# 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.

_______________________________________________________________________________

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 $\beta_Y$.

_______________________________________________________________________________

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:
$x \in X$

$\downarrow$

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

$\downarrow$

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

_________________________________
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. $\blacksquare$

_______________________________________________________________________________

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 $\alpha_1 X \le \alpha_2 X$. $\blacksquare$

_______________________________________________________________________________

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 $\le$.

$\text{ }$

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 $\alpha X \approx \beta X$.

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 $\beta X$. $\blacksquare$

_______________________________________________________________________________

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 $\beta X$. $\blacksquare$

_______________________________________________________________________________

Blog Posts on Stone-Cech Compactification

_______________________________________________________________________________

Reference

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

_______________________________________________________________________________

$\copyright \ \ 2012$