Revisiting example 106 from Steen and Seebach

Posted on December 25, 2021 by Dan Ma

The example 106 from Counterexamples in Topology by Steen and Seebach [3] is the space $\omega_1 \times I^I$ where the first factor $\omega_1$ is the space of countable ordinals with the usual order topology and the second factor $I^I$ is the product of continuum many copies of the unit interval $I=[0,1]$ .

This space was previously discussed in this site. One of the key results from that discussion is that $\omega_1 \times I^I$ is not normal, a result not shown in Steen and Seebach. The proof that was given in this site (see here) is based on an article published in 1976 [1], long before the publication date of the first edition of Steen and Seebach in 1970. It turns out that the non-normality of $\omega_1 \times I^I$ was given as an exercise in Steen and Seebach in the problem section at the end of the book (problem 127 in page 211, Dover edition). Problem 127: Show that $[0, \Omega) \times I^I$ is not normal. This indicates that the result not shown in Steen and Seebach was because it was given as a problem and not because the tool for solving it was not yet available. The fact that it is given as an exercise also means that there is a more basic proof of the non-normality of $\omega_1 \times I^I$ . So, once this is realized, I set out to find a simpler proof or at least one that does not rely on the result from [1]. Interestingly, this proof brings out a broader discussion that is worthwhile and goes beyond the example at hand. The goal here is to examine the more basic proof and the broader discussion.

A Classic Example

Before talking about the promised proof, we consider the product of $\omega_1$ and its immediate successor.

As noted at the beginning, the space $\omega_1$ is the set of all countable ordinals with the order topology. The ordinal $\omega_1+1$ is the immediate successor of $\omega_1$ . It can be regarded as the result of adding one more point to $\omega_1$ . The extra point is $\omega_1$ , i.e., $\omega_1 +1=\omega_1 \cup \{ \omega_1 \}$ with $\omega_1$ greater than all points $\beta < \omega_1$ . The ordinal $\omega_1+1$ with the order topology is a compact space. Using interval notation, $\omega_1=[0, \omega_1)$ and $\omega_1+1=[0, \omega_1]$ . As ordinals, $\omega_1$ is the first uncountable ordinal and $\omega_1+1$ is the first uncountable successor ordinal. For more information, see here.

The product $[0, \omega_1) \times [0, \omega_1]$ is a classic example of a product of a normal space (the first factor) and a compact space (the second factor) that is not normal. This example and others like it show that normality is easily broken upon taking product even if one of the factors is as nice as a compact space. The non-normality of $[0, \omega_1) \times [0, \omega_1]$ is discussed here. In that proof, two disjoint closed sets $H$ and $K$ are given such that they cannot be separated by disjoint open sets. The $H$ and $K$ are:

$H=\{ (\alpha, \alpha): \alpha < \omega_1 \}$

$K=\{(\alpha, \omega_1): \alpha < \omega_1 \}$

The Basic Proof

To show that $\omega_1 \times I^I$ is not normal, we show that one of its closed subspaces is not normal. That closed subspace is $[0, \omega_1) \times [0, \omega_1]$ . To this end, we show that $[0, \omega_1]$ can be embedded in the product space $I^I$ . With a non-normal closed subspace, it follows that $\omega_1 \times I^I$ is not normal. The remainder of the proof is to give the embedding.

We show that $[0, \omega_1]$ can be embedded as a closed subspace of $I^{\omega_1}$ , the product of $\omega_1$ many copies of $I$ . This means that $[0, \omega_1]$ is also a closed subspace of $I^I$ .

For each $\beta < \omega_1$ , define $T_\beta: \omega_1 \rightarrow I$ as follows:

$T_\beta(\gamma) = \begin{cases} 1 & \ \ \ \mbox{if } \gamma < \beta \\ 0 & \ \ \ \mbox{if } \gamma \ge \beta \end{cases}$

Furthermore, define $T: \omega_1 \rightarrow I$ by letting $T(\beta)=1$ for all $\beta < \omega_1$ . Consider the correspondence $\beta \rightarrow T_\beta$ with $\beta < \omega_1$ and $\omega_1 \rightarrow T$ . The mapping is clearly one-to-one from $[0, \omega_1]$ onto $\{ T_\beta: \beta < \omega_1 \} \cup \{ T \}$ . Upon closer inspection, the mapping in each direction is continuous (this is a good exercise to walk through). Thus, the mapping is a homeomorphism. It follows that $[0, \omega_1]$ can be considered a subspace of $I^{\omega_1}$ . Since $[0, \omega_1]$ is compact, it must be a closed subspace. With the cardinality of $\omega_1$ being less than or equal to continuum, it follows that $[0, \omega_1]$ can be embedded as a closed subspace of $I^I$ .

Stone-Cech Compactification

The first broader discussion is that of Stone-Cech compactification. More specifically, $\beta \omega_1=\omega_1+1$ , i.e., the Stone-Cech compactification of the first uncountable ordinal is its immediate successor.

To see that $\beta \omega_1=\omega_1+1$ , note that every continuous function defined on $[0,\omega_1)$ is bounded and is eventually constant (see result B here). As a result, every continuous function defined on $[0,\omega_1)$ can be extended to a continuous function defined on $[0, \omega_1]$ . For any continuous function $f: \omega_1 \rightarrow \mathbb{R}$ , we can simply define $f(\omega_1)$ to be the eventual constant value. A subspace $W$ of a space $Y$ is $C^*$ -embedded in $Y$ if every bounded continuous real-valued function on $W$ can be extended to $Y$ . According to theorem 19.12 in [4], if $Y$ is a compactification of $X$ and if $X$ is $C^*$ -embedded in $Y$ , then $Y$ is the Stone-Cech compactification of $X$ . Thus $[0,\omega_1)$ is $C^*$ -embedded in $[0,\omega_1]$ and $[0,\omega_1]$ is the Stone-Cech compactification of $[0,\omega_1)$ . In this instance, the Stone-Cech compactification agrees with the one-point compactification. Consider the following class theorem about normality in product space. The theorem is Corollary 3.4 in the chapter on products of normal spaces in the handbook of set-theoretic topology [2].

Theorem 1
Let $X$ be a space. The following conditions are equivalent.

The space $X$ is paracompact.
The product space $X \times \beta X$ is normal.

Based on the discussion presented above, the non-normality of $\omega_1 \times I^I$ is due to the non-normality of $[0, \omega_1) \times [0, \omega_1]$ . Based on this theorem, the non-normality of $[0, \omega_1) \times [0, \omega_1]$ is due to the non-paracompactness of $[0, \omega_1)$ . See result G here for a proof that $[0, \omega_1)$ is not paracompact.

The discussion up to this point points to two ways to prove that $\omega_1 \times I^I$ is not normal. One way is the basic proof indicated above. The other way is to use Theorem 1, along with the homeomorphic embedding from $[0, \omega_1]$ into $I^I$ , the fact that $\beta \omega_1=\omega_1+1$ and the fact that $[0, \omega_1)$ is not paracompact. Both are valuable. The first way is basic and is a constructive proof. Because it is more hands-on, it is a better proof to learn from. The second way provides a broader perspective that is informative but requires quoting a couple of fairly deep results. Perhaps it is best used as a second proof for perspective.

Countable Tightness

The essence of the basic proof above goes like this: if the space $Y$ contains a copy of $\omega_1+1=[0, \omega_1]$ , then the product space $[0, \omega_1) \times Y$ is not normal. The contrapositive statement would be the following:

Corollary
Let $Y$ be a space. If the product space $\omega_1 \times Y$ is normal, then $Y$ cannot contain a copy of $\omega_1+1$ .

In the space of $\omega_1+1=[0, \omega_1]$ , note the following about the last point: $\omega_1 \in \overline{[0, \omega_1)}$ but $\omega_1 \notin \overline{C}$ for any countable $C \subset [0, \omega_1)$ , i.e., the last point is the limit point of the set of all the points preceding it but is not in the closure of any countable set. This means that the space $\omega_1+1=[0, \omega_1]$ does not have countable tightness (or is not countably tight). See here for definition. The property of countable tightness is hereditary. If $Y$ contains a copy of $\omega_1+1$ , then $Y$ is not countably tight (or is uncountably tight). This brings us to the following theorem.

Theorem 2
Let $Y$ be an infinite compact space. Then $\omega_1 \times Y$ is normal if and only if $Y$ has countable tightness.

Whenever we consider the normality of a product with the first factor being $\omega_1$ and the second factor being a compact space, the real story is the tightness of that compact space. If the tightness is countable, the product is normal. Otherwise, the product is not normal. The theorem is another reason that $\omega_1 \times I^I$ is not normal. Instead of embedding $[0, \omega_1]$ into $I^I$ , we can actually show that $I^I$ does not have countable tightness. This is the approach that was taken in this previous post.

Theorem 2 is the result from 1976 alluded to earlier [1]. A proof of Theorem 2 is found in this previous post. For results concerning normality in a product space with a compact factor (the other factor does not have to be $\omega_1$ ), see the chapter on products of normal spaces in the handbook of set-theoretic topology [2].

Reference

Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.
Przymusinski T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

$\text{ }$

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$\copyright$ 2021 – Dan Ma

Counterexample 106 from Steen and Seebach

Posted on August 18, 2015 by Dan Ma

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

The counterexample 106 is the space $X=\omega_1 \times I^I$ , which is the product of $\omega_1$ with the interval topology (ordered 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 basic topological properties of the space $X=\omega_1 \times I^I$ that are covered in [2] are:

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.

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 tool in [1] is this theorem: for any compact space $Y$ , the product $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight. For a proof of this theorem, see here. 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 define $S$ as follows. For each finite $A \subset I=[0,1]$ , define $f_A: I \rightarrow I$ such that $f_A$ maps $A$ to 1 and maps $I \backslash A$ to 0. Let $S$ be the set of $f_A$ for all possible finite $A \subset I$ . Let $f:I \rightarrow I$ be defined by $f(x)=1$ for all $x \in I$ .

It follows that $f \in \overline{S}$ . We claim that for any countable $B \subset S$ , $f \notin \overline{B}$ . Let $B=\{f_{A_1},f_{A_2}, \cdots \} \subset S$ be countable where each $A_j$ is finite. Then choose $a \in I \backslash \bigcup_{j} A_j$ . Consider the open set $U=\prod_{t \in I} W_t$ where $W_t=I$ for $t \ne a$ and $W_a=(0.5,1]$ . Then $f \in U$ and $U \cap B=\varnothing$ . Thus $f \notin \overline{B}$ . 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.

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 $S=\Sigma_{t \in I} I$ where $S$ is the space of all $f \in I^I$ such that for each $f$ , $f(x) \ne 0$ for at most countably many $x \in I$ .

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 (see here). 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 $S \times S$ . Note that $S \times S \cong S$ . Since $\omega_1 \times S$ can be embedded as a closed subspace of the normal space $S$ , the space $\omega_1 \times S$ is normal.

Reference

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

$\copyright$ 2015 Dan Ma

Revised January 28, 2021

Tietze-Urysohn-like theorems for completely regular spaces

Posted on January 22, 2014 by Dan Ma

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

$X$

$A \subset X$

$B \subset X$

$A$

$f:X \rightarrow [0,1]$

$f(x)=0$ for all $x \in A$ ,
$f(x)=1$ for all $x \in B$ .

Theorem 2

$X$

$A \subset X$

$f:A \rightarrow \mathbb{R}$

$X$

$\hat{f}:X \rightarrow \mathbb{R}$

$\hat{f}(x)=f(x)$

$x \in A$

$\hat{f} \upharpoonright A=f$

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

$X$

$A \subset X$

$B \subset X$

$f:X \rightarrow [0,1]$

$f(x)=0$ for all $x \in A$ ,
$f(x)=1$ for all $x \in B$ .

Tietze’s Extension Theorem

$X$

$A \subset X$

$f:A \rightarrow \mathbb{R}$

$X$

$\hat{f}:X \rightarrow \mathbb{R}$

$\hat{f}(x)=f(x)$

$x \in A$

$\hat{f} \upharpoonright A=f$

____________________________________________________________________

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

Posted on October 11, 2012 by Dan Ma

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

$\left\{k_1,k_2 \right\} \subset \cap \mathcal{W}(t)$

$k_1 \ne k_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

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

Embedding Completely Regular Spaces into a Cube

Post #1: A Beginning Look at Stone-Cech Compactification

Post #2: Two Characterizations of Stone-Cech Compactification

Post #3: C*-Embedding Property and Stone-Cech Compactification

Post #4: Stone-Cech Compactification is Maximal

Post #5: Stone-Cech Compactification of the Integers – Basic Facts

Post #6 (this post): Stone-Cech Compactifications – Another Two Characterizations

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Reference

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

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$\copyright \ \ 2012$

Stone-Cech Compactification of the Integers – Basic Facts

Posted on October 1, 2012 by Dan Ma

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.

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

$X$

$f:X \rightarrow Y$

$X$

$Y$

$F: \beta X \rightarrow Y$

$F \circ \beta=f$

$\text{ }$

Theorem U1

$K$

$X$

$K$

$\beta X$

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

Theorem C2

$X$

$\beta X$

$X$

$\le$

$\text{ }$

Theorem U2

$\beta X$

$X$

$\alpha X$

$\le$

$\alpha X \approx \beta X$

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

Theorem C3

$X$

$C^*$

$\beta X$

$\text{ }$

Theorem U3.1

$X$

$I=[0,1]$

$\alpha X$

$X$

$f:X \rightarrow I$

$\hat{f}:\alpha X \rightarrow I$

$\alpha X$

$\beta X$

$\text{ }$

Theorem U3.2

$\alpha X$

$X$

$C^*$

$\alpha X$

$\beta X$

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

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

$X$

$H$

$K$

$X$

$H$

$K$

$\beta X$

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

$X$

$H$

$X$

$\overline{H}$

$H$

$\beta X$

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

$A \subset \omega$

$\overline{A}$

$A$

$\beta \omega$

Note that Lemma 3 is a corollary of Theorem 2.

Lemma 4

$O \subset \beta \omega$

$\beta \omega$

$O=\overline{A}$

$A= O \cap \omega$

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

$A \subset \omega$

$\overline{A}$

$A$

$\beta \omega$

$2^c$

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

$A \subset \beta \omega-\omega$

$A$

$\overline{A}$

$A$

$\beta \omega$

$2^c$

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

Embedding Completely Regular Spaces into a Cube

Post #1: A Beginning Look at Stone-Cech Compactification

Post #2: Two Characterizations of Stone-Cech Compactification

Post #3: C*-Embedding Property and Stone-Cech Compactification

Post #4: Stone-Cech Compactification is Maximal

Post #5 (this post): Stone-Cech Compactification of the Integers – Basic Facts

Post #6: Stone-Cech Compactifications – Another Two Characterizations

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Reference

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

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$\copyright \ \ 2012$

Stone-Cech Compactification is Maximal

Posted on September 24, 2012 by Dan Ma

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

$\lvert \beta X \lvert \le 2^{2^{d(X)}}$ .
$w(\beta X) \le 2^{d(X)}$ .

Stone-Cech Compactification is Maximal

For every compactification $\alpha X$ of the space $X$ , $\lvert \alpha X \lvert \le \lvert \beta X \lvert$ .
For every compactification $\alpha X$ of the space $X$ , $w(\alpha X) \le w(\beta X)$ .

Upper Bounds for all Compactifications

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

$X$

$C(X)$

$f:X \rightarrow \mathbb{R}$

$\lvert C(X) \lvert \le 2^{d(X)}$

Lemma 2

$S$

$T$

$S$

$w(T) \le w(S)$

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

$X$

$\beta X$

$X$

$\le$

$\text{ }$

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

Embedding Completely Regular Spaces into a Cube

Post #1: A Beginning Look at Stone-Cech Compactification

Post #2: Two Characterizations of Stone-Cech Compactification

Post #3: C*-Embedding Property and Stone-Cech Compactification

Post #4 (this post): Stone-Cech Compactification is Maximal

Post #5: Stone-Cech Compactification of the Integers – Basic Facts

Post #6: Stone-Cech Compactifications – Another Two Characterizations

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Reference

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

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$\copyright \ \ 2012$

C*-Embedding Property and Stone-Cech Compactification

Posted on September 23, 2012 by Dan Ma

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

$X$

$I=[0,1]$

$\alpha X$

$X$

$f:X \rightarrow I$

$\hat{f}:\alpha X \rightarrow I$

$\alpha X$

$\beta X$

$\text{ }$

Theorem U3.2

$\alpha X$

$X$

$C^*$

$\alpha X$

$\beta X$

$\text{ }$

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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

$X$

$f:X \rightarrow Y$

$X$

$Y$

$F: \beta X \rightarrow Y$

$F \circ \beta=f$

$\text{ }$

Theorem U1

$K$

$X$

$K$

$\beta X$

$\text{ }$

Theorem C2

$X$

$\beta X$

$X$

$\text{ }$

Theorem U2

$\beta X$

$\alpha X$

$X$

$\alpha X$

$\beta X$

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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:

$t \in \alpha X$

$H(t)=a=< a_f >_{f \in C(X,I)}$

$a_f=\hat{f}(t)$

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

Embedding Completely Regular Spaces into a Cube

Post #1: A Beginning Look at Stone-Cech Compactification

Post #2: Two Characterizations of Stone-Cech Compactification

Post #3 (this post): C*-Embedding Property and Stone-Cech Compactification

Post #4: Stone-Cech Compactification is Maximal

Post #5: Stone-Cech Compactification of the Integers – Basic Facts

Post #6: Stone-Cech Compactifications – Another Two Characterizations

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Reference

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

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Two Characterizations of Stone-Cech Compactification

Posted on September 21, 2012 by Dan Ma

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

$X$

$f:X \rightarrow Y$

$X$

$Y$

$F: \beta X \rightarrow Y$

$F \circ \beta=f$

$\text{ }$

Theorem U1

$K$

$X$

$K$

$\beta X$

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:

$x \in X$

$\beta_X(x)=t=< t_g >_{g \in C(X,I)}$

$t \in \prod \limits_{g \in C(X,I)} I_g$

$t_g=g(x)$

$g \in C(X,I)$

$g^{th}$

$t$

$g(x)$

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

Definition

$\beta_X$

$\beta_X(X)$

$X$

$\prod \limits_{f \in C(X,I)} I_f$

$X$

$\beta_X(X)$

$\prod \limits_{f \in C(X,I)} I_f$

$\beta_X X=\overline{\beta_X(X)}$

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.

$x \in X$

$\downarrow$

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

$\downarrow$

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

_________________________________

$x \in X$

$\downarrow$

$f(x) \in Y$

$\downarrow$

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

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

$\alpha_1 X$

$\alpha_2 X$

$X$

$\alpha_1 X \le \alpha_2 X$

$\alpha_2 X \le \alpha_1 X$

$\alpha_1 X \approx \alpha_2 X$

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

$X$

$\beta X$

$X$

$\le$

$\text{ }$

Theorem U2

$\beta X$

$X$

$\alpha X$

$\le$

$\alpha X \approx \beta X$

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

Embedding Completely Regular Spaces into a Cube

Post #1: A Beginning Look at Stone-Cech Compactification

Post #2 (this post): Two Characterizations of Stone-Cech Compactification

Post #3: C*-Embedding Property and Stone-Cech Compactification

Post #4: Stone-Cech Compactification is Maximal

Post #5: Stone-Cech Compactification of the Integers – Basic Facts

Post #6: Stone-Cech Compactifications – Another Two Characterizations

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Reference

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

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Embedding Completely Regular Spaces into a Cube

Posted on August 30, 2012 by Dan Ma

This is a continuation of a discussion on completely regular spaces (continuing from these two posts: Completely Regular Spaces and Pseudocompact Spaces, Completely Regular Spaces and Function Spaces). This post gives another reason (one of the most important ones) why the class of completely regular spaces occupies a central place in general topology, which is that completely regular spaces are precisely the spaces that can be embedded in a cube (the product of copies of the unit interval). This theorem was proved by Tychonoff in 1930 . The tool that makes this theorem possible also allowed Stone and Cech in 1937 to construct for any completely regular space $X$ , a compact Hausdorff space $\beta X$ that contains $X$ as a dense set ( $\beta X$ is called the Stone-Cech compactification of $X$ ). In this post we discuss the role played by complete regularity in this construction. We discuss the following theorem.

Theorem 1
Let $X$ be a space. Then $X$ is completely regular if and only if $X$ is homeomorphic to a subspace of a cube.

Original articles are [1], [4] and [5]. The Stone-Cech compactification and the related concepts are classic topics that are covered in standard texts (see [2] and [6]).

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Completely Regular Spaces

A space $X$ is said to be completely regular if $X$ is a $T_0$ space and for each $x \in X$ and for each closed subset $A$ of $X$ with $x \notin A$ , there is a continuous function $f:X \rightarrow [0,1]$ such that $f(A) \subset \left\{0 \right\}$ and $f(x)=1$ . Note that the $T_0$ axiom and the existence of the continuous function imply the $T_1$ axiom, which is equivalent to the property that single points are closed sets. Completely regular spaces are also called Tychonoff spaces.

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The Evaluation Map

The evaluation map can be defined in a more general setting. We define it here just to deal with the task at hand, namely to discuss Theorem 1.

Let $X$ be a completely regular space. Let $I=[0,1]$ be the unit interval in the real line $\mathbb{R}$ . A cube is of the form $I^\mathcal{K}$ , i.e., the product of $\mathcal{K}$ many copies of $I$ where $\mathcal{K}$ is some cardinal. Let $C(X,I)$ be the set of all continuous real-valued functions defined on the space $X$ . Consider the product space $\prod \limits_{f \in C(X,I)} I_f$ where each $I_f=I$ . This is the cube where $X$ is embedded as a subspace. We can represent each point in the cube $\prod \limits_{f \in C(X,I)} I_f$ as a function $H:C(X,I) \rightarrow I$ or as a sequence $< H_f >_{f \in C(X,I)}$ such that each term (or coordinate) $H_f \in I=[0,1]$ .

The key to embed $X$ into a cube is through the evaluation map, which is a map $E$ from $X$ into the product space $\prod \limits_{f \in C(X,I)} I_f$ . Thus we have:

$\displaystyle (1) \ \ \ \ \ \ E:X \longrightarrow \prod \limits_{f \in C(X,I)} I_f$

We now define the map $E$ . For each $x \in X$ , $E(x)=< H_f >_{f \in C(X,I)}$ such that $H_f=f(x)$ for each $f \in C(X,I)$ . In other words, $E(x)$ is the point in the product space $\prod \limits_{f \in C(X,I)} I_f$ whose $f^{th}$ coordinate is $f(x)$ .

We show that because $X$ is completely regular, the evaluation map $E$ is a homeomorphism. We show the following:

The map $E$ is continuous.
The map $E$ is one-to-one.
The map $E^{-1}$ is continuous.

The continuity of the map $E$ follows from the fact that each $f \in C(X,I)$ is continuous. The map $E$ being one-to-one follows from the fact that for each pair $x,y \in X$ with $x \ne y$ , there is an $f \in C(X,I)$ such that $f(x) \ne f(y)$ .

We now show $E^{-1}$ , the inverse of $E$ , is continuous. This is where $X$ must be completely regular. Let $U$ be open in $X$ such that $x =E^{-1}(E(x)) \in U$ . Since $X$ is completely regular, there is a continuous $g:X \rightarrow I$ such that $g(X-U) \subset \left\{0 \right\}$ and $g(x)=1$ . Let $V=\prod \limits_{f \in C(X,I)} I_f$ where $I_f=I$ for all $f \ne g$ and $I_g=(0,1]$ . Then $V_0=V \cap E(X)$ is open in $E(X)$ and $E(x) \in V_0$ . We claim that $E^{-1}(V_0) \subset U$ . To see this, pick $T=E(y) \in V_0$ . Note that $T_g \in (0,1]$ since $T \in V_0$ . If $y \in X-U$ , then $T_g =0$ . Thus we must have $y =E^{-1}(E(y)) \in U$ .

The above discussion shows that any completely regular space $X$ is homeomorphic to a subspace of the product space $[0,1]^\mathcal{K}$ where $\mathcal{K}$ is the cardinality of $C(X,I)$ . This establishes one direction of Theorem 1. The other direction is clear. Note that any cube is a compact Hausdorff space and any subspace of a compact Hausdorff space is completely regular.

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The Stone-Cech Compactification

Let $X$ be a completely regular space. Let $E$ be the evaluation as defined in $(1)$ above. According to the above discussion $E$ is a homeomorphism. Hence $E(X)$ is a topological copy of $X$ as a subspace of the product space $\prod \limits_{f \in C(X,I)} I_f$ . Consider $\overline{E(X)}$ where the closure is taken in the product space. The Stone-Cech compactification of $X$ is denoted by $\beta X$ and is defined to be this closure $\beta X=\overline{E(X)}$ .

See [2] and [6] for more information.

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Reference

Cech, E., On bicompact spaces, Ann. Math. (2) 38, 823-844, 1937.
Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.
Stone, M. H., Applications of the Theorey of Boolean Rings to General Topology, Trans. Amer. Math. Soc., 41, 375-481, 1937.
Tychonoff, A., Uber die topologische von Raumen, Math., Ann., 102, 544-561, 1930.
Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

Examples of Lindelof Spaces that are not Hereditarily Lindelof

Posted on April 15, 2012 by Dan Ma

We observe from the following statement two examples of Lindelof spaces that are not hereditarily Lindelof.

Any product space contains a discrete subspace having the same cardinality as the number of factor spaces.

Using the above observation, by choosing the factor spaces judiciously, the product of uncountably many spaces is a handy way of obtaining Lindelof spaces (in some cases $\sigma$ -compact spaces) that are not hereditarily Lindelof. For definition and basic information about product spaces, see this previous post.

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All spaces under consideration are at least Hausdorff. For each $\alpha \in A$ , let $X_\alpha$ be a space with at least two points. For each $\alpha \in A$ , fix two points $p_\alpha, q_\alpha \in X_\alpha$ . Then the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ contains a discrete subspace $Y$ that has the same cardinality as the cardinality of the index set $A$ .

For each $\alpha \in A$ , define $y_\alpha \in \prod \limits_{\alpha \in A} X_\alpha$ by the following:

$\displaystyle (1) \ \ \ \ \ \ y_\alpha(\gamma)=\left\{\begin{matrix}p_\alpha&\ \gamma=\alpha\\{q_\alpha}&\ \gamma \ne \alpha \end{matrix}\right.$

Let $Y=\left\{ y_\alpha: \alpha \in A\right\}$ . It follows that $\lvert Y \lvert = \lvert A \lvert$ and that $Y$ is a discrete space.

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Whenever the index set $A$ is uncountable, the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ contains an uncountable discrete subspace. Thus even if the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is Lindelof, one of its subspace $Y$ cannot be Lindelof. Taking the product of uncountably many factor spaces is a handy way to obtain Lindelof space that is not hereditarily Lindelof. Some examples are shown below.

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Examples

Let the index set $A$ be uncountable. To make the product space Lindelof, we can make every one of its factor $X_\alpha$ compact. Thus the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is compact and not hereditarily Lindelof.

Thus the product space $[0,1]^{\omega_1}$ , the product of $\omega_1$ many copies of the unit interval, is compact and not hereditarily Lindelof. Another example is $\left\{ 0,1 \right\}^{\omega_1}$ , the product of $\omega_1$ many copies of $\left\{ 0,1 \right\}$

Another way to make the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ Lindelof is to make some of the factors compact such that the product of the remaining non-compact factors is Lindelof. Then the product space is essentially the product of a compact space and a Lindelof space, which is always Lindelof.

For example, let $X_0=\mathbb{R}$ and let $X_\alpha=[0,1]$ for all $\alpha$ with $0<\alpha<\omega_1$ . Then the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is Lindelof since it is essentially the product of a compact space and a Lindelof space. However, the product $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is not hereditarily Lindelof.

In fact, the product space in the previous paragraph is $\sigma$ -compact (i.e. the union of countably many compact sets). To make the example not $\sigma$ -compact, simply make the first factor space a non-locally compact Lindelof space. For example, use the Sorgenfrey line or the space of the irrational numbers.

Dan Ma's Topology Blog

Expository Articles In General Topology

Category Archives: Product of unit interval

Revisiting example 106 from Steen and Seebach

Counterexample 106 from Steen and Seebach

Tietze-Urysohn-like theorems for completely regular spaces

Stone-Cech Compactifications – Another Two Characterizations

Stone-Cech Compactification of the Integers – Basic Facts

Stone-Cech Compactification is Maximal

C*-Embedding Property and Stone-Cech Compactification

Two Characterizations of Stone-Cech Compactification

Embedding Completely Regular Spaces into a Cube

Examples of Lindelof Spaces that are not Hereditarily Lindelof