Looking for non-normal subspaces of the square of a compact X

Posted on April 14, 2015 by Dan Ma

A theorem of Katetov states that if $X$ is compact with a hereditarily normal cube $X^3$ , then $X$ is metrizable (discussed in this previous post). This means that for any non-metrizable compact space $X$ , Katetov’s theorem guarantees that some subspace of the cube $X^3$ is not normal. Where can a non-normal subspace of $X^3$ be found? Is it in $X$ , in $X^2$ or in $X^3$ ? In other words, what is the “dimension” in which the hereditary normality fails for a given compact non-metrizable $X$ (1, 2 or 3)? Katetov’s theorem guarantees that the dimension must be at most 3. Out of curiosity, we gather a few compact non-metrizable spaces. They are discussed below. In this post, we motivate an independence result using these examples.

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Katetov’s theorems

First we state the results of Katetov for reference. These results are proved in this previous post.

Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

The factor $X$ is perfectly normal.
Every countable and infinite subset of the factor $Y$ is closed.

Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.

Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.

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Examples of compact non-metrizable spaces

The set-theoretic result presented here is usually motivated by looking at Theorem 3. The question is: Can $X^3$ in Theorem 3 be replaced by $X^2$ ? We take a different angle of looking at some standard compact non-metric spaces and arrive at the same result. The following is a small listing of compact non-metrizable spaces. Each example in this list is defined in ZFC alone, i.e. no additional axioms are used beyond the generally accepted axioms of set theory.

One-point compactification of the Tychonoff plank.
One-point compactification of $\psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$ .
The first compact uncountable ordinal, i.e. $\omega_1+1$ .
The one-point compactification of an uncountable discrete space.
Alexandroff double circle.
Double arrow space.
Unit square with the lexicographic order.

Since each example in the list is compact and non-metrizable, the cube of each space must not be hereditarily normal according to Theorem 3 above. Where does the hereditary normality fail? For #1 and #2, $X$ is a compactification of a non-normal space and thus not hereditarily normal. So the dimension for the failure of hereditary normality is 1 for #1 and #2.

For #3 through #7, $X$ is hereditarily normal. For #3 through #5, each $X$ has a closed subset that is not a $G_\delta$ set (hence not perfectly normal). In #3 and #4, the non- $G_\delta$ -set is a single point. In #5, the the non- $G_\delta$ -set is the inner circle. Thus the compact space $X$ in #3 through #5 is not perfectly normal. By Theorem 2, the dimension for the failure of hereditary normality is 2 for #3 through #5.

For #6 and #7, each $X^2$ contains a copy of the Sorgenfrey plane. Thus the dimension for the failure of hereditary normality is also 2 for #6 and #7.

In the small sample of compact non-metrizable spaces just highlighted, the failure of hereditary normality occurs in “dimension” 1 or 2. Naturally, one can ask:

Question

$X$

$X^2$

$X^3$

Such a space $X$ would be an example to show that the condition “ $X^3$ is hereditarily normal” in Theorem 3 is necessary. In other words, the hypothesis in Theorem 3 cannot be weakened if the example just described were to exist.

The above list of compact non-metrizable spaces is a small one. They are fairly standard examples for compact non-metrizable spaces. Could there be some esoteric example out there that fits the description? It turns out that there are such examples. In [1], Gruenhage and Nyikos constructed a compact non-metrizable $X$ such that $X^2$ is hereditarily normal. The construction was done using MA + not CH (Martin’s Axiom coupled with the negation of the continuum hypothesis). In that same paper, they also constructed another another example using CH. With the examples from [1], one immediate question was whether the additional set-theoretic axioms of MA + not CH (or CH) was necessary. Could a compact non-metrizable $X$ such that $X^2$ is hereditarily normal be still constructed without using any axioms beyond ZFC, the generally accepted axioms of set theory? For a relatively short period of time, this was an open question.

In 2001, Larson and Todorcevic [3] showed that it is consistent with ZFC that every compact $X$ with hereditarily normal $X^2$ is metrizable. In other words, there is a model of set theory that is consistent with ZFC in which Theorem 3 can be improved to assuming $X^2$ is hereditarily normal. Thus it is impossible to settle the above question without assuming additional axioms beyond those of ZFC. This means that if a compact non-metrizable $X$ is constructed without using any axiom beyond ZFC (such as those in the small list above), the hereditary normality must fail at dimension 1 or 2. Numerous other examples can be added to the above small list. Looking at these ZFC examples can help us appreciate the results in [1] and [3]. These ZFC examples are excellent training ground for general topology.

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Reference

Gruenhage G., Nyikos P. J., Normality in $X^2$ for Compact $X$ , Trans. Amer. Math. Soc., Vol 340, No 2 (1993), 563-586
Katetov M., Complete normality of Cartesian products, Fund. Math., 35 (1948), 271-274
Larson P., Todorcevic S., KATETOV’S PROBLEM, Trans. Amer. Math. Soc., Vol 354, No 5 (2001), 1783-1791

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$\copyright \ 2015 \text{ by Dan Ma}$

When a product space is hereditarily normal

Posted on April 14, 2015 by Dan Ma

When the spaces $X$ and $Y$ are normal spaces, the product space $X \times Y$ is not necessarily normal. Even if one of the factors is metrizable, there is still no guarantee that the product is normal. So it is possible that the normality of each of the factors $X$ and $Y$ can have no influence on the normality of the product $X \times Y$ . The dynamics in the other direction are totally different. When the product $X \times Y$ is hereditarily normal, the two factors $X$ and $Y$ are greatly impacted. In this post, we discuss a theorem of Katetov, which shows that the hereditary normality of the product can impose very strict conditions on the factors, which lead to many interesting results. This theorem also leads to an interesting set-theoretic result, and thus can possibly be a good entry point to the part of topology that deals with consistency and independence results – statements that cannot be proved true or false based on the generally accepted axioms of set theory (ZFC). In this post, we discuss Katetov’s theorem and its consequences. In the next post, we discuss examples that further motivate the set-theoretic angle.

A subset $W$ of a space $X$ is said to be a $G_\delta$ -set in $X$ if $W$ is the intersection of countably many open subsets of $X$ . A space $X$ is perfectly normal if it is normal and that every closed subset of $X$ is a $G_\delta$ -set. Some authors use other statements to characterize perfect normality (here is one such characterization). Perfect normality implies hereditarily normal (see Theorem 6 in this previous post). The implication cannot be reversed. Katetov’s theorem implies that the hereditary normality of the product $X \times Y$ will in many cases make one or both of the factors perfectly normal. Thus the hereditary normality in the product $X \times Y$ is a very strong property.

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Katetov’s theorems

Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

The factor $X$ is perfectly normal.
Every countable and infinite subset of the factor $Y$ is closed.

Proof of Theorem 1
The strategy we use is to define a subspace of $X \times Y$ that is not normal after assuming that none of the two conditions is true. So assume that $X$ has a closed subspace $W$ that is not a $G_\delta$ -set and assume that $T=\left\{t_n: n=1,2,3,\cdots \right\}$ is an infinite subset of $Y$ that is not closed. Let $p \in Y$ be a limit point of $T$ such that $p \notin T$ . The candidate for a non-normal subspace of $X \times Y$ is:

$M=X \times Y-W \times \left\{p \right\}$

Note that $M$ is an open subspace of $X \times Y$ since it is the result of subtracting a closed set from $X \times Y$ . The following are the two closed sets that demonstrate that $M$ is not normal.

$H=W \times (Y-\left\{p \right\})$

$K=(X-W) \times \left\{p \right\}$

It is clear that $H$ and $K$ are closed subsets of $M$ . Let $U$ and $V$ be open subsets of $M$ such that $H \subset U$ and $K \subset V$ . We show that $U \cap V \ne \varnothing$ . To this end, define $U_j=\left\{x \in X: (x,t_j) \in U \right\}$ for each $j$ . It follows that for each $j$ , $W \subset U_j$ . Furthermore each $U_j$ is an open subspace of $X$ . Thus $W \subset \bigcap_j U_j$ . Since $W$ is not a $G_\delta$ -set in $X$ , there must exist $t \in \bigcap_j U_j$ such that $t \notin W$ . Then $(t, p) \in K$ and $(t, p) \in V$ .

Since $V$ is open in the product $X \times Y$ , choose open sets $A \subset X$ and $B \subset Y$ such that $(t,p) \in A \times B$ and $A \times B \subset V$ . With $p \in B$ , there exists some $j$ such that $t_j \in B$ . First, $(t,t_j) \in V$ . Since $t \in U_j$ , $(t,t_j) \in U$ . Thus $U \cap V \ne \varnothing$ . This completes the proof that the subspace $M$ is not normal and that $X \times Y$ is not hereditarily normal. $\blacksquare$

Let’s see what happens in Theorem 1 when both factors are compact. If both $X$ and $Y$ are compact and if $X \times Y$ is hereditarily normal, then both $X$ and $Y$ must be perfect normal. Note that in any infinite compact space, not every countably infinite subset is closed. Thus if compact spaces satisfy the conclusion of Theorem 1, they must be perfectly normal. Hence we have the following theorem.

Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.

Moe interestingly, Theorem 1 leads to a metrization theorem for compact spaces.

Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.

Proof of Theorem 3
Suppose that $X^3$ is hereditarily normal. By Theorem 2, the compact spaces $X^2$ and $X$ are perfectly normal. In particular, the following subset of $X^2$ is a $G_\delta$ -set in $X^2$ .

$\Delta=\left\{(x,x): x \in X \right\}$

The set $\Delta$ is said to be the diagonal of the space $X$ . It is a well known result that any compact space whose diagonal is a $G_\delta$ -set in the square is metrizable (discussed here). $\blacksquare$

The results discussed here make it clear that hereditary normality in product spaces is a very strong property. One obvious question is whether Theorem 3 can be improved by assuming only the hereditary normality of $X^2$ . This was indeed posted by Katetov himself. This leads to the discussion in the next post.

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Reference

Engelking R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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$\copyright \ 2015 \text{ by Dan Ma}$

Compact metrizable scattered spaces

Posted on April 1, 2015 by Dan Ma

A scattered space is one in which there are isolated points found in every subspace. Specifically, a space $X$ is a scattered space if every non-empty subspace $Y$ of $X$ has a point $y \in Y$ such that $y$ is an isolated point in $Y$ , i.e. the singleton set $\left\{y \right\}$ is open in the subspace $Y$ . A handy example is a space consisting of ordinals. Note that in a space of ordinals, every non-empty subset has an isolated point (e.g. its least element). In this post, we discuss scattered spaces that are compact metrizable spaces.

Here’s what led the author to think of such spaces. Consider Theorem III.1.2 found on page 91 of Arhangelskii’s book on topological function space [1], which is Theorem 1 stated below:

Thereom 1
For any compact space $X$ , the following conditions are equivalent:

The function space $C_p(X)$ is a Frechet-Urysohn space.
The function space $C_p(X)$ is a k space.
$X$ is a scattered space.

Let’s put aside the Frechet-Urysohn property and the k space property for the moment. For any Hausdorff space $X$ , let $C(X)$ be the set of all continuous real-valued functions defined on the space $X$ . Since $C(X)$ is a subspace of the product space $\mathbb{R}^X$ , a natural topology that can be given to $C(X)$ is the subspace topology inherited from the product space $\mathbb{R}^X$ . Then $C_p(X)$ is simply the set $C(X)$ with the product subspace topology (also called the pointwise convergence topology).

Let’s say the compact space $X$ is countable and infinite. Then the function space $C_p(X)$ is metrizable since it is a subspace of $\mathbb{R}^X$ , a product of countably many lines. Thus the function space $C_p(X)$ has the Frechet-Urysohn property (being metrizable implies Frechet-Urysohn). This means that the compact space $X$ is scattered. The observation just made is a proof that any infinite compact space that is countable in cardinality must be scattered. In particular, every infinite compact and countable space must have an isolated point. There must be a more direct proof of this same fact without taking the route of a function space. The indirect argument does not reveal the essential nature of compact metric spaces. The essential fact is that any uncountable compact metrizable space contains a Cantor set, which is as unscattered as any space can be. Thus the only scattered compact metrizable spaces are the countable ones.

The main part of the proof is the construction of a Cantor set in a compact metrizable space (Theorem 3). The main result is Theorem 4. In many settings, the construction of a Cantor set is done in the real number line (e.g. the middle third Cantor set). The construction here is in a more general setting. But the idea is still the same binary division process – the splitting of a small open set with compact closure into two open sets with disjoint compact closure. We also use that fact that any compact metric space is hereditarily Lindelof (Theorem 2).

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Compact metrizable spaces

We first define some notions before looking at compact metrizable spaces in more details. Let $X$ be a space. Let $A \subset X$ . Let $p \in X$ . We say that $p$ is a limit point of $A$ if every open subset of $X$ containing $p$ contains a point of $A$ distinct from $p$ . So the notion of limit point here is from a topology perspective and not from a metric perspective. In a topological space, a limit point does not necessarily mean that it is the limit of a convergent sequence (however, it does in a metric space). The proof of the following theorem is straightforward.

Theorem 2
Let $X$ be a hereditarily Lindelof space (i.e. every subspace of $X$ is Lindelof). Then for any uncountable subset $A$ of $X$ , all but countably many points of $A$ are limit points of $A$ .

We now discuss the main result.

Theorem 3
Let $X$ be a compact metrizable space such that every point of $X$ is a limit point of $X$ . Then there exists an uncountable closed subset $C$ of $X$ such that every point of $C$ is a limit point of $C$ .

Proof of Theorem 3
Note that any compact metrizable space is a complete metric space. Consider a complete metric $\rho$ on the space $X$ . One fact that we will use is that if there is a sequence of closed sets $X \supset H_1 \supset H_2 \supset H_3 \supset \cdots$ such that the diameters of the sets $H$ (based on the complete metric $\rho$ ) decrease to zero, then the sets $H_n$ collapse to one point.

The uncountable closed set $C$ we wish to define is a Cantor set, which is constructed from a binary division process. To start, pick two points $p_0,p_1 \in X$ such that $p_0 \ne p_1$ . By assumption, both points are limit points of the space $X$ . Choose open sets $U_0,U_1 \subset X$ such that

$p_0 \in U_0$ ,
$p_1 \in U_1$ ,
$K_0=\overline{U_0}$ and $K_1=\overline{U_1}$ ,
$K_0 \cap K_1 = \varnothing$ ,
the diameters for $K_0$ and $K_1$ with respect to $\rho$ are less than 0.5.

Note that each of these open sets contains infinitely many points of $X$ . Then we can pick two points in each of $U_0$ and $U_1$ in the same manner. Before continuing, we set some notation. If $\sigma$ is an ordered string of 0’s and 1’s of length $n$ (e.g. 01101 is a string of length 5), then we can always extend it by tagging on a 0 and a 1. Thus $\sigma$ is extended as $\sigma 0$ and $\sigma 1$ (e.g. 01101 is extended by 011010 and 011011).

Suppose that the construction at the $n$ th stage where $n \ge 1$ is completed. This means that the points $p_\sigma$ and the open sets $U_\sigma$ have been chosen such that $p_\sigma \in U_\sigma$ for each length $n$ string of 0’s and 1’s $\sigma$ . Now we continue the picking for the $(n+1)$ st stage. For each $\sigma$ , an $n$ -length string of 0’s and 1’s, choose two points $p_{\sigma 0}$ and $p_{\sigma 1}$ and choose two open sets $U_{\sigma 0}$ and $U_{\sigma 1}$ such that

$p_{\sigma 0} \in U_{\sigma 0}$ ,
$p_{\sigma 1} \in U_{\sigma 1}$ ,
$K_{\sigma 0}=\overline{U_{\sigma 0}} \subset U_{\sigma}$ and $K_{\sigma 1}=\overline{U_{\sigma 1}} \subset U_{\sigma}$ ,
$K_{\sigma 0} \cap K_{\sigma 1} = \varnothing$ ,
the diameters for $K_{\sigma 0}$ and $K_{\sigma 1}$ with respect to $\rho$ are less than $0.5^{n+1}$ .

For each positive integer $m$ , let $C_m$ be the union of all $K_\sigma$ over all $\sigma$ that are $m$ -length strings of 0’s and 1’s. Each $C_m$ is a union of finitely many compact sets and is thus compact. Furthermore, $C_1 \supset C_2 \supset C_3 \supset \cdots$ . Thus $C=\bigcap \limits_{m=1}^\infty C_m$ is non-empty. To complete the proof, we need to show that

$C$ is uncountable (in fact of cardinality continuum),
every point of $C$ is a limit point of $C$ .

To show the first point, we define a one-to-one function $f: \left\{0,1 \right\}^N \rightarrow C$ where $N=\left\{1,2,3,\cdots \right\}$ . Note that each element of $\left\{0,1 \right\}^N$ is a countably infinite string of 0’s and 1’s. For each $\tau \in \left\{0,1 \right\}^N$ , let $\tau \upharpoonright n$ denote the string of the first $n$ digits of $\tau$ . For each $\tau \in \left\{0,1 \right\}^N$ , let $f(\tau)$ be the unique point in the following intersection:

$\displaystyle \bigcap \limits_{n=1}^\infty K_{\tau \upharpoonright n} = \left\{f(\tau) \right\}$

This mapping is uniquely defined. Simply conceptually trace through the induction steps. For example, if $\tau$ are 01011010…., then consider $K_0 \supset K_{01} \supset K_{010} \supset \cdots$ . At each next step, always pick the $K_{\tau \upharpoonright n}$ that matches the next digit of $\tau$ . Since the sets $K_{\tau \upharpoonright n}$ are chosen to have diameters decreasing to zero, the intersection must have a unique element. This is because we are working in a complete metric space.

It is clear that the map $f$ is one-to-one. If $\tau$ and $\gamma$ are two different strings of 0’s and 1’s, then they must differ at some coordinate, then from the way the induction is done, the strings would lead to two different points. It is also clear to see that the map $f$ is reversible. Pick any point $x \in C$ . Then the point $x$ must belong to a nested sequence of sets $K$ ‘s. This maps to a unique infinite string of 0’s and 1’s. Thus the set $C$ has the same cardinality as the set $\left\{0,1 \right\}^N$ , which has cardinality continuum.

To see the second point, pick $x \in C$ . Suppose $x=f(\tau)$ where $\tau \in \left\{0,1 \right\}^N$ . Consider the open sets $U_{\tau \upharpoonright n}$ for all positive integers $n$ . Note that $x \in U_{\tau \upharpoonright n}$ for each $n$ . Based on the induction process described earlier, observe these two facts. This sequence of open sets has diameters decreasing to zero. Each open set $U_{\tau \upharpoonright n}$ contains infinitely many other points of $C$ (this is because of all the open sets $U_{\tau \upharpoonright k}$ that are subsets of $U_{\tau \upharpoonright n}$ where $k \ge n$ ). Because the diameters are decreasing to zero, the sequence of $U_{\tau \upharpoonright n}$ is a local base at the point $x$ . Thus, the point $x$ is a limit point of $C$ . This completes the proof. $\blacksquare$

Theorem 4
Let $X$ be a compact metrizable space. It follows that $X$ is scattered if and only if $X$ is countable.

Proof of Theorem 4
$\Longleftarrow$
In this direction, we show that if $X$ is countable, then $X$ is scattered (the fact that can be shown using the function space argument pointed out earlier). Here, we show the contrapositive: if $X$ is not scattered, then $X$ is uncountable. Suppose $X$ is not scattered. Then every point of $X$ is a limit point of $X$ . By Theorem 3, $X$ would contain a Cantor set $C$ of cardinality continuum.

$\Longrightarrow$
In this direction, we show that if $X$ is scattered, then $X$ is countable. We also show the contrapositive: if $X$ is uncountable, then $X$ is not scattered. Suppose $X$ is uncountable. By Theorem 2, all but countably many points of $X$ are limit points of $X$ . After discarding these countably many isolated points, we still have a compact space. So we can just assume that every point of $X$ is a limit point of $X$ . Then by Theorem 3, $X$ contains an uncountable closed set $C$ such that every point of $C$ is a limit point of $C$ . This means that $X$ is not scattered. $\blacksquare$

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Remarks

A corollary to the above discussion is that the cardinality for any compact metrizable space is either countable (including finite) or continuum (the cardinality of the real line). There is nothing in between or higher than continuum. To see this, the cardinality of any Lindelof first countable space is at most continuum according to a theorem in this previous post (any compact metric space is one such). So continuum is an upper bound on the cardinality of compact metric spaces. Theorem 3 above implies that any uncountable compact metrizable space has to contain a Cantor set, hence has cardinality continuum. So the cardinality of a compact metrizable space can be one of two possibilities – countable or continuum. Even under the assumption of the negation of the continuum hypothesis, there will be no uncountable compact metric space of cardinality less than continuum. On the other hand, there is only one possibility for the cardinality of a scattered compact metrizable, which is countable.

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Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2015 \text{ by Dan Ma}$

Dan Ma's Topology Blog

Expository Articles In General Topology

Monthly Archives: April 2015

Looking for non-normal subspaces of the square of a compact X

When a product space is hereditarily normal

Compact metrizable scattered spaces