# Counterexample 106 from Steen and Seebach

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

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

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

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

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

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

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

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

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The space $X$ is not normal

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

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

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

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

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

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

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

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The space $X$ is a dense subspace that is normal

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

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Reference

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

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

# Normality in the powers of countably compact spaces

Let $\omega_1$ be the first uncountable ordinal. The topology on $\omega_1$ we are interested in is the ordered topology, the topology induced by the well ordering. The space $\omega_1$ is also called the space of all countable ordinals since it consists of all ordinals that are countable in cardinality. It is a handy example of a countably compact space that is not compact. In this post, we consider normality in the powers of $\omega_1$. We also make comments on normality in the powers of a countably compact non-compact space.

Let $\omega$ be the first infinite ordinal. It is well known that $\omega^{\omega_1}$, the product space of $\omega_1$ many copies of $\omega$, is not normal (a proof can be found in this earlier post). This means that any product space $\prod_{\alpha<\kappa} X_\alpha$, with uncountably many factors, is not normal as long as each factor $X_\alpha$ contains a countable discrete space as a closed subspace. Thus in order to discuss normality in the product space $\prod_{\alpha<\kappa} X_\alpha$, the interesting case is when each factor is infinite but contains no countable closed discrete subspace (i.e. no closed copies of $\omega$). In other words, the interesting case is that each factor $X_\alpha$ is a countably compact space that is not compact (see this earlier post for a discussion of countably compactness). In particular, we would like to discuss normality in $X^{\kappa}$ where $X$ is a countably non-compact space. In this post we start with the space $X=\omega_1$ of the countable ordinals. We examine $\omega_1$ power $\omega_1^{\omega_1}$ as well as the countable power $\omega_1^{\omega}$. The former is not normal while the latter is normal. The proof that $\omega_1^{\omega}$ is normal is an application of the normality of $\Sigma$-product of the real line.

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The uncountable product

Theorem 1
The product space $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal.

Theorem 1 follows from Theorem 2 below. For any space $X$, a collection $\mathcal{C}$ of subsets of $X$ is said to have the finite intersection property if for any finite $\mathcal{F} \subset \mathcal{C}$, the intersection $\cap \mathcal{F} \ne \varnothing$. Such a collection $\mathcal{C}$ is called an f.i.p collection for short. It is well known that a space $X$ is compact if and only collection $\mathcal{C}$ of closed subsets of $X$ satisfying the finite intersection property has non-empty intersection (see Theorem 1 in this earlier post). Thus any non-compact space has an f.i.p. collection of closed sets that have empty intersection.

In the space $X=\omega_1$, there is an f.i.p. collection of cardinality $\omega_1$ using its linear order. For each $\alpha<\omega_1$, let $C_\alpha=\left\{\beta<\omega_1: \alpha \le \beta \right\}$. Let $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$. It is a collection of closed subsets of $X=\omega_1$. It is an f.i.p. collection and has empty intersection. It turns out that for any countably compact space $X$ with an f.i.p. collection of cardinality $\omega_1$ that has empty intersection, the product space $X^{\omega_1}$ is not normal.

Theorem 2
Let $X$ be a countably compact space. Suppose that there exists a collection $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$ of closed subsets of $X$ such that $\mathcal{C}$ has the finite intersection property and that $\mathcal{C}$ has empty intersection. Then the product space $X^{\omega_1}$ is not normal.

Proof of Theorem 2
Let’s set up some notations on product space that will make the argument easier to follow. By a standard basic open set in the product space $X^{\omega_1}=\prod_{\alpha<\omega_1} X$, we mean a set of the form $O=\prod_{\alpha<\omega_1} O_\alpha$ such that each $O_\alpha$ is an open subset of $X$ and that $O_\alpha=X$ for all but finitely many $\alpha<\omega_1$. Given a standard basic open set $O=\prod_{\alpha<\omega_1} O_\alpha$, the notation $\text{Supp}(O)$ refers to the finite set of $\alpha$ for which $O_\alpha \ne X$. For any set $M \subset \omega_1$, the notation $\pi_M$ refers to the projection map from $\prod_{\alpha<\omega_1} X$ to the subproduct $\prod_{\alpha \in M} X$. Each element $d \in X^{\omega_1}$ can be considered a function $d: \omega_1 \rightarrow X$. By $(d)_\alpha$, we mean $(d)_\alpha=d(\alpha)$.

For each $t \in X$, let $f_t: \omega_1 \rightarrow X$ be the constant function whose constant value is $t$. Consider the following subspaces of $X^{\omega_1}$.

$H=\prod_{\alpha<\omega_1} C_\alpha$

$\displaystyle K=\left\{f_t: t \in X \right\}$

Both $H$ and $K$ are closed subsets of the product space $X^{\omega_1}$. Because the collection $\mathcal{C}$ has empty intersection, $H \cap K=\varnothing$. We show that $H$ and $K$ cannot be separated by disjoint open sets. To this end, let $U$ and $V$ be open subsets of $X^{\omega_1}$ such that $H \subset U$ and $K \subset V$.

Let $d_1 \in H$. Choose a standard basic open set $O_1$ such that $d_1 \in O_1 \subset U$. Let $S_1=\text{Supp}(O_1)$. Since $S_1$ is the support of $O_1$, it follows that $\pi_{S_1}^{-1}(\pi_{S_1}(d_1)) \subset O_1 \subset U$. Since $\mathcal{C}$ has the finite intersection property, there exists $a_1 \in \bigcap_{\alpha \in S_1} C_\alpha$.

Define $d_2 \in H$ such that $(d_2)_\alpha=a_1$ for all $\alpha \in S_1$ and $(d_2)_\alpha=(d_1)_\alpha$ for all $\alpha \in \omega_1-S_1$. Choose a standard basic open set $O_2$ such that $d_2 \in O_2 \subset U$. Let $S_2=\text{Supp}(O_2)$. It is possible to ensure that $S_1 \subset S_2$ by making more factors of $O_2$ different from $X$. We have $\pi_{S_2}^{-1}(\pi_{S_2}(d_2)) \subset O_2 \subset U$. Since $\mathcal{C}$ has the finite intersection property, there exists $a_2 \in \bigcap_{\alpha \in S_2} C_\alpha$.

Now choose a point $d_3 \in H$ such that $(d_3)_\alpha=a_2$ for all $\alpha \in S_2$ and $(d_3)_\alpha=(d_2)_\alpha$ for all $\alpha \in \omega_1-S_2$. Continue on with this inductive process. When the inductive process is completed, we have the following sequences:

• a sequence $d_1,d_2,d_3,\cdots$ of point of $H=\prod_{\alpha<\omega_1} C_\alpha$,
• a sequence $S_1 \subset S_2 \subset S_3 \subset \cdots$ of finite subsets of $\omega_1$,
• a sequence $a_1,a_2,a_3,\cdots$ of points of $X$

such that for all $n \ge 2$, $(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_{n-1}$ and $\pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$. Let $A=\left\{a_1,a_2,a_3,\cdots \right\}$. Either $A$ is finite or $A$ is infinite. Let’s examine the two cases.

Case 1
Suppose that $A$ is infinite. Since $X$ is countably compact, $A$ has a limit point $a$. That means that every open set containing $a$ contains some $a_n \ne a$. For each $n \ge 2$, define $y_n \in \prod_{\alpha< \omega_1} X$ such that

• $(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$,
• $(y_n)_\alpha=a$ for all $\alpha \in \omega_1-S_n$

From the induction step, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$. Let $t=f_a \in K$, the constant function whose constant value is $a$. It follows that $t$ is a limit of $\left\{y_1,y_2,y_3,\cdots \right\}$. This means that $t \in \overline{U}$. Since $t \in K \subset V$, $U \cap V \ne \varnothing$.

Case 2
Suppose that $A$ is finite. Then there is some $m$ such that $a_m=a_j$ for all $j \ge m$. For each $n \ge 2$, define $y_n \in \prod_{\alpha< \omega_1} X$ such that

• $(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$,
• $(y_n)_\alpha=a_m$ for all $\alpha \in \omega_1-S_n$

As in Case 1, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$. Let $t=f_{a_m} \in K$, the constant function whose constant value is $a_m$. It follows that $t=y_n$ for all $n \ge m+1$. Thus $U \cap V \ne \varnothing$.

Both cases show that $U \cap V \ne \varnothing$. This completes the proof the product space $X^{\omega_1}$ is not normal. $\blacksquare$

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The countable product

Theorem 3
The product space $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is normal.

Proof of Theorem 3
The proof here actually proves more than normality. It shows that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is collectionwise normal, which is stronger than normality. The proof makes use of the $\Sigma$-product of $\kappa$ many copies of $\mathbb{R}$, which is the following subspace of the product space $\mathbb{R}^{\kappa}$.

$\Sigma(\kappa)=\left\{x \in \mathbb{R}^{\kappa}: x(\alpha) \ne 0 \text{ for at most countably many } \alpha<\kappa \right\}$

It is well known that $\Sigma(\kappa)$ is collectionwise normal (see this earlier post). We show that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is a closed subspace of $\Sigma(\kappa)$ where $\kappa=\omega_1$. Thus $\omega_1^{\omega}$ is collectionwise normal. This is established in the following claims.

Claim 1
We show that the space $\omega_1$ is embedded as a closed subspace of $\Sigma(\omega_1)$.

For each $\beta<\omega_1$, define $f_\beta:\omega_1 \rightarrow \mathbb{R}$ such that $f_\beta(\gamma)=1$ for all $\gamma<\beta$ and $f_\beta(\gamma)=0$ for all $\beta \le \gamma <\omega_1$. Let $W=\left\{f_\beta: \beta<\omega_1 \right\}$. We show that $W$ is a closed subset of $\Sigma(\omega_1)$ and $W$ is homeomorphic to $\omega_1$ according to the mapping $f_\beta \rightarrow W$.

First, we show $W$ is closed by showing that $\Sigma(\omega_1)-W$ is open. Let $y \in \Sigma(\omega_1)-W$. We show that there is an open set containing $y$ that contains no points of $W$.

Suppose that for some $\gamma<\omega_1$, $y_\gamma \in O=\mathbb{R}-\left\{0,1 \right\}$. Consider the open set $Q=(\prod_{\alpha<\omega_1} Q_\alpha) \cap \Sigma(\omega_1)$ where $Q_\alpha=\mathbb{R}$ except that $Q_\gamma=O$. Then $y \in Q$ and $Q \cap W=\varnothing$.

So we can assume that for all $\gamma<\omega_1$, $y_\gamma \in \left\{0, 1 \right\}$. There must be some $\theta$ such that $y_\theta=1$. Otherwise, $y=f_0 \in W$. Since $y \ne f_\theta$, there must be some $\delta<\gamma$ such that $y_\delta=0$. Now choose the open interval $T_\theta=(0.9,1.1)$ and the open interval $T_\delta=(-0.1,0.1)$. Consider the open set $M=(\prod_{\alpha<\omega_1} M_\alpha) \cap \Sigma(\omega_1)$ such that $M_\alpha=\mathbb{R}$ except for $M_\theta=T_\theta$ and $M_\delta=T_\delta$. Then $y \in M$ and $M \cap W=\varnothing$. We have just established that $W$ is closed in $\Sigma(\omega_1)$.

Consider the mapping $f_\beta \rightarrow W$. Based on how it is defined, it is straightforward to show that it is a homeomorphism between $\omega_1$ and $W$.

Claim 2
The $\Sigma$-product $\Sigma(\omega_1)$ has the interesting property it is homeomorphic to its countable power, i.e.

$\Sigma(\omega_1) \cong \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \ \ \ \ \ \ \ \ \ \ \ \text{(countably many times)}$.

Because each element of $\Sigma(\omega_1)$ is nonzero only at countably many coordinates, concatenating countably many elements of $\Sigma(\omega_1)$ produces an element of $\Sigma(\omega_1)$. Thus Claim 2 can be easily verified. With above claims, we can see that

$\displaystyle \omega_1^{\omega}=\omega_1 \times \omega_1 \times \omega_1 \times \cdots \subset \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \cong \Sigma(\omega_1)$

Thus $\omega_1^{\omega}$ is a closed subspace of $\Sigma(\omega_1)$. Any closed subspace of a collectionwise normal space is collectionwise normal. We have established that $\omega_1^{\omega}$ is normal. $\blacksquare$

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The normality in the powers of $X$

We have established that $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal. Hence any higher uncountable power of $\omega_1$ is not normal. We have also established that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$, the countable power of $\omega_1$ is normal (in fact collectionwise normal). Hence any finite power of $\omega_1$ is normal. However $\omega_1^{\omega}$ is not hereditarily normal. One of the exercises below is to show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Theorem 2 can be generalized as follows:

Theorem 4
Let $X$ be a countably compact space has an f.i.p. collection $\mathcal{C}$ of closed sets such that $\bigcap \mathcal{C}=\varnothing$. Then $X^{\kappa}$ is not normal where $\kappa=\lvert \mathcal{C} \lvert$.

The proof of Theorem 2 would go exactly like that of Theorem 2. Consider the following two theorems.

Theorem 5
Let $X$ be a countably compact space that is not compact. Then there exists a cardinal number $\kappa$ such that $X^{\kappa}$ is not normal and $X^{\tau}$ is normal for all cardinal number $\tau<\kappa$.

By the non-compactness of $X$, there exists an f.i.p. collection $\mathcal{C}$ of closed subsets of $X$ such that $\bigcap \mathcal{C}=\varnothing$. Let $\kappa$ be the least cardinality of such an f.i.p. collection. By Theorem 4, that $X^{\kappa}$ is not normal. Because $\kappa$ is least, any smaller power of $X$ must be normal.

Theorem 6
Let $X$ be a space that is not countably compact. Then $X^{\kappa}$ is not normal for any cardinal number $\kappa \ge \omega_1$.

Since the space $X$ in Theorem 6 is not countably compact, it would contain a closed and discrete subspace that is countable. By a theorem of A. H. Stone, $\omega^{\omega_1}$ is not normal. Then $\omega^{\omega_1}$ is a closed subspace of $X^{\omega_1}$.

Thus between Theorem 5 and Theorem 6, we can say that for any non-compact space $X$, $X^{\kappa}$ is not normal for some cardinal number $\kappa$. The $\kappa$ from either Theorem 5 or Theorem 6 is at least $\omega_1$. Interestingly for some spaces, the $\kappa$ can be much smaller. For example, for the Sorgenfrey line, $\kappa=2$. For some spaces (e.g. the Michael line), $\kappa=\omega$.

Theorems 4, 5 and 6 are related to a theorem that is due to Noble.

Theorem 7 (Noble)
If each power of a space $X$ is normal, then $X$ is compact.

A proof of Noble’s theorem is given in this earlier post, the proof of which is very similar to the proof of Theorem 2 given above. So the above discussion the normality of powers of $X$ is just another way of discussing Theorem 7. According to Theorem 7, if $X$ is not compact, some power of $X$ is not normal.

The material discussed in this post is excellent training ground for topology. Regarding powers of countably compact space and product of countably compact spaces, there are many topics for further discussion/investigation. One possibility is to examine normality in $X^{\kappa}$ for more examples of countably compact non-compact $X$. One particular interesting example would be a countably compact non-compact $X$ such that the least power $\kappa$ for non-normality in $X^{\kappa}$ is more than $\omega_1$. A possible candidate could be the second uncountable ordinal $\omega_2$. By Theorem 2, $\omega_2^{\omega_2}$ is not normal. The issue is whether the $\omega_1$ power $\omega_2^{\omega_1}$ and countable power $\omega_2^{\omega}$ are normal.

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Exercises

Exercise 1
Show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Exercise 2
Show that the mapping $f_\beta \rightarrow W$ in Claim 3 in the proof of Theorem 3 is a homeomorphism.

Exercise 3
The proof of Theorem 3 shows that the space $\omega_1$ is a closed subspace of the $\Sigma$-product of the real line. Show that $\omega_1$ can be embedded in the $\Sigma$-product of arbitrary spaces.

For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Let $p \in \prod_{\alpha<\omega_1} X_\alpha$. The $\Sigma$-product of the spaces $X_\alpha$ is the following subspace of the product space $\prod_{\alpha<\omega_1} X_\alpha$.

$\Sigma(X_\alpha)=\left\{x \in \prod_{\alpha<\omega_1} X_\alpha: x(\alpha) \ne p(\alpha) \ \text{for at most countably many } \alpha<\omega_1 \right\}$

The point $p$ is the center of the $\Sigma$-product. Show that the space $\Sigma(X_\alpha)$ contains $\omega_1$ as a closed subspace.

Exercise 4
Find a direct proof of Theorem 3, that $\omega_1^{\omega}$ is normal.

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

# The product of uncountably many factors is never hereditarily normal

The space $Y=\prod_{\alpha<\omega_1} \left\{0,1 \right\}=\left\{0,1 \right\}^{\omega_1}$ is the product of $\omega_1$ many copies of the two-element set $\left\{0,1 \right\}$ where $\omega_1$ is the first uncountable ordinal. It is a compact space by Tychonoff’s theorem. It is a normal space since every compact Hausdorff space is normal. A space is hereditarily normal if every subspace is normal. Is the space $Y$ hereditarily normal? In this post, we give two proofs that it is not hereditarily normal. It then follows that any product space $\prod X_\alpha$ cannot be hereditarily normal as long as there are uncountably many factors and every factor has at least two point.

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The connection with a theorem of Katetov

It turns out that there is a connection with a theorem of Katetov. For any compact space, knowing hereditary normality of the first several self product spaces can reveal a great deal of information about the compact space. More specifically, for any compact space $X$, knowing whether $X$, $X^2$ and $X^3$ are hereditarily normal can tell us whether $X$ is metrizable. If all three are hereditarily normal, then $X$ is metrizable. If one of the three self products is not hereditarily normal, then $X$ is not metrizable. This fact is based on a theorem of Katetov (see this previous post). The space $Y=\left\{0,1 \right\}^{\omega_1}$ is not metrizable since it is not first countable (see Problem 1 below). Thus one of its first three self products must fail to be hereditarily normal.

These two proofs are not direct proof in the sense that a non-normal subspace is not explicitly produced. Instead the proofs use other theorem or basic but important background results. One of the two proofs (#2) uses a theorem of Katetov on hereditarily normal spaces. The other proof (#1) uses the fact that the product of uncountably many copies of a countable discrete space is not normal. We believe that these two proofs and the required basic facts are an important training ground for topology. We list out these basic facts as exercises. Anyone who wishes to fill in the gaps can do so either by studying the links provided or by consulting other sources.

The theorem of Katetov mentioned earlier provides a great exercise – for any non-metrizable compact space $X$, determine where the hereditary normality fails. Does it fail in $X$, $X^2$ or $X^3$? This previous post examines a small list of compact non-metrizable spaces. In all the examples in this list, the hereditary normality fails in $X$ or $X^2$. The space $Y=\left\{0,1 \right\}^{\omega_1}$ can be added to this list. All the examples in this list are defined using no additional set theory axioms beyond ZFC. A natural question: does there exist an example of compact non-metrizable space $X$ such that the hereditary normality holds in $X^2$ and fails in $X^3$? It turns out that this was a hard problem and the answer is independent of ZFC. This previous post provides a brief discussion and has references for the problem.

All spaces under consideration are Hausdorff spaces.

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Exercises

Problem 1
Let $X$ be a compact space. Show that $X$ is normal.

Problem 2
For each $\alpha<\omega_1$, let $A_\alpha$ be a set with cardinality $\le \omega_1$. Show that $\lvert \bigcup_{\alpha<\omega_1} A_\alpha \lvert \le \omega_1$.

Problem 2 holds for any infinite cardinal, not just $\omega_1$. One reference for Problem 2 is Lemma 10.21 on page 30 of Set Theorey, An Introduction to Independence Proofs by Kenneth Kunen.

Problem 3
For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Show that for every point $p \in \prod_{\alpha<\omega_1} X_\alpha$, there does not exist a countable base at the point $p$. In other words, the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not first countable at every point. It follows that product space $\prod_{\alpha<\omega_1} X_\alpha$ is not metrizable.

Problem 4
In any space, a $G_\delta$-set is a set that is the intersection of countably many open sets. When a singleton set $\left\{ x \right\}$ is a $G_\delta$-set, we say the point $x$ is a $G_\delta$-point. For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Show that every point $p$ in the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not a $G_\delta$-point.

Note that Problem 4 implies Problem 3.

For Problem 3 and Problem 4, use the fact that there are uncountably many factors and that a basic open set in the product space is of the form $\prod_{\alpha<\omega_1} O_\alpha$ and that it has only finitely many coordinates at which $O_\alpha \ne X_\alpha$.

Problem 5
For each $\alpha<\omega_1$, let $X_\alpha=\left\{0,1,2,\cdots \right\}$ be the set of non-negative integers with the discrete topology. Show that the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not normal.

See here for a discussion of Problem 5.

Problem 6
Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. Show that $Y$ has a countably infinite subspace

$W=\left\{y_0,y_1,y_2,y_3\cdots \right\}$

such that $W$ is relatively discrete. In other words, $W$ is discrete in the subspace topology of $W$. However $W$ is not discrete in the product space $Y$ since $Y$ is compact.

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Proof #1

Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. We show that $Y$ is not hereditarily normal.

Note that the product space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$ can be written as the product of $\omega_1$ many copies of itself:

$\displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

The fact (1) follows from the fact that the union of $\omega_1$ many pairwise disjoint sets, each of which has cardinality $\omega_1$, has cardinality $\omega_1$ (see Problem 2). The space $\left\{0,1 \right\}^{\omega_1}$ has a countably infinite subspace that is relatively discrete (see Problem 6). In other words, it has a subspace that is homemorphic to $\omega=\left\{0,1,2,\cdots \right\}$ where $\omega$ has the discrete topology. Thus the following is homeomorphic to a subspace of $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$.

$\displaystyle \omega^{\omega_1} = \omega \times \omega \times \omega \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

By Problem 5, the space $\omega^{\omega_1}$ is not normal. Hence the compact space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$ contains the non-normal space $\omega^{\omega_1}$ and is thus not hereditarily normal. $\blacksquare$

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Proof #2

Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. We show that $Y$ is not hereditarily normal. This proof uses a theorem of Katetov, discussed in this previous post and stated below.

Theorem 1
If $X_1 \times X_2$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

• The factor $X_1$ is perfectly normal.
• Every countable and infinite subset of the factor $X_2$ is closed.

First, $Y$ can be written as the product of two copies of itself:

$\displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

This is because the union of two disjoints sets, each of which has cardinality $\omega_1$, has carinality $\omega_1$. Note that the countably infinite subset $W$ from Problem 6 is not a closed subset of $Y$. If it were, the compact space $Y$ would contain an infinite set with no limit point. Thus the second condition of Theorem 1 is not satisfied. If $Y \cong Y \times Y$ were to be hereditarily normal, then the first condition must be satisfied, i.e. $Y$ is perfectly normal (meaning that $Y$ is normal and that every closed subset of it is a $G_\delta$-set). However, Problem 4 indicates that no point in $Y$ can be a $G_\delta$ point. Therefore $Y$ cannot be hereditarily normal. $\blacksquare$

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Corollary

The product of uncountably many spaces, each one of which has at least two points, contains a homeomorphic copy of the space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. Thus such a product space can never be hereditarily normal. We state this more formally below.

Theorem 2
Let $\kappa$ be any uncountable cardinal. For each $\alpha<\kappa$, let $X_\alpha$ be a space with at least two points. Then $\prod_{\alpha<\kappa} X_\alpha$ is not hereditarily normal.

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

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

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.

1. One-point compactification of the Tychonoff plank.
2. One-point compactification of $\psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$.
3. The first compact uncountable ordinal, i.e. $\omega_1+1$.
4. The one-point compactification of an uncountable discrete space.
5. Alexandroff double circle.
6. Double arrow space.
7. 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. Is there an example of a compact non-metrizable space $X$ such that the failure of hereditary nornmality occurs in “dimension” 3? Specifically, is there a compact non-metrizable $X$ such that $X^2$ is hereditarily normal but $X^3$ is not hereditarily normal?

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

1. Gruenhage G., Nyikos P. J., Normality in $X^2$ for Compact $X$, Trans. Amer. Math. Soc., Vol 340, No 2 (1993), 563-586
2. Katetov M., Complete normality of Cartesian products, Fund. Math., 35 (1948), 271-274
3. 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

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

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

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

# Comparing two function spaces

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. Furthermore consider these ordinals as topological spaces endowed with the order topology. It is a well known fact that any continuous real-valued function $f$ defined on either $\omega_1$ or $\omega_1+1$ is eventually constant, i.e., there exists some $\alpha<\omega_1$ such that the function $f$ is constant on the ordinals beyond $\alpha$. Now consider the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$. Thus individually, elements of these two function spaces appear identical. Any $f \in C_p(\omega_1)$ matches a function $f^* \in C_p(\omega_1+1)$ where $f^*$ is the result of adding the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$. This fact may give the impression that the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$ are identical topologically. The goal in this post is to demonstrate that this is not the case. We compare the two function spaces with respect to some convergence properties (countably tightness and Frechet-Urysohn property) as well as normality.

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Tightness

One topological property that is different between $C_p(\omega_1)$ and $C_p(\omega_1+1)$ is that of tightness. The function space $C_p(\omega_1+1)$ is countably tight, while $C_p(\omega_1)$ is not countably tight.

Let $X$ be a space. The tightness of $X$, denoted by $t(X)$, is the least infinite cardinal $\kappa$ such that for any $A \subset X$ and for any $x \in X$ with $x \in \overline{A}$, there exists $B \subset A$ for which $\lvert B \lvert \le \kappa$ and $x \in \overline{B}$. When $t(X)=\omega$, we say that $X$ has countable tightness or is countably tight. When $t(X)>\omega$, we say that $X$ has uncountable tightness or is uncountably tight.

First, we show that the tightness of $C_p(\omega_1)$ is greater than $\omega$. For each $\alpha<\omega_1$, define $f_\alpha: \omega_1 \rightarrow \left\{0,1 \right\}$ such that $f_\alpha(\beta)=0$ for all $\beta \le \alpha$ and $f_\alpha(\beta)=1$ for all $\beta>\alpha$. Let $g \in C_p(\omega_1)$ be the function that is identically zero. Then $g \in \overline{F}$ where $F$ is defined by $F=\left\{f_\alpha: \alpha<\omega_1 \right\}$. It is clear that for any countable $B \subset F$, $g \notin \overline{B}$. Thus $C_p(\omega_1)$ cannot be countably tight.

The space $\omega_1+1$ is a compact space. The fact that $C_p(\omega_1+1)$ is countably tight follows from the following theorem.

Theorem 1
Let $X$ be a completely regular space. Then the function space $C_p(X)$ is countably tight if and only if $X^n$ is Lindelof for each $n=1,2,3,\cdots$.

Theorem 1 is a special case of Theorem I.4.1 on page 33 of [1] (the countable case). One direction of Theorem 1 is proved in this previous post, the direction that will give us the desired result for $C_p(\omega_1+1)$.

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The Frechet-Urysohn property

In fact, $C_p(\omega_1+1)$ has a property that is stronger than countable tightness. The function space $C_p(\omega_1+1)$ is a Frechet-Urysohn space (see this previous post). Of course, $C_p(\omega_1)$ not being countably tight means that it is not a Frechet-Urysohn space.

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Normality

The function space $C_p(\omega_1+1)$ is not normal. If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ would have countable extent. However, there exists an uncountable closed and discrete subset of $C_p(\omega_1+1)$ (see this previous post). On the other hand, $C_p(\omega_1)$ is Lindelof. The fact that $C_p(\omega_1)$ is Lindelof is highly non-trivial and follows from [2]. The author in [2] showed that if $X$ is a space consisting of ordinals such that $X$ is first countable and countably compact, then $C_p(X)$ is Lindelof.

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Embedding one function space into the other

The two function space $C_p(\omega_1+1)$ and $C_p(\omega_1)$ are very different topologically. However, one of them can be embedded into the other one. The space $\omega_1+1$ is the continuous image of $\omega_1$. Let $g: \omega_1 \longrightarrow \omega_1+1$ be a continuous surjection. Define a map $\psi: C_p(\omega_1+1) \longrightarrow C_p(\omega_1)$ by letting $\psi(f)=f \circ g$. It is shown in this previous post that $\psi$ is a homeomorphism. Thus $C_p(\omega_1+1)$ is homeomorphic to the image $\psi(C_p(\omega_1+1))$ in $C_p(\omega_1)$. The map $g$ is also defined in this previous post.

The homeomposhism $\psi$ tells us that the function space $C_p(\omega_1)$, though Lindelof, is not hereditarily normal.

On the other hand, the function space $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. Note that $C_p(\omega_1+1)$ is countably tight, which is a hereditary property.

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Remark

There is a mapping that is alluded to at the beginning of the post. Each $f \in C_p(\omega_1)$ is associated with $f^* \in C_p(\omega_1+1)$ which is obtained by appending the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$. It may be tempting to think of the mapping $f \rightarrow f^*$ as a candidate for a homeomorphism between the two function spaces. The discussion in this post shows that this particular map is not a homeomorphism. In fact, no other one-to-one map from one of these function spaces onto the other function space can be a homeomorphism.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Buzyakova, R. Z., In search of Lindelof $C_p$‘s, Comment. Math. Univ. Carolinae, 45 (1), 145-151, 2004.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Cp(omega 1 + 1) is not normal

In this and subsequent posts, we consider $C_p(X)$ where $X$ is a compact space. Recall that $C_p(X)$ is the space of all continuous real-valued functions defined on $X$ and that it is endowed with the pointwise convergence topology. One of the compact spaces we consider is $\omega_1+1$, the first compact uncountable ordinal. There are many interesting results about the function space $C_p(\omega_1+1)$. In this post we show that $C_p(\omega_1+1)$ is not normal. An even more interesting fact about $C_p(\omega_1+1)$ is that $C_p(\omega_1+1)$ does not have any dense normal subspace [1].

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. The set $\omega_1$ is the first uncountable ordinal. Furthermore consider these ordinals as topological spaces endowed with the order topology. As mentioned above, the space $\omega_1+1$ is the first compact uncountable ordinal. In proving that $C_p(\omega_1+1)$ is not normal, a theorem that is due to D. P. Baturov is utilized [2]. This theorem is also proved in this previous post.

For the basic working of function spaces with the pointwise convergence topology, see the post called Working with the function space Cp(X).

The fact that $C_p(\omega_1+1)$ is not normal is established by the following two points.

• If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ has countable extent, i.e. every closed and discrete subspace of $C_p(\omega_1+1)$ is countable.
• There exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$.

We discuss each of the bullet points separately.

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The function space $C_p(\omega_1+1)$ is a dense subspace of $\mathbb{R}^{\omega_1}$, the product of $\omega_1$ many copies of $\mathbb{R}$. According to a result of D. P. Baturov [2], any dense normal subspace of the product of $\omega_1$ many separable metric spaces has countable extent (also see Theorem 1a in this previous post). Thus $C_p(\omega_1+1)$ cannot be normal if the second bullet point above is established.

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Now we show that there exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$. For each $\alpha$ with $0<\alpha<\omega_1$, define $h_\alpha:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by:

$h_\alpha(\gamma) = \begin{cases} 1 & \mbox{if } \gamma \le \alpha \\ 0 & \mbox{if } \alpha<\gamma \le \omega_1 \end{cases}$

Clearly, $h_\alpha \in C_p(\omega_1 +1)$ for each $\alpha$. Let $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$. We show that $H$ is a closed and discrete subspace of $C_p(\omega_1 +1)$. The fact that $H$ is closed in $C_p(\omega_1 +1)$ is establish by the following claim.

Claim 1
Let $h \in C_p(\omega_1 +1) \backslash H$. There exists an open subset $U$ of $C_p(\omega_1 +1)$ such that $h \in U$ and $U \cap H=\varnothing$.

First we get some easy cases out of the way. Suppose that there exists some $\alpha<\omega_1$ such that $h(\alpha) \notin \left\{0,1 \right\}$. Then let $U=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in \mathbb{R} \backslash \left\{0,1 \right\} \right\}$. Clearly $h \in U$ and $U \cap H=\varnothing$.

Another easy case: If $h(\alpha)=0$ for all $\alpha \le \omega_1$, then consider the open set $U$ where $U=\left\{f \in C_p(\omega_1 +1): f(0) \in \mathbb{R} \backslash \left\{1 \right\} \right\}$. Clearly $h \in U$ and $U \cap H=\varnothing$.

From now on we can assume that $h(\omega_1+1) \subset \left\{0,1 \right\}$ and that $h$ is not identically the zero function. Suppose Claim 1 is not true. Then $h \in \overline{H}$. Next observe the following:

Observation.
If $h(\beta)=1$ for some $\beta \le \omega_1$, then $h(\alpha)=1$ for all $\alpha \le \beta$.

To see this, if $h(\alpha)=0$, $h(\beta)=1$ and $\alpha<\beta$, then define the open set $V$ by $V=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (-0.1,0.1) \text{ and } f(\beta) \in (0.9,1.1) \right\}$. Note that $h \in V$ and $V \cap H=\varnothing$, contradicting that $h \in \overline{H}$. So the above observation is valid.

Now either $h(\omega_1)=1$ or $h(\omega_1)=0$. We claim that $h(\omega_1)=1$ is not possible. Suppose that $h(\omega_1)=1$. Let $V=\left\{f \in C_p(\omega_1 +1): f(\omega_1) \in (0.9,1.1) \right\}$. Then $h \in V$ and $V \cap H=\varnothing$, contradicting that $h \in \overline{H}$. It must be the case that $h(\omega_1)=0$.

Because of the continuity of $h$ at the point $x=\omega_1$, of all the $\gamma<\omega_1$ for which $h(\gamma)=1$, there is the largest one, say $\beta$. Now $h(\beta)=1$. According to the observation made above, $h(\alpha)=1$ for all $\alpha \le \beta$. This means that $h=h_\beta$. This is a contradiction since $h \notin H$. Thus Claim 1 must be true and the fact that $H$ is closed is established.

Next we show that $H$ is discrete in $C_p(\omega_1 +1)$. Fix $h_\alpha$ where $0<\alpha<\omega_1$. Let $W=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (0.9,1.1) \text{ and } f(\alpha+1) \in (-0.1,0.1) \right\}$. It is clear that $h_\alpha \in W$. Furthermore, $h_\gamma \notin W$ for all $\alpha < \gamma$ and $h_\gamma \notin W$ for all $\gamma <\alpha$. Thus $W$ is open such that $\left\{h_\alpha \right\}=W \cap H$. This completes the proof that $H$ is discrete.

We have established that $H$ is an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$. This implies that $C_p(\omega_1 +1)$ is not normal.

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Remarks

The set $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$ as defined above is closed and discrete in $C_p(\omega_1 +1)$. However, the set $H$ is not discrete in a larger subspace of the product space. The set $H$ is also a subset of the following $\Sigma$-product:

$\Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}$

Because $\Sigma(\omega_1)$ is the $\Sigma$-product of separable metric spaces, it is normal (see here). By Theorem 1a in this previous post, $\Sigma(\omega_1)$ would have countable extent. Thus the set $H$ cannot be closed and discrete in $\Sigma(\omega_1)$. We can actually see this directly. Let $\alpha<\omega_1$ be a limit ordinal. Define $t:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by $t(\beta)=1$ for all $\beta<\alpha$ and $t(\beta)=0$ for all $\beta \ge \alpha$. Clearly $t \notin C_p(\omega_1 +1)$ and $t \in \Sigma(\omega_1)$. Furthermore, $t \in \overline{H}$ (the closure is taken in $\Sigma(\omega_1)$).

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Reference

1. Arhangel’skii, A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.

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$\copyright \ 2014 \text{ by Dan Ma}$