Counterexample 106 from Steen and Seebach

As the title suggests, this post discusses counterexample 106 in Steen and Seebach [2]. We extend the discussion by adding two facts not found in [2].

The counterexample 106 is the space X=\omega_1 \times I^I, which is the product of \omega_1 with the interval topology and the product space I^I=\prod_{t \in I} I where I is of course the unit interval [0,1]. The notation of \omega_1, the first uncountable ordinal, in Steen and Seebach is [0,\Omega).

Another way to notate the example X is the product space \prod_{t \in I} X_t where X_0 is \omega_1 and X_t is the unit interval I for all t>0. Thus in this product space, all factors except for one factor is the unit interval and the lone non-compact factor is the first uncountable ordinal. The factor of \omega_1 makes this product space an interesting example.

The following lists out the basic topological properties of the space that X=\omega_1 \times I^I are covered in [2].

  • The space X is Hausdorff and completely regular.
  • The space X is countably compact.
  • The space X is neither compact nor sequentially compact.
  • The space X is neither separable, Lindelof nor \sigma-compact.
  • The space X is not first countable.
  • The space X is locally compact.

All the above bullet points are discussed in Steen and Seebach. In this post we add the following two facts.

  • The space X is not normal.
  • The space X has a dense subspace that is normal.

It follows from these bullet points that the space X is an example of a completely regular space that is not normal. Not being a normal space, X is then not metrizable. Of course there are other ways to show that X is not metrizable. One is that neither of the two factors \omega_1 or I^I is metrizable. Another is that X is not first countable.

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The space X is not normal

Now we are ready to discuss the non-normality of the example. It is a natural question to ask whether the example X=\omega_1 \times I^I is normal. The fact that it was not discussed in [2] could be that the tool for answering the normality question was not yet available at the time [2] was originally published, though we do not know for sure. It turns out that the tool became available in the paper [1] published a few years after the publication of [2]. The key to showing the normality (or the lack of) in the example X=\omega_1 \times I^I is to show whether the second factor I^I is a countably tight space.

The main result in [1] is discussed in this previous post. Theorem 1 in the previous post states that for any compact space Y, the product \omega_1 \times Y is normal if and only if Y is countably tight. Thus the normality of the space X (or the lack of) hinges on whether the compact factor I^I=\prod_{t \in I} I is countably tight.

A space Y is countably tight (or has countable tightness) if for each S \subset Y and for each x \in \overline{S}, there exists some countable B \subset S such that x \in \overline{B}. The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space I^I=\prod_{t \in I} I is not countably tight, we let S be the subspace of I^I consisting of points, each of which is non-zero on at most countably many coordinates. Specifically S is defined as follows:

    S=\Sigma_{t \in I} I=\left\{y \in I^I: y(t) \ne 0 \text{ for at most countably many } t \in I \right\}

The set S just defined is also called the \Sigma-product of copies of unit interval I. Let g \in I^I be defined by g(t)=1 for all t \in I. It follows that g \in \overline{S}. It can also be verified that g \notin \overline{B} for any countable B \subset S. This shows that the product space I^I=\prod_{t \in I} I is not countably tight.

By Theorem 1 found in this link, the space X=\omega_1 \times I^I is not normal.

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The space X has a dense subspace that is normal

Now that we know X=\omega_1 \times I^I is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace \omega_1 \times S where S is the \Sigma-product S=\Sigma_{t \in I} I defined in the preceding section. The subspace S is dense in the product space I^I. Thus \omega_1 \times S is dense in X=\omega_1 \times I^I. The space S is normal since the \Sigma-product of separable metric spaces is normal. Furthermore, \omega_1 can be embedded as a closed subspace of S=\Sigma_{t \in I} I. Then \omega_1 \times S is homeomorphic to a closed subspace of S \times S. Since S \times S \cong S, the space \omega_1 \times S is normal.

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Reference

  1. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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\copyright \ 2015 \text{ by Dan Ma}

Tietze-Urysohn-like theorems for completely regular spaces

Completely regular spaces (also called Tychonoff spaces) are topological spaces that come with a guarantee of having continuous real-valued functions in sufficient quantity. Thus the class of completely regular spaces is an ideal setting for many purposes that require the use of continuous real-valued functions (one example is working with function spaces). In a completely regular space X, for any closed set B and for any point x not in the closed set B, there always exists a continuous function f:X \rightarrow [0,1] such that f(x)=0 and f maps B to 1. It turns out that in such a space, we can replace the point x with any compact set A that is disjoint from the closed set B. This is a useful tool for proving theorems as well as for constructing objects. In this post, we discuss and prove this result (which resembles Urysohn’s lemma) and another useful fact about completely regular spaces that works very much like Tietze’s extension theorem. Specifically we prove the following results.

    Theorem 1

      Let X be a completely regular space. For any compact set A \subset X and for any closed set B \subset X that is disjoint from A, there exists a continuous function f:X \rightarrow [0,1] such that

      • f(x)=0 for all x \in A,
      • f(x)=1 for all x \in B.
    Theorem 2

      Let X be a completely regular space. For any compact set A \subset X, any continuous function f:A \rightarrow \mathbb{R} can be continuously extended over X, i.e., there exists a continuous function \hat{f}:X \rightarrow \mathbb{R} such that \hat{f}(x)=f(x) for all x \in A (in symbol we write \hat{f} \upharpoonright A=f).

A space X is normal if any two disjoint closed sets A \subset X and B \subset X can be separated by disjoint open sets, i.e., A \subset U and B \subset V for some disjoint open subsets U and V of X. Normal spaces are usually have the additional requirement that singleton sets are closed (i.e. T_1 spaces).

These two theorems remind us of two important tools for normal spaces, namely Urysohn’s lemma and Tietze’s extension theorem.

Urysohn’s lemma indicates that for any two disjoint closed sets in a normal space, the space can be mapped continuously to the closed unit interval [0,1] such that one closed set is mapped to 0 and the other closed set is mapped to 1. Theorem 1 is like a weakened version of Urysohn’s lemma in that one of the two disjoint closed sets must be compact.

Tietze’s extension theorem indicates that in a normal space, any continuous real-valued function defined on a closed subspace can be extended to the entire space. Theorem 2 is like a weakened version of Tiezte’s extension theorem in that the continuous extension only works for continuous functions defined on a compact subspace.

So if one only works in a completely regular space, one can still apply these two theorems about normal spaces (the weakened versions of course). For the sake of completeness, we state these two theorems about normal spaces.

    Urysohn’s Lemma

      Let X be a normal space. For any two disjoint closed sets A \subset X and B \subset X, there exists a continuous function f:X \rightarrow [0,1] such that

      • f(x)=0 for all x \in A,
      • f(x)=1 for all x \in B.
    Tietze’s Extension Theorem

      Let X be a normal space. For any closed set A \subset X, any continuous function f:A \rightarrow \mathbb{R} can be continuously extended over X, i.e., there exists a continuous function \hat{f}:X \rightarrow \mathbb{R} such that \hat{f}(x)=f(x) for all x \in A (in symbol we write \hat{f} \upharpoonright A=f).

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Proof of Theorem 1

We now prove Theorem 1. Let X be a completely regular space. Let A \subset X and B \subset X be two disjoint closed sets where A is compact. For each x \in A, there exists a continuous function f_x:X \rightarrow [0,1] such that f_x(x)=0 and f_x(B) \subset \left\{1 \right\}. The following collection is an open cover of the compact set A.

    \left\{f_x^{-1}([0,\frac{1}{10})): x \in A \right\}

Finitely many sets in this collection would cover A since A is compact. Choose x_1,x_2,\cdots,x_n \in A such that A \subset \bigcup \limits_{j=1}^n f_{x_n}^{-1}([0,\frac{1}{10})). Define h:X \rightarrow [0,1] by, for each x \in X, letting h(x) be the minimum of f_{x_1}(x),\cdots,f_{x_n}(x). It can be shown that the function h, being the minimum of finitely many continuous real-valued functions, is continuous. Furthermore, we have:

  • A \subset h^{-1}([0,\frac{1}{10})), and
  • h(B) \subset \left\{1 \right\}

Now define w:X \rightarrow [0,1] by, for each x \in X, letting w(x) be as follows:

    \displaystyle w(x)=\frac{10}{9} \cdot \biggl[ \text{max} \left\{h(x)-\frac{1}{10},0\right\} \biggr]

It can be shown that the function w is continuous. It is clear that w(x)=0 for all x \in A and w(x)=1 for all x \in B. \blacksquare

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Proof of Theorem 2

Interestingly, the proof of Theorem 2 given here uses Tietze’s extension theorem even Theorem 2 is described earlier as a weakened version of Tietze’s extension theorem. Beside using Tietze’s extension theorem, we also use the fact that any completely regular space can be embedded in a cube (see the previous post called Embedding Completely Regular Spaces into a Cube).

The proof is quite short once all the deep results that are used are understood. Let X be a completely regular space. Then X can be embedded in a cube, which is a product of the closed unit interval [0,1]. Thus X is homeomorphic to a subspace of the following product space

    Y=\prod \limits_{a \in S} I_a

for some index set S where I_a=[0,1] for all a \in S. We can now regard X as a subspace of the compact space Y. Let A \subset X be a compact subset of X. Let f:A \rightarrow \mathbb{R} be a continuous function.

The set A is a subset of X and can also be regarded as a subspace of the compact space Y, which is normal. Hence Tietze’s extension theorem is applicable in Y. Let \bar{f}:Y \rightarrow \mathbb{R} be a continuous extension of f. Let \hat{f}=\bar{f} \upharpoonright X. Then \hat{f} is the required continuous extension. \blacksquare

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\copyright \ 2014 \text{ by Dan Ma}

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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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

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Theorem C4 and Theorem U4

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

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

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

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

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Theorem C5 and Theorem U5

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

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

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

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Blog Posts on Stone-Cech Compactification

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Reference

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

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\copyright \ \ 2012

Stone-Cech Compactification of the Integers – Basic Facts

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

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

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

    \text{ }

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

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Characterization Theorems

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

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

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

    \text{ }

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

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

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

    \text{ }

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

    \text{ }

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

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

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

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Result 1 and Result 2

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

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

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

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Result 3

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

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

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

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

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

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

Note that Lemma 3 is a corollary of Theorem 2.

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

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

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

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

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

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Result 4 and Result 5

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

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

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

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

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

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

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

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

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

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

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

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Result 6

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

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Result 7

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

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

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

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Result 8

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

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Result 9

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

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

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

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Blog Posts on Stone-Cech Compactification

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Reference

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

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\copyright \ \ 2012

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)

    _________________________________
    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.

    \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

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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