# 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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