $C^*$-embedded in its Stone-Cech compactification $\beta X$ and that any compactification of $X$ in which $X$ is $C^*$-embedded must be $\beta X$. In other words, this property of $C^*$-embedding is unique to Stone-Cech compactification. We prove the following two theorems (U3 has two versions).

The links for other posts on Stone-Cech compactification can be found toward the end of this post.

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Definition. Let $Y$ be a space. Let $A \subset Y$. The subspace $A$ is $C^*$-embedded in $Y$ if every bounded continuous function $f:A \rightarrow \mathbb{R}$ is extendable to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$.

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 $\beta X$.
$\text{ }$

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Other Characterizations

Two other characterizations of $\beta X$ are proved in the previous post (Two Characterizations of Stone-Cech Compactification).

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 $\beta X$.
$\text{ }$

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 the largest compactification.

$\text{ }$

Theorem U2
The property in Theorem C2 is unique to $\beta X$. That is, if $\alpha X$ is a compactification of $X$, then $\alpha X$ must be equivalent to $\beta X$.

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Remark

The C theorems and the U theorems are a great tool to determine whether a given compactification is $\beta X$. Whenever a compactification $\alpha X$ of a space $X$ satisfies the property belonging to a C theorem, based on the corresponding U theorem, we know that this compactification $\alpha X$ must be $\beta X$. For example, any compactification $\alpha X$ that satisfies the function extension property in Theorem C1 must be $\beta X$. Th $C^*$-embedding property in Theorem C3 and Theorem U3 (both versions) is also a function extension property much like that in Theorems C1 and U1, but is easier to use. The reason being that we only need to extend a smaller class of continuous functions (i.e., to check whether functions from $X$ into $I=[0,1]$ can be extended), rather than checking all continuous functions from $X$ to arbitrary compact spaces. As the following example below about $\beta \omega_1$ illustrates that the $C^*$-embedding in Theorem C3 and U3.1 can be used to describe $\beta X$ explicitly.

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Proving Theorem U3.1 and Theorem U3.2

Let $Y$ be a space. Let $A$ be a subspace of $X$. Recall that $A$ is $C^*$-embedded in $Y$ if every bounded continuous function $f:A \rightarrow \mathbb{R}$ can be extended to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$.

Any bounded continuous function $f: X \rightarrow \mathbb{R}$ can be regarded as $f: X \rightarrow I_f$ where $I_f$ is some closed and bounded interval. The $C^*$-embedding property in Theorem C3 is a function extension property like the one in Theorem C1, except that it deals with function from $X$ into a specific type of compact spaces $Y$, namely the closed and bounded intervals in $\mathbb{R}$. Theorem C3 is a corollary of Theorem C1 (see below). So we only need to prove Theorem U3.1 and Theorem U3.2. Theorem U3.2 is a corollary of Theorem U3.1.

Proof of Theorem U3.1
By Theorem C2, we have $\alpha X \le \beta X$. So we only need to show $\beta X \le \alpha X$. To this end, we need to produce a continuous function $H: \alpha X \rightarrow \beta X$ such that $H \circ \alpha=\beta$.

Let $C(X,I)$ be the set of all continuous functions from $X$ into $I$. For each $f \in C(X,I)$, let $I_f=I$. Recall that $\beta X$ is embedded in the cube $\prod \limits_{f \in C(X,I)} I_f$ by the mapping $\beta$. For each $f \in C(X,I)$, let $\pi_f$ be the projection map from this cube into $I_f$.

Each $f \in C(X,I)$ can be expressed as $f=\pi_f \circ \beta$. Thus by assumption, each $f$ can be extended by $\hat{f}: \alpha X \rightarrow I$. Now define $H: \alpha X \rightarrow \prod \limits_{f \in C(X,I)} I_f$ by the following:

For each $t \in \alpha X$, $H(t)=a=< a_f >_{f \in C(X,I)}$ such that $a_f=\hat{f}(t)$

For each $x \in \alpha(X)$, we have $H(\alpha(x))=\beta(x)$. Note that $\hat{f}$ agrees with $f$ on $\alpha(X)$ since $\hat{f}$ extends $f$. So we have $H(\alpha(x))=a$ where $a_f=\hat{f}(\alpha(x))=f(x)$ for each $f \in C(X,I)$. On the other hand, by definition of $\beta$, we have $\beta(x)=a$ where $a_f=f(x)$ for each $f \in C(X,I)$. Thus we have $H \circ \alpha=\beta$ and the following:

$H(\alpha(X)) \subset \beta(X) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

It is straightforward to verify that $H$ is continuous. Note that $\alpha(X)$ is dense in $\alpha X$. Since $H$ is continuous, $H(\alpha(X))$ is dense in $H(\alpha X)$. Thus we have:

$H(\alpha X)=\overline{H(\alpha(X))} \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Putting $(1)$ and $(2)$ together, we have the following:

$H(\alpha X)=\overline{H(\alpha(X))} \subset \overline{\beta(X)}=\beta X$

Thus we can describe the map $H$ as $H: \alpha X \rightarrow \beta X$. As noted before, we have $H \circ \alpha=\beta$. Thus $\beta X \le \alpha X$. $\blacksquare$

Proof of Theorem U3.2
Suppose $\alpha X$ is a compactification of $X$ such that $X$ is $C^*$-embedded in $\alpha X$. Then every bounded continuous $f:X \rightarrow I_f$ can be extended to $\hat{f}:\alpha X \rightarrow I_f$ where $I_f$ is some closed and bounded interval containing the range. In particular, this means every continuous $f:X \rightarrow I$ can be extended. By Theorem U3.1, we have $\alpha X \approx \beta X$. $\blacksquare$

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Example

This is one example where we can use $C^*$-embedding to describe $\beta X$ explicitly.

Let $\omega_1$ be the first uncountable ordinal. Let $\omega_1+1$ be the successor ordinal of $\omega_1$ (i.e. $\omega_1$ with one additional point at the end). Consider $X=\omega_1$ and $Y=\omega_1+1$ as topological spaces with the order topology derived from the well ordering of the ordinals. The space $Y$ is a compactification of $X$. In fact $Y$ is the one-point compactification of $X$.

It is well known that every continuous real-valued function on $X$ is bounded (note that $X$ here is countably compact and hence pseudocompact). Furthermore, every continuous real-valued function on $X$ is eventually constant. This means that if $f:X \rightarrow \mathbb{R}$ is continuous, for some $\alpha < \omega_1$, $f$ is constant on the final segment $X_\alpha=\left\{\rho < \omega_1: \rho>\alpha \right\}$ (see result B in The First Uncountable Ordinal). As a result, every continuous bounded real-valued function $f:X \rightarrow \mathbb{R}$ can be extended to a continuous $\hat{f}:Y \rightarrow \mathbb{R}$. Then according to Theorem U3.2, $\beta X=\beta \omega_1=Y=\beta \omega_1+1$.

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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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$\copyright \ \ 2012$

# Two Characterizations of Stone-Cech Compactification

This is the second post on Stone-Cech compactification (continuing from A Beginning Look at Stone-Cech Compactification). In this post, we establish two characterizations of Stone-Cech compactification. The first one is represented in the following diagram. The second one is that Stone-Cech compactification is maximal with respect to a certain partial order.

The first characterization is a central characteristic of Stone-Cech compactification. It is a function extension property that uniquely characterizes the Stone-Cech compactification of a completely regularly space. Here’s the diagram.

Figure 1

In this diagram, $X$ is a completely regular space and $\beta X$ is the Stone-Cech compactification of $X$ where $\beta$ is the homeomorphism mapping $X$ onto $\beta(X)$, which is dense in $\beta X$. The function $f: X \rightarrow Y$ is an arbitrary continuous function where $Y$ is compact. Then there exists a continuous function $F:\beta X \rightarrow Y$ such that $F$ restricted to $\beta(X)$ is identical to the function $f$. In other words, if we think of $X$ as a subset of $\beta X$, any continuous function from $X$ to a compact space can be extended to all of $\beta X$. This function extension property is stated in Theorem C1 below.

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 $F \circ \beta=f$. See Figure 1 above.
$\text{ }$
Theorem U1
If $K$ is any compactification of $X$ that satisfies condition in Theorem C1, then $K$ must be equivalent to $\beta X$.

Theorem C1 is the statement of the extension property described at the beginning. Theorem U1 states that this property is unique to $\beta X$. That is, of all the possible compactifications of $X$, only $\beta X$ can satisfy Theorem C1.

For the other characterization, see Theorem C2 and Theorem U2 below.

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Defining Stone-Cech Compactification

The definition of $\beta X=\beta_X X$ is given in this previous post (A Beginning Look at Stone-Cech Compactification) and is repeated here again for the sake of completeness. Let $C(X,I)$ be the set of all continuous functions from $X$ into $I=[0,1]$. For each $g \in C(X,I)$, $I_g=[0,1]$. The map $\beta_X:X \rightarrow \prod \limits_{g \in C(X,I)} I_g$ is defined by:

For each $x \in X$, $\beta_X(x)=t=< t_g >_{g \in C(X,I)}$ is the point $t \in \prod \limits_{g \in C(X,I)} I_g$ such that $t_g=g(x)$ for each $g \in C(X,I)$ (i.e. the $g^{th}$ coordinate of the point $t$ is $g(x)$).

For the proof that $\beta_X$ is a homeomorphism, see A Beginning Look at Stone-Cech Compactification. We have the following definition.

Definition
Under the map $\beta_X$, $\beta_X(X)$ is the topological copy of $X$ within the cube $\prod \limits_{f \in C(X,I)} I_f$. The Stone-Cech compactification of $X$ is defined to be the closure of $\beta_X(X)$ in the cube $\prod \limits_{f \in C(X,I)} I_f$, i.e., set $\beta_X X=\overline{\beta_X(X)}$.

When there is no ambiguity as to what the space $X$ is, the embedding $\beta_X$ is written as $\beta$ and the compactification $\beta_X X$ is written as $\beta X$ (as in Figure 1 above). When more than one space is involved, we use subscripts to distinguish the embeddings, e.g., $\beta_X$ and $\beta_Y$.

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Proof of Theorem U1

Let $f:X \rightarrow Y$ be a continuous function from $X$ into a compact Hausdorff space $Y$. Let $\beta_X X$ be the Stone-Cech compactification of $X$ where $\beta_X$ is the homeomorphic embedding that defines $\beta_X X$. Since $Y$ is a completely regular space, it has a Stone-Cech compactification $\beta_Y Y$, where $\beta_Y$ is the homeomorphic embedding. We also define a map $W$ from $\prod \limits_{g \in C(X,I)} I_g$ into $\prod \limits_{g \in C(Y,I)} I_k$. We have the following diagram.

Figure 2

The desired function $F$ will be defined by $F=\beta_Y^{-1} \circ (W \upharpoonright \beta_X X)$. The rest of the proof is to define $W$ and to show that this definition of $F$ makes sense.

To define the function $W$, for each $t \in \prod \limits_{g \in C(X,I)} I_g$, let $W(t)=a$ such that $a_k=t_{k \circ f}$ (i.e. the $k^{th}$ coordinate of $W(t)=a$ is the $(k \circ f)^{th}$ coordinate of $t$). With the definition of $W$, the diagram in Figure 2 commutes, i.e.,

$W \circ \beta_X=\beta_Y \circ f \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Starting with a point $x \in X$ (the upper left corner of the diagram), we can reach the same point in the lower right corner regardless the path we take ($W \circ \beta_X$ or $\beta_Y \circ f$). The following shows the derivation.

One direction:
$x \in X$

$\downarrow$

$\beta_X(x)=t \text{ where } t_g=g(x) \ \forall \ g \in C(X,I)$

$\downarrow$

$W(t)=a \text{ where } a_k=t_{k \circ f}=(k \circ f)(x)=k(f(x)) \ \forall \ k \in C(Y,I)$

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The other direction:
$x \in X$

$\downarrow$

$f(x) \in Y$

$\downarrow$

$\beta_Y(f(x))=a \text{ where } a_k=k(f(x)) \ \forall \ k \in C(Y,I)$

It is straightforward to verify that the map $W$ is continuous. Based on $(1)$ above, note that $W(\beta_X(X)) \subset \beta_Y(Y)$. The following derivation shows that $W(\beta_X X) \subset \beta_Y(Y)$.

\displaystyle \begin{aligned} W(\beta_X X)&=W(\overline{\beta_X(X)}) \\&\subset \overline{W(\beta_X(X))} \ \ \ \ \text{ based on the continuity of } W\\&\subset \overline{\beta_Y(Y)} \ \ \ \ \ \ \ \ \ \ \text{ based on (1)}\\&=\beta_Y Y \\&=\beta_Y(Y) \ \ \ \ \ \ \ \ \ \ \text{ based on the compactness of Y} \end{aligned}

With the above derivation, we now know that the function $W$ maps points of $\beta_X X$ to points of $\beta_Y(Y)$. So it makes sense to define $F=\beta_Y^{-1} \circ (W \upharpoonright \beta_X X)$. Note that for each $x \in X$, we have:

\displaystyle \begin{aligned} F(\beta_X(x))&=\beta_Y^{-1}(W(\beta_X(x)) \\&=\beta_Y^{-1}(\beta_Y(f(x))) \\&=f(x) \end{aligned}

Then we have $F \circ \beta_X=f$ and $F$ is the desired function. $\blacksquare$

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Compactifications

In order to prove Theorem U1, we first have a basic discussion on compactifications. Most importantly, we pin down what we mean when we say two compactifications of $X$ are equivalent. In the process, we produce another characterization of Stone-Cech compactification (see Theorem C2 and Theorem U2 below).

Let $X$ be a completely regular space. A pair $(T,\alpha)$ is said to be a compactification of the space $X$ if $T$ is a compact Hausdorff space and $\alpha:X \rightarrow T$ is a homeomorphism from $X$ into $T$ such that $\alpha(X)$ is dense in $T$. More informally, a compactification of the space $X$ can also be thought of as a compact space $T$ containing a topological copy of the space $X$ as a dense subspace.

Given a compactification $(T,\alpha)$, we use the notation $\alpha X$ rather than the pair $(T,\alpha)$. By saying that $\alpha X$ is a compactification of $X$, we mean $\alpha X$ is the compact space $T$ where $\alpha$ is the homeomorphism embedding $X$ onto $\alpha(X)$.

The Stone-Cech compactification construction above is an example of a compactification. There can be more than one compactification of a given space $X$. For example, for $X=\mathbb{R}$, we have the Stone-Cech compactification $\beta \mathbb{R}$, which is a subspace of the cube $\prod \limits_{f \in C(\mathbb{R},I)} I_f$. The circle $S^1=\left\{(x,y) \in \mathbb{R}^2: x^2+y^2=1 \right\}$ contains a copy of the real line $\mathbb{R}$ as a dense subspace, as does the unit interval $[0,1]$. Thus both $S^1$ and $I=[0,1]$ are also compactifications of $\mathbb{R}$. See A Beginning Look at Stone-Cech Compactification for a discussion of these examples.

We say that compactifications $\alpha_1 X$ and $\alpha_2 X$ are equivalent (we write $\alpha_1 X \approx \alpha_2 X$) if there exists a homeomorphism $f: \alpha_1 X \rightarrow \alpha_2 X$ such that $f \circ \alpha_1= \alpha_2$. In other words, the following diagram commutes.

Figure 3

Essentially, two compactifications $\alpha_1 X$ and $\alpha_2 X$ of $X$ are equivalent if there is a homeomorphism $f$ between the two and if each $x \in X$ is mapped by $f$ to itself, i.e., $\alpha_1(x)$ is mapped to $\alpha_2(x)$.

For a given completely regular space $X$, let $\mathcal{C}(X)$ be the class of all compactifications of $X$. We define a partial order $\le$ on $\mathcal{C}(X)$. For $\alpha_1 X$ and $\alpha_2 X$, both in $\mathcal{C}(X)$, we say that $\alpha_2 X \le \alpha_1 X$ if there is a continuous function $f:\alpha_1 X \rightarrow \alpha_2 X$ such that $f \circ \alpha_1=\alpha_2$. See Figure 4 below.

Figure 4

The following theorem ties the partial order $\le$ to the equivalence relation $\approx$ for compactifications.

Theorem 1
Let $\alpha_1 X$ and $\alpha_2 X$ be two compactifications of $X$. Then $\alpha_1 X \le \alpha_2 X$ and $\alpha_2 X \le \alpha_1 X$ if and only if $\alpha_1 X \approx \alpha_2 X$.

Proof of Theorem 1
$\Rightarrow$ With $\alpha_2 X \le \alpha_1 X$, there exists continuous $f_1:\alpha_1 X \rightarrow \alpha_2 X$ such that

$f_1 \circ \alpha_1=\alpha_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (A1)$

With $\alpha_1 X \le \alpha_2 X$, there exists continuous $f_2:\alpha_2 X \rightarrow \alpha_1 X$ such that

$f_2 \circ \alpha_2=\alpha_1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (A2)$

Applying $f_2$ to $(A1)$, we have $f_2 \circ f_1 \circ \alpha_1=f_2 \circ \alpha_2$. Applying $(A2)$ to this result, we have

$f_2 \circ f_1 \circ \alpha_1=\alpha_1 \ \ \ \ \ \ \ \ \ \ \ (A3)$

Note that $f_2 \circ f_1$ is a map from $\alpha_1 X$ into $\alpha_1 X$. The equation $(A3)$ indicates that when $f_2 \circ f_1$ is restricted to $\alpha_1(X)$, it is the identity map. Thus $f_2 \circ f_1$ agrees with the identity map on the dense set $\alpha_1(X)$. This implies that $\alpha_1(X)$ must agree with the identity map on all of $\alpha_1 X$.

Likewise we can see that $f_1 \circ f_2$ must equal to the identity map on $\alpha_2 X$. So $f_1:\alpha_1 X \rightarrow \alpha_2 X$ is a homeomorphism and it follows that $\alpha_1 X$ and $\alpha_2 X$ are equivalent compactifications of $X$.

$\Leftarrow$ This direction is straightforward. Let $f:\alpha_1 X \rightarrow \alpha_2 X$ a homeomorphism that makes $\alpha_1 X$ and $\alpha_2 X$ equivalent (as described by Figure 3). Then the map $f$ implies $\alpha_2 X \le \alpha_1 X$ and the map $f^{-1}$ implies $\alpha_1 X \le \alpha_2 X$. $\blacksquare$

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Another Characterization of the Stone-Cech Compactification

The next theorem says that the Stone-Cech compactification is the maximal compactification with respect to the partial order $\le$ defined here. Furthermore, this property is unique (there is only one maximal compactification up to equivalence). This result will simplify the work when we need to show that a given compactification is equivalent to $\beta X$.

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 $\le$.

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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 $\alpha X \approx \beta X$.

Proof Theorem C2
Let $\alpha X$ be any compactification of $X$. Consider the continuous map $\alpha:X \rightarrow \alpha X$. By Theorem C1, $\alpha$ can be extended to $\beta X$. In other words, there exists a continuous $F: \beta X \rightarrow \alpha X$ such that $F \circ \beta = \alpha$. The existence of the map $F$ implies that $\alpha X \le \beta X$. $\blacksquare$

Proof Theorem U2
Let $\alpha X$ be another maximal compactification of $X$. This implies that $\beta X \le \alpha X$. By Theorem C2, we have $\alpha X \le \beta X$. By Theorem 1, $\alpha X$ must be equivalent to $\beta X$. $\blacksquare$

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Proof of Theorem U1

We are now ready to prove Theorem U1.

Proof of Theorem U1
Let $\alpha X$ be a compactification of $X$ that satisfies the extension property in Theorem C1. In light of Theorem C2, we have $\alpha X \le \beta X$. So we only need to show $\beta X \le \alpha X$. Consider the map $\beta: X \rightarrow \beta X$. By the assumption that $\alpha X$ satisfies the extension property in Theorem C1, there exists a continuous function $F:\alpha X \rightarrow \beta X$ such that $F \circ \alpha=\beta$. The existence of $F$ implies that $\beta X \le \alpha X$. By Theorem 1, $\alpha X$ must be equivalent to $\beta X$. $\blacksquare$

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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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$\copyright \ \ 2012$

# A Beginning Look at Stone-Cech Compactification

For every completely regular space $X$, there exists a unique compact space $\beta X$ containing $X$ such that (1) $\beta X$ has a dense subspace that is a topological copy of $X$ and that (2) if $X$ is considered as a subspace of $\beta X$, any continuous function $f: X \rightarrow Y$ from $X$ into a compact space $Y$ can be extended to all of $\beta X$. The compact space $\beta X$ is said to be the Stone-Cech compactification of $X$. We indicate how $\beta X$ is constructed. The construction is done by embedding a completely regular space into a cube. To provide a glimpse of what $\beta X$ might look like, we also take a brief look at $\beta \mathbb{R}$.

The links for other posts on Stone-Cech compactification can be found toward the end of this post.

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Construction

According to a theorem by Tychonoff from 1930, every completely regular space can be embedded in a cube. Embedding into a cube is one way Stone-Cech compactification is constructed. Let $X$ be a completely regular space. Let $I=[0,1]$ be the unit interval in the real line $\mathbb{R}$. Let $C(X,I)$ be the set of all continuous real-valued functions defined on the space $X$. The cube for which $X$ is embedded is $I^{\mathcal{K}}$, the product of $\mathcal{K}$ copies of $I$ where $\mathcal{K}$ is the cardinality of $C(X,I)$. We can also view $I^{\mathcal{K}}$ as the product space $\prod \limits_{f \in C(X,I)} I_f$ where each $I_f=I$. We can represent each point in the cube as a function $t:C(X,I) \rightarrow I$ or as a sequence $< t_f >_{f \in C(X,I)}$ such that each term (or coordinate) $t_f \in I=[0,1]$.

The embedding from $X$ into the cube $\prod \limits_{f \in C(X,I)} I_f$ is the evaluation map $\beta_X$ which is defined as: for each $x \in X$, $\beta_X(x)=t=< t_f >_{f \in C(X,I)}$ is the point in the cube such that for each $f \in C(X,I)$, $t_f=f(x)$. This map $\beta_X$ is shown to be a homeomorphism from $X$ into $\prod \limits_{f \in C(X,I)} I_f$. What makes $\beta_X$ a homeomorphism stems from the fact that in a completely regular space, there are enough bounded real-valued continuous functions to separate points from closed sets (see Embedding Completely Regular Spaces into a Cube). We have the following definition.

Definition
Under the map $\beta_X$, $\beta_X(X)$ is the topological copy of $X$ within the cube $\prod \limits_{f \in C(X,I)} I_f$. The Stone-Cech compactification of $X$ is defined to be the closure of $\beta_X(X)$ in the cube $\prod \limits_{f \in C(X,I)} I_f$, i.e., set $\beta_X X=\overline{\beta_X(X)}$.

According to the Tychonoff theorem, the cube $\prod \limits_{f \in C(X,I)} I_f$, being a product of compact spaces, is a compact space. Thus $\beta_X X$, being a closed subspace of the cube, is a compact space. Furthermore, $\beta_X X$ contains a topological copy of $X$ as a dense subspace.

When there is no ambiguity as to what the space $X$ is, the embedding $\beta_X$ is written as $\beta$ and the compactification $\beta_X X$ is written as $\beta X$.

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Characterization

Let $X$ be a completely regular space. A compactification of $X$ is a compact space $K$ that has a dense subspace that is a topological copy of $X$. For a given $X$, there can be many compactifications of $X$. The Stone-Cech compactification $\beta X$ has characteristics that are not shared by other compactifications of $X$. One such characteristic is stated at the beginning of the post and is repeated in the following two theorems.

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 $F \circ \beta=f$. See Figure 1 below.
$\text{ }$
Theorem U1
If $K$ is any compactification of $X$ that satisfies condition in Theorem C1, then $K$ must be equivalent to $\beta X$.

Figure 1

When the continuous function $F$ in Theorem C1 is restricted to to $\beta(X)$, it is identical to the function $f$. If we think of $X$ as a subset of $\beta X$, $F$ extends $f$. Thus Theorem U1 essentially says that any continuous function from $X$ into any compact space can be extended to all of $\beta X$. Theorem U1 says that the property stated in Theorem C1 uniquely characterizes $\beta X$. This means that any compactification of $X$ that satisfies Theorem C1 must be equivalent to $\beta X$.

The proofs of Theorem C1 and Theorem U1 are found in the next post (Two Characterizations of Stone-Cech Compactification).

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Examples

We now look at $\beta \mathbb{R}$. We rely on Theorem C1 to provide some insight about $\beta \mathbb{R}$.

Note that $S^1$ and $[0,1]$ are also compactifications of $\mathbb{R}$, where $S^1$ is the unit circle $S^1=\left\{(x,y) \in \mathbb{R}^2: x^2+y^2=1 \right\}$. First, we show the extension property described in Theorem C1 does not hold for both $S^1$ and $[0,1]$. Thus these two compactifications of $\mathbb{R}$ cannot be $\beta \mathbb{R}$.

Consider the function $\Phi(x):\mathbb{R} \rightarrow [0,1]$ defined by:

$\displaystyle \Phi(x)=\int_{-\infty}^x \ \frac{1}{\sqrt{2 \pi}} \ e^{\frac{- t^2}{2}} \ dt$

Note that $\Phi(x)$ is the cumulative distribution function (CDF) of the standard normal distribution. The following figure is the graph.

Figure 2

Note that $S^1$ is the one-point compactification of $\mathbb{R}$. Since $\lim \limits_{x \rightarrow +\infty} \Phi(x)=1$ and $\lim \limits_{x \rightarrow -\infty} \Phi(x)=0$, we cannot extend the function $\Phi$ to a continuous function defined on $S^1$. Thus, $S^1$ as a compactification of $\mathbb{R}$ cannot satisfy Theorem C1. It is clear that $\beta \mathbb{R}$ cannot be $S^1$.

The closed unit interval $[0,1]$ is the two-point compactification of $\mathbb{R}$. Now consider the function $s(x)=sin(\frac{1}{x})$ defined on the open interval $(0,1)$, which is a topological copy of $\mathbb{R}$. It is impossible to extend $s(x)$ to a continuous function defined on $[0,1]$. So to construct $\beta \mathbb{R}$, it take more than adding two additional points. In fact, the following discussion shows that $\beta \mathbb{R}$ has uncountably many points in the remainder $\beta \mathbb{R}-\mathbb{R}$.

Consider the function the sine function $w(x)=\text{sin}(x)$, which maps $\mathbb{R}$ onto $[-1,1]$. Based on Theorem C1, $w(x)$ can be extended to a function $G: \beta \mathbb{R} \rightarrow [-1,1]$.

For each $t \in [-1,1]$, let $A_t=\left\{x \in \mathbb{R}: w(x)=\text{sin}(x)=t \right\}$. For example, $A_0$ is the set $\displaystyle \left\{n \pi: n=0, \pm 1, \pm 2, \pm 3, \cdots \right\}$. Furthermore, for each $t \in [-1,1]$, let $B_t=G^{-1}(t)$, which is a compact set in $\beta \mathbb{R}$.

Consider $\mathbb{R}$ as a subset of $\beta \mathbb{R}$. We have $A_t \subset B_t$ for each $t \in [-1,1]$. Each $A_t$ is a discrete set in $\mathbb{R}$. However, each $A_t$ is not discrete in $\beta \mathbb{R}$ ($A_t$ is an infinite subset of a compact set in $\beta \mathbb{R}$). Thus $B_t-A_t \ne \varnothing$ for each $t \in [-1,1]$.

If we think of constructing a compactification as the process of adding points to $X$ to form a compact space, the non-empty sets $B_t-A_t$ show that at minimum we are adding continuum many points to form $\beta \mathbb{R}$ (adding as many points as there are points in $[-1,1]$ but in reality more than continuum many points are added). Note that for $t \ne p$, $B_t$ and $B_p$ are disjoint (they are separated by disjoint open sets in $\beta \mathbb{R}$ through inverse images of the function $G$).

In fact, the cardinality of $\beta \mathbb{R}$ is larger than continuum. Specifically, $\lvert \beta \mathbb{R} \lvert=2^c$, where $c$ is the cardinality of $\mathbb{R}$ and $2^c$ is the cardinality of the set of all subsets of $\mathbb{R}$. For this fact, see Corollary 3.6.12 in [1] or Exercise 19.H in [2].

The large size of $\beta \mathbb{R}$ also tells us something about it topologically. It is a well known theorem in general topology that the cardinality of every first countable compact Hausdorff space is at most continuum (see Corollary 3.1.30 in [1] or see The cardinality of compact first countable spaces, I). Thus $\beta \mathbb{R}$ cannot be first countable. The points at which it is not first countable are the points in the remainder $\beta \mathbb{R}-\mathbb{R}$. Not being a first countable, $\beta \mathbb{R}$ is not metrizable.

Much more can be said about $\beta \mathbb{R}$ and other Stone-Cech compactification $\beta X$. This brief walk in $\beta \mathbb{R}$ shows that the Stone-Cech compactification can be quite large and quite different topologically even when the starting space is the familiar Euclidean space. In subsequent discussions, we will see that nice properties such as first countability, second countability and metrizability do not carry over from $X$ to $\beta X$.

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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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$\copyright \ \ 2012$

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# Sorgenfrey Line is not a Moore Space

We found an incorrect statement about the Sorgenfrey line in an entry in Wikipedia about Moore space (link). This statement opens up a discussion on the question of whether the Sorgenfrey line is a Moore space as well as a discussion on Moore space. The following is the incorrect statement found in Wikipedia by the author.

The Sorgenfrey line is the space whose underlying set is the real line $S=\mathbb{R}$ where the topology is generated by a base consisting the half open intervals of the form $[a,b)$. The Sorgenfrey plane is the square $S \times S$.

Even though the Sorgenfrey line is normal, the Sorgenfrey plane is not normal. In fact, the Sorgenfrey line is the classic example of a normal space whose square is not normal. Both the Sorgenfrey line and the Sorgenfrey plane are not Moore space but not for the reason given. The statement seems to suggest that any normal Moore space is second countable. But this flies in the face of all the profound mathematics surrounding the normal Moore space conjecture, which is also discussed in the Wikipedia entry.

The statement indicated above is only a lead-in to a discussion of Moore space. We are certain that it will be corrected. We always appreciate readers who kindly alert us to errors found in this blog.

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Moore Spaces

Let $X$ be a regular space. A development for $X$ is a sequence $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$ of open covers of $X$ such that for each $x \in X$, and for each open subset $U$ of $X$ with $x \in U$, there exists one cover $\mathcal{G}_n$ satisfying the condition that for any open set $V \in \mathcal{G}_n$, $x \in V \Rightarrow V \subset U$. When $X$ has a development, $X$ is said to be a Moore space (also called developable space). A Note On The Sorgenfrey Line is an introductory note on the Sorgenfrey line.

Moore spaces can be viewed as a generalization of metrizable spaces. Moore spaces are first countable (having a countable base at each point). For a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$, the open sets in $\mathcal{G}_n$ are considered “smaller” as the index $n$ increases. In fact, this is how a development is defined for a metric space, where $\mathcal{G}_n$ consists of all open balls with diameters less than $\frac{1}{n}$. Thus metric spaces are developable. There are plenty of non-metrizable Moore space. One example is the Niemytzki’s Tangent Disc space.

In a Moore space, every closed set is a $G_\delta$-set. Thus if a Moore space is normal, it is perfectly normal. Any Moore space has a $G_\delta$-diagonal (the diagonal $\Delta=\left\{(x,x): x \in X \right\}$ is a $G_\delta$-set in $X \times X$). It is a well known theorem that every compact space with a $G_\delta$-diagonal is metrizable. Thus any compact Moore space is metrizable.

The last statement can be shown more directly. Suppose that $X$ is compact and has a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$. Then each $\mathcal{G}_n$ has a finite subcover $\mathcal{H}_n$. Then $\bigcup_{n=1}^\infty \mathcal{H}_n$ is a countable base for $X$. Thus any compact Moore space is second countable and hence metrizable.

What about paracompact Moore space? Suppose that $X$ is paracompact and has a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$. Then each $\mathcal{G}_n$ has a locally finite open refinement $\mathcal{H}_n$. Then $\bigcup_{n=1}^\infty \mathcal{H}_n$ is a $\sigma$-locally finite base for $X$. The Smirnov-Nagata metrization theorem states that a space is metrizable if and only if it has a $\sigma$-locally finite base (see Theorem 23.9 on page 170 of [2]). Thus any paracompact Moore space has a $\sigma$-locally finite base and is thus metrizable (after using the big gun of the Smirnov-Nagata metrization theorem).

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Sorgenfrey Line

The Sorgenfrey line is regular and Lindelof. Hence it is paracompact. Since the Sorgenfrey line is not metrizable, by the above discussion it cannot be a Moore space. The Sorgenfrey plane is also not a Moore space. Note that being a Moore space is a hereditary property. So if the Sorgenfrey plane is a Moore space, then every subspace of the Sorgenfrey plane (including the Sorgenfrey line) is a Moore space.

The following theorem is another way to show that the Sorgenfrey line is not a Moore space.

Bing’s Metrization Theorem
A topological space is metrizable if and only if it is a collectionwise normal Moore space.

Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [1]). Thus the Sorgenfrey line is collectionwise normal and hence cannot be a Moore space. A space $X$ is said to be collectionwise normal if $X$ is a $T_1$-space and for every discrete collection $\left\{W_\alpha: \alpha \in A \right\}$ of closed sets in $X$, there exists a discrete collection $\left\{V_\alpha: \alpha \in A \right\}$ of open subsets of $X$ such that $W_\alpha \subset V_\alpha$. For a proof of Bing’s metrization theorem, see page 329 of [1].

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Remark

The normal Moore space conjecture is the statement that every normal Moore space is metrizable. This conjecture had been one of the key motivating questions for many set theorists and topologists during a large part of the twentieth century. The bottom line is that this statement cannot not be decided just on the basis of the set of generally accepted axioms called Zermelo–Fraenkel set theory with the axiom of choice, commonly abbreviated ZFC. But Bing’s metrization theorem states that if we strengthen normality to collectionwise normality, we have a definite answer.

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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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$\copyright \ \ 2012$

# An Example of a Completely Regular Space that is not Normal

We present an example of a space $S$ that is Hausdorff, locally compact, zero-dimensional (having a base consisting of closed and open sets), metacompact, completely regular but not normal. This space $S$ is a building block for defining other spaces. In a previous post, we show that the space $S$ can be turned into a space that is regular but not completely regular by adding one point (see Regular but not Completely Regular). Even though $S$ is discussed in the previous post, the space $S$ is an interesting example by itself and deserves a separate post.

Let $S$ be the set of all points $(x,y) \in \mathbb{R} \times \mathbb{R}$ such that $y \ge 0$. For each real number $x$, define $V_x$ to be the set $V_x=\left\{(x,y) \in S: 0 \le y \le 2 \right\}$, define $D_x$ to be the set $D_x=\left\{(s,s-x) \in S: x \le s \le x+2 \right\}$, and define $O_x=V_x \cup D_x$. The topology on $S$ is defined by the following:

• Each point $(x,y) \in S$ where $y>0$ is isolated.
• For each point $(x,0) \in S$, a basic open set is of the form $O_x - F$ where $(x,0) \notin F$ and $F$ is a finite subset of $O_x$.

It is straightforward to verify that the basic open sets defined above form a base for a topology on the set $S$ and that the resulting topology is Hausdorff. One important observation to make is that this base consists of sets that are both closed and open. Whenever a space has a base consisting of closed and open sets, it is said to be a zero-dimensional space. It is straightforward to show that any zero-dimensional space is completely regular. Another interesting point about the space $S$ is that it is metacompact. Recall that a space $X$ is metacompact if every open cover of $X$ has a point-finite open refinement.

Now we discuss why $S$ is not normal. Note that the x-axis in $S$ is a closed and discrete set of cardinality continuum. Jones’ lemma states that in a normal and separable space, the cardinality of any closed and discrete set must not equal to or exceed continuum. But Jones’ lemma is of no use here since $S$ is not separable. However, the two disjoint closed sets that destroyed normality are from the x-axis, namely $H=\left\{(x,0): x \in \mathbb{Q} \right\}$ and $K=\left\{(x,0): x \in \mathbb{R}-\mathbb{Q} \right\}$.

Note that $H$ and $K$ are disjoint closed sets in $S$. If $S$ were normal, there would be a continuous $f:S \rightarrow [0,1]$ such that $f(a)=0$ for all $a \in H$ and $f(a)=1$ for all $a \in K$ (using Urysohn lemma). But this function is not possible. It can be shown that any continuous function $g:S \rightarrow [0,1]$ that maps $H$ to zero would have to map the entire x-axis to zero except for a countable subset of x-axis. This fact follows from the Main Result presented in Regular but not Completely Regular.

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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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$\copyright \ \ 2012$

# One way to get CCC spaces that are not separable

A space $X$ is said to have countable chain condition (ccc) if every disjoint collection of open subsets of $X$ is countable. For convenience, we refer to spaces that have countable chain condition as ccc spaces. It is easy to verify that separable spaces are ccc spaces. We present a specific way of generating spaces that always have ccc but are not separable. These spaces are the sigma products of separable spaces.

The product of separable spaces always have ccc (see Product of Spaces with Countable Chain Condition). However, the product of separable spaces is not separable when the number of factors is greater than continuum. Thus one way to get an example of ccc but not separable space is to take the product of more than continuum many separable spaces. For example, if $c$ is the cardinality of continuum, $\left\{0,1 \right\}^{2^c}$, the product of $2^c$ many copies of $\left\{0,1 \right\}$, is a space that has ccc but is not separable.

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Another example is obtained from taking sigma product of separable spaces. Let $\left\{X_\alpha: \alpha \in A \right\}$ be a collection of spaces where $A$ is some index set. Consider the product space $\prod \limits_{\alpha \in A} X_\alpha$. Fix a point $p$ in the product. The sigma-product about the point $p$ is denoted by $\sum \limits_{\alpha \in A} X_\alpha$ and is the following subspace of the product space $\prod \limits_{\alpha \in A} X_\alpha$:

$\sum \limits_{\alpha \in A} X_\alpha=\left\{x \in \prod \limits_{\alpha \in A} X_\alpha: x(\alpha) \ne p(\alpha) \text{ for at most countably many } \alpha \right\}$

To obtain the desired example, let $A$ be an uncountable index set and let each $X_\alpha$ be separable. The product space $\prod \limits_{\alpha \in A} X_\alpha$ has ccc. Note that $\sum \limits_{\alpha \in A} X_\alpha$ is always dense in the product space $\prod \limits_{\alpha \in A} X_\alpha$. Thus the sigma-product $\sum \limits_{\alpha \in A} X_\alpha$ has ccc since it is a dense subspace of a ccc space. On the other hand, $\sum \limits_{\alpha \in A} X_\alpha$ is never separable as long as there are uncountably many spaces $X_\alpha$.

As specific example, take $X_\alpha=\left\{0,1 \right\}$ for each $\alpha < \omega_1$ and let the fixed point $p$ be such that $p(\alpha)=0$ for all $\alpha < \omega_1$. The resulting $\sum \limits_{\alpha < \omega_1} \left\{0,1 \right\}$ is a ccc space that is not separable. Of course, $\sum \limits_{\alpha < \omega_1} X_\alpha$ in this case is the set of all $x \in \left\{0,1 \right\}^{\omega_1}$ such that $x(\alpha) \ne 0$ for at most countably many $\alpha < \omega_1$.

One interesting note about the sigma-product $\sum \limits_{\alpha < \omega_1} \left\{0,1 \right\}$ is that the overall product space $\left\{0,1 \right\}^{\omega_1}$ is an example of a separable but not hereditarily separable space. Another interesting point is that $\sum \limits_{\alpha < \omega_1} \left\{0,1 \right\}$ is a countably compact non-compact space (see A note about sigma-product of compact spaces).

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Previous discussion of CCC spaces in this blog:

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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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$\copyright \ \ 2012$