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.
___________________________________________________________________________________

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).

___________________________________________________________________________________

Some Helpful Results

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

___________________________________________________________________________________

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

___________________________________________________________________________________

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

___________________________________________________________________________________

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

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s