# Every Corson compact space has a dense first countable subspace

In any topological space $X$, a point $x \in X$ is a $G_\delta$ point if the one-point set $\left\{ x \right\}$ is the intersection of countably many open subsets of $X$. It is well known that any compact Hausdorff space is first countable at every $G_\delta$ point, i.e., if a point of a compact space is a $G_\delta$ point, then there is a countable local base at that point. It is also well known that uncountable power of first countable spaces can fail to be first countable at every point. For example, no point of the compact space $[0,1]^{\omega_1}$ can be a $G_\delta$ point. In this post, we show that any Corson compact space has a dense set of $G_\delta$ point. Therefore, any Corson compact space is first countable on a dense set (see Corollary 4 below). However, it is not true that every Corson compact space has a dense metrizable subspace. See Theorem 9.14 in [2] for an example of a first countable Corson compact space with no dense metrizable subspace. A list of other blog posts on Corson compact spaces is given at the end of this post.

The fact that every Corson compact space has a dense first countable subspace is taken as a given in the literature. For one example, see chapter c-16 of [1]. Even though Corollary 4 is a basic fact of Corson compact spaces, the proof involves much more than a direct application of the relevant definitions. The proof given here is intended to be an online resource for any one interested in knowing more about Corson compact spaces.

For any infinite cardinal number $\kappa$, the $\Sigma$-product of $\kappa$ many copies of $\mathbb{R}$ is the following subspace of $\mathbb{R}^\kappa$:

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

A compact space is said to be a Corson compact space if it can be embedded in $\Sigma(\kappa)$ for some infinite cardinal $\kappa$.

For each $x \in \Sigma(\kappa)$, let $S(x)$ denote the support of the point $x$, i.e., $S(x)$ is the set of all $\alpha<\kappa$ such that $x_\alpha \ne 0$.

Proposition 1
Let $Y$ be a Corson compact space. Then $Y$ has a $G_\delta$ point.

Proof of Proposition 1
If $Y$ is finite, then every point is isolated and is thus a $G_\delta$ point. Assume $Y$ is infinite. Let $\kappa$ be an infinite cardinal number such that $Y \subset \Sigma(\kappa)$. For $f,g \in Y$, define $f \le g$ if the following holds:

$\forall \ \alpha \in S(f)$, $f(\alpha)=g(\alpha)$

It is relatively straightforward to verify that the following three properties are satisfied:

• $f \le f$ for all $f \in Y$. (reflexivity)
• For all $f,g \in Y$, if $f \le g$ and $g \le f$, then $f=g$. (antisymmetry)
• For all $f,g,h \in Y$, if $f \le g$ and $g \le h$, then $f \le h$. (transitivity)

Thus $\le$ as defined here is a partial order on the compact space $Y$. Let $C \subset Y$ such that $C$ is a chain with respect to $\le$, i.e., for all $f,g \in C$, $f \le g$ or $g \le f$. We show that $C$ has an upper bound (in $Y$) with respect to the partial order $\le$. We need this for an argument using Zorn’s lemma.

Let $W=\bigcup_{f \in C} S(f)$. For each $\alpha \in W$, choose some $f \in C$ such that $\alpha \in S(f)$ and define $u_\alpha=f_\alpha$. For all $\alpha \notin W$, define $u_\alpha=0$. Because $C$ is a chain, the point $u$ is well-defined. It is also clear that $f \le u$ for all $f \in C$. If $u \in Y$, then $u$ is a desired upper bound of $C$. So assume $u \notin Y$. It follows that $u$ is a limit point of $C$, i.e., every open set containing $u$ contains a point of $C$ different from $u$. Hence $u$ is a limit point of $Y$ too. Since $Y$ is compact, $u \in Y$, a contradiction. Thus it must be that $u \in Y$. Thus every chain in the partially ordered set $(Y,\le)$ has an upper bound. By Zorn’s lemma, there exists at least one maximal element with respect to the partial order $\le$, i.e., there exists $t \in Y$ such that $f \le t$ for all $f \in Y$.

We now show that $t$ is a $G_\delta$ point in $Y$. Let $S(t)=\left\{\alpha_1,\alpha_2,\alpha_3,\cdots \right\}$. For each $p \in \mathbb{R}$ and for each positive integer $n$, let $B_{p,n}$ be the open interval $B_{p,n}=(p-\frac{1}{n},p+\frac{1}{n})$. For each positive integer $n$, define the open set $O_n$ as follows:

$O_n=(B_{t_{\alpha_1},n} \times \cdots \times B_{t_{\alpha_n},n} \times \prod_{\alpha<\kappa,\alpha \notin \left\{ \alpha_1,\cdots,\alpha_n \right\}} \mathbb{R}) \cap Y$

Note that $t \in \bigcap_{n=1}^\infty O_n$. Because $t$ is a maximal element, note that if $g \in Y$ such that $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$, then it must be the case that $g=t$. Thus if $g \in \bigcap_{n=1}^\infty O_n$, then $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$. We have $\left\{t \right\}= \bigcap_{n=1}^\infty O_n$. $\blacksquare$

Lemma 2
Let $Y$ be a compact space such that for every non-empty compact subspace $K$ of $Y$, there exists a $G_\delta$ point in $K$. Then every non-empty open subset of $Y$ contains a $G_\delta$ point.

Proof of Lemma 2
Let $U_1$ be a non-empty open subset of the compact space $Y$. If there exists $y \in U_1$ such that $\left\{y \right\}$ is open in $Y$, then $y$ is a $G_\delta$ point. So assume that every point of $U_1$ is a non-isolated point of $Y$. By regularity, choose an open subset $U_2$ of $Y$ such that $\overline{U_2} \subset U_1$. Continue in the same manner and obtain a decreasing sequence $U_1,U_2,U_3,\cdots$ of open subsets of $Y$ such that $\overline{U_{n+1}} \subset U_n$ for each positive integer $n$. Then $K=\bigcap_{n=1}^\infty \overline{U_n}$ is a non-empty closed subset of $Y$ and thus compact. By assumption, $K$ has a $G_\delta$ point, say $p \in K$.

Then $\left\{p \right\}=\bigcap_{n=1}^\infty W_n$ where each $W_n$ is open in $K$. For each $n$, let $V_n$ be open in $Y$ such that $W_n=V_n \cap K$. For each $n$, let $V_n^*=V_n \cap U_n$, which is open in $Y$. Then $\left\{p \right\}=\bigcap_{n=1}^\infty V_n^*$. This means that $p$ is a $G_\delta$ point in the compact space $Y$. Note that $p \in U_1$, the open set we start with. This completes the proof that every non-empty open subset of $Y$ contains a $G_\delta$ point. $\blacksquare$

Proposition 3
Let $Y$ be a Corson compact space. Then $Y$ has a dense set of $G_\delta$ points.

Proof of Proposition 3
Note that Corson compactness is hereditary with respect to closed sets. Thus every compact subspace of $Y$ is also Corson compact. By Proposition 1, every compact subspace of $Y$ has a $G_\delta$ point. By Lemma 2, $Y$ has a dense set of $G_\delta$ points. $\blacksquare$

Corollary 4
Every Corson compact space has a dense first countable subspace.

____________________________________________________________________

Blog posts on Corson compact spaces

____________________________________________________________________

Reference

1. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
2. Todorcevic, S., Trees and Linearly Ordered Sets, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 235-293, 1984.

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

# The Sorgenfrey plane is subnormal

The Sorgenfrey line is the real line with the topology generated by the base of half-open intervals of the form $[a,b)$. The Sorgenfrey line is one of the most important counterexamples in general topology. One of the often recited facts about this counterexample is that the Sorgenfrey plane (the square of the Sorgengfrey line) is not normal. We show that, though far from normal, the Sorgenfrey plane is subnormal.

A subset $M$ of a space $Y$ is a $G_\delta$ subset of $Y$ (or a $G_\delta$-set in $Y$) if $M$ is the intersection of countably many open subsets of $Y$. A subset $M$ of a space $Y$ is a $F_\sigma$ subset of $Y$ (or a $F_\sigma$-set in $Y$) if $Y-M$ is a $G_\delta$-set in $Y$ (equivalently if $M$ is the union of countably many closed subsets of $Y$).

A space $Y$ is normal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. A space $Y$ is subnormal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint $G_\delta$ subsets $V_H$ and $V_K$ of $Y$ such that $H \subset V_H$ and $K \subset V_K$. Clearly any normal space is subnormal. The Sorgenfrey plane is an example of a subnormal space that is not normal.

In the proof of the non-normality of the Sorgenfrey plane in this previous post, one of the two disjoint closed subsets of the Sorgenfrey plane that cannot be separated by disjoint open sets is countable. Thus the Sorgenfrey plane is not only not normal; it is not pseudonormal (also discussed in this previous post). A space $Y$ is pseudonormal if for any disjoint closed subsets $H$ and $K$ of $Y$ (one of which is countable), there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. The examples of the Sorgenfrey plane and $\omega_1 \times (\omega_1+1)$ show that these two weak forms of normality (pseudonormal and subnormal) are not equivalent. The space $\omega_1 \times (\omega_1+1)$ is pseudonormal but not subnormal (see this previous post for the non-subnormality).

A space $Y$ is said to be a perfect space if every closed subset of $Y$ is a $G_\delta$ subset of $Y$ (equivalently, every open subset of $Y$ is an $F_\sigma$-subset of $Y$). It is clear that any perfect space is subnormal. We show that the Sorgenfrey plane is perfect. There are subnormal spaces that are not perfect (see the example below).

____________________________________________________________________

The Sorgenfrey plane is perfect

Let $S$ denote the Sorgenfrey line, i.e., the real line $\mathbb{R}$ topologized using the base of half-open intervals of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x . The Sorgenfrey plane is the product space $S \times S$. We show the following:

Proposition 1
The Sorgenfrey line $S$ is perfect.

Proof of Proposition 1
Let $U$ be a non-empty subset of $S$. We show that $U$ is a $F_\sigma$-set. Let $U_0$ be the interior of $U$ in the usual topology. In other words, $U_0$ is the following set:

$U_0=\left\{x \in U: \exists \ (a,b) \text{ such that } x \in (a,b) \text{ and } (a,b) \subset U \right\}$

The real line with the usual topology is perfect. Thus $U_0=\bigcup_{n=1}^\infty F_n$ where each $F_n$ is a closed subset of the real line $\mathbb{R}$. Since the Sorgenfrey topology is finer than the usual topology, each $F_n$ is also closed in the Sorgenfrey line.

Consider $Y=U-U_0$. We claim that $Y$ is countable. Suppose $Y$ is uncountable. Since the Sorgenfrey line is hereditarily Lindelof, there exists $y \in Y$ such that $y$ is a limit point of $Y$ (see Corollary 2 in this previous post). Since $y \in Y \subset U$, $[y,t) \subset U$ for some $t$. Note that $(y,t) \subset U_0$, which means that no point of the open interval $(y,t)$ can belong to $Y$. On the other hand, since $y$ is a limit point of $Y$, $y for some $w \in Y$, a contradiction. Thus $Y$ must be countable. It follows that $U$ is the union of countably many closed subsets of $S$. $\blacksquare$

Proposition 2
If $X$ is perfect and $Y$ is metrizable, then $X \times Y$ is perfect.

Proof of Proposition 2
Let $X$ be perfect. Let $Y$ be a space with a base $\mathcal{B}=\bigcup_{n=1}^\infty \mathcal{B}_n$ such that each $\mathcal{B}_n$, in addition to being a collection of basic open sets, is a discrete collection. The existence of such a base is equivalent to metrizability, a well known result called Bing’s metrization theorem (see Theorem 4.4.8 in [1]). Let $U$ be a non-empty open subset of $X \times Y$. We show that it is an $F_\sigma$-set in $X \times Y$. For each $x \in U$, there is some open subset $V$ of $X$ and there is some $W \in \mathcal{B}$ such that $x \in V \times W$ and $V \times \overline{W} \subset U$. Thus $U$ is the union of a collection of sets of the form $V \times \overline{W}$. Thus we have:

$U=\bigcup \mathcal{O} \text{ where } \mathcal{O}=\left\{ V_\alpha \times \overline{W_\alpha}: \alpha \in A \right\}$

for some index set $A$. For each positive integer $m$, let $\mathcal{O}_m$ be defined by

$\mathcal{O}_m=\left\{V_\alpha \times \overline{W_\alpha} \in \mathcal{O}: W_\alpha \in \mathcal{B}_m \right\}$

For each $\alpha \in A$, let $V_\alpha=\bigcup_{n=1}^\infty V_{\alpha,n}$ where each $V_{\alpha,n}$ is a closed subset of $X$. For each pair of positive integers $n$ and $m$, define $\mathcal{O}_{n,m}$ by

$\mathcal{O}_{n,m}=\left\{V_{\alpha,n} \times \overline{W_\alpha}: V_\alpha \times \overline{W_\alpha} \in \mathcal{O}_m \right\}$

We claim that each $\mathcal{O}_{n,m}$ is a discrete collection of sets in the space $X \times Y$. Let $(a,b) \in X \times Y$. Since $\mathcal{B}_m$ is discrete, there exists some open subset $H_b$ of $Y$ with $b \in H_b$ such that $H_b$ can intersect at most one $\overline{W}$ where $W \in \mathcal{B}_m$. Then $X \times H_b$ is an open subset of $X \times Y$ with $(a,b) \in X \times H_b$ such that $X \times H_b$ can intersect at most one set of the form $V_{\alpha,n} \times \overline{W_\alpha}$. Then $C_{n,m}=\bigcup \mathcal{O}_{n,m}$ is a closed subset of $X \times Y$. It is clear that $U$ is the union of $C_{n,m}$ over all countably many possible pairs $n,m$. Thus $U$ is an $F_\sigma$-set in $X \times Y$. $\blacksquare$

Proposition 3
The Sorgenfrey plane $S \times S$ is perfect.

Proof of Proposition 3
To get ready for the proof, consider the product spaces $X_1=\mathbb{R} \times S$ and $X_2=S \times \mathbb{R}$ where $\mathbb{R}$ has the usual topology. By both Proposition 1 and Proposition 2, both $X_1$ and $X_2$ are perfect. Also note that the Sorgenfrey plane topology is finer than the topologies for both $X_1$ and $X_2$. Thus a closed set in $X_1$ (in $X_2$) is also a closed set in $S \times S$. It follows that any $F_\sigma$-set in $X_1$ (in $X_2$) is also an $F_\sigma$-set in $S \times S$.

Let $U$ be a non-empty subset of $S \times S$. We show that $U$ is a $F_\sigma$-set. We assume that $U$ is the union of basic open sets of the form $[a,b) \times [a,b)$. Consider the sets $U_1$ and $U_2$ defined by:

$U_1=\left\{x \in U: \exists \ (a,b) \times [a,b) \text{ such that } x \in (a,b) \times [a,b) \text{ and } (a,b) \times [a,b) \subset U \right\}$

$U_2=\left\{x \in U: \exists \ [a,b) \times (a,b) \text{ such that } x \in [a,b) \times (a,b) \text{ and } [a,b) \times (a,b) \subset U \right\}$

Note that $U_1$ is the interior of $U$ when $U$ is considered as a subspace of $X_1$. Likewise, $U_2$ is the interior of $U$ when $U$ is considered as a subspace of $X_2$. Since both $X_1$ and $X_2$ are perfect, $U_1$ and $U_2$ are $F_\sigma$ in $X_1$ and $X_2$, respectively. Hence both $U_1$ and $U_2$ are $F_\sigma$-sets in $S \times S$.

Let $Y=U-(U_1 \cup U_2)$. We claim that $Y$ is an $F_\sigma$-set in $S \times S$. Proposition 3 is established when this claim is proved. To get ready to prove this claim, for each $x=(x_1,x_2) \in S \times S$, and for each positive integer $k$, let $B_k(x)$ be the half-open square $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$. Then $\mathcal{B}(x)=\left\{B_k(x): k=1,2,3,\cdots \right\}$ is a local base at the point $x$. For each positive integer $k$, define $Y_k$ by

$Y_k=\left\{y=(y_1,y_2) \in Y: B_k(y) \subset U \right\}$

Clearly $Y=\bigcup_{k=1}^\infty Y_k$. We claim that each $Y_k$ is closed in $S \times S$. Suppose $x=(x_1,x_2) \in S \times S-Y_k$. In relation to the point $x$, $Y_k$ can be broken into several subsets as follows:

$Y_{k,1}=\left\{y=(y_1,y_2) \in Y_k: y_1=x_1 \text{ and } y_2 \ne x_2 \right\}$

$Y_{k,2}=\left\{y=(y_1,y_2) \in Y_k: y_1 \ne x_1 \text{ and } y_2 = x_2 \right\}$

$Y_{k,\varnothing}=\left\{y=(y_1,y_2) \in Y_k: y_1 \ne x_1 \text{ and } y_2 \ne x_2 \right\}$

Since $x \notin Y_k$, it follows that $Y_k=Y_{k,1} \cup Y_{k,2} \cup Y_{k,\varnothing}$. We show that for each of these three sets, there is an open set containing the point $x$ that is disjoint from the set.

Consider $Y_{k,1}$. If $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ is disjoint from $Y_{k,1}$, then we are done. So assume $B_k(x) \cap Y_{k,1} \ne \varnothing$. Let $t=(t_1,t_2) \in B_k(x) \cap Y_{k,1}$. Note that $t_1=x_1$ and $t_2 > x_2$. Now consider the following open set:

$G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_2

The set $G$ is an open set containing the point $x$. We claim that $G \cap Y_{k,1}=\varnothing$. Suppose $g \in G \cap Y_{k,1}$. Then $g_1=x_1$ and $x_2. Consider the following set:

$H=B_k(g) \cap \left\{h=(h_1,h_2) \in S \times S: g_2

Note that $H$ is an open subset of $X_2=S \times \mathbb{R}$. Since $g \in Y_k$, it follows that $H \subset B_k(g) \subset U$. Thus $H$ is a subset of the interior of $U$ (as a subspace of $X_2$). We have $H \subset U_2$. It follows that $t \in H$ since

$x_1=g_1=t_1$

$x_2

On the other hand, $t \in Y_{k,1} \subset Y_k \subset Y$. Hence $t \notin U_2$, a contradiction. Thus the claim that $G \cap Y_{k,1}=\varnothing$ must be true.

The case $Y_{k,2}$ is symmetrical to the case $Y_{k,1}$. Thus by applying a similar argument, there is an open set containing the point $x$ that is disjoint from the set $Y_{k,2}$.

Now consider the case $Y_{k,\varnothing}$. If $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ is disjoint from $Y_{k,\varnothing}$, then we are done. So assume $B_k(x) \cap Y_{k,\varnothing} \ne \varnothing$. Let $t=(t_1,t_2) \in B_k(x) \cap Y_{k,\varnothing}$. Note that $t_1>x_1$ and $t_2 > x_2$. Now consider the following open set:

$G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_1

The set $G$ is an open set containing the point $x$. We claim that $G \cap Y_{k,\varnothing}=\varnothing$. Suppose $g \in G \cap Y_{k,\varnothing}$. Then $x_1 and $x_2. Consider the following set:

$H=B_k(g) \cap \left\{h=(h_1,h_2) \in S \times S: g_2

As in the previous case, $H$ is an open subset of $X_2=S \times \mathbb{R}$. Since $g \in Y_k$, it follows that $H \subset B_k(g) \subset U$. As before, $H \subset U_2$. We also have a contradiction in that $t \in H$ (based on the following)

$x_1

$x_2

and on the one hand and $t \in Y_{k,\varnothing} \subset Y=U-(U_1 \cup U_2)$. Thus the claim that $G \cap Y_{k,\varnothing}=\varnothing$ is true. Take the intersection of the three open sets from the three cases, we have an open set containing $x$ that is disjoint from $Y_k$. Thus $Y_k$ is closed in $S \times S$ and $Y=\bigcup_{k=1}^\infty Y_k$ is $F_\sigma$ in $S \times S$ . $\blacksquare$

Remarks
The authors of [2] showed that any finite power of the Sorgenfrey line is perfect. The proof in [2] is an inductive proof: if $S^n$ is perfect, then $S^{n+1}$ is perfect. We take the inductive proof in [2] and adapt it for the Sorgenfrey plane. The authors in [2] also proved that for a sequence of spaces $X_1,X_2,X_3,\cdots$ such that the product of any finite number of these spaces is perfect, the product $\prod_{n=1}^\infty X_n$ is perfect. Then $S^\omega$ is perfect.

____________________________________________________________________

A non-perfect example

Any perfect space is subnormal. Subnormal spaces do not have to be perfect. In fact subnormal non-normal spaces do not have to be perfect. From a perfect space that is not normal (e.g. the Sorgenfrey plane), one can generate a subnormal and non-normal space that is not perfect. Let $X$ be a subnormal and non-normal space. Let $Y$ be a normal space that is not perfectly normal. There are many possible choices for $Y$. If a specific example is needed, one can take $Y=\omega_1$ with the order topology. Let $X \bigoplus Y$ be the disjoint sum (union) of $X$ and $Y$. The presence of $Y$ destroys the perfectness. It is clear that any two disjoint closed sets can be separated by disjoint $G_\delta$-sets.

____________________________________________________________________

Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Heath, R. W., Michael, E., A property of the Sorgenfrey line, Compositio Math., 23, 185-188, 1971.

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

# Normal x compact needs not be subnormal

In this post, we revisit a counterexample that was discussed previously in this blog. A previous post called “Normal x compact needs not be normal” shows that the Tychonoff product of two normal spaces needs not be normal even when one of the factors is compact. The example is $\omega_1 \times (\omega_1+1)$. In this post, we show that $\omega_1 \times (\omega_1+1)$ fails even to be subnormal. Both $\omega_1$ and $\omega_1+1$ are spaces of ordinals. Thus they are completely normal (equivalent to hereditarily normal). The second factor is also a compact space. Yet their product is not only not normal; it is not even subnormal.

A subset $M$ of a space $Y$ is a $G_\delta$ subset of $Y$ (or a $G_\delta$-set in $Y$) if $M$ is the intersection of countably many open subsets of $Y$. A subset $M$ of a space $Y$ is a $F_\sigma$ subset of $Y$ (or a $F_\sigma$-set in $Y$) if $Y-M$ is a $G_\delta$-set in $Y$ (equivalently if $M$ is the union of countably many closed subsets of $Y$).

A space $Y$ is normal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. A space $Y$ is subnormal if for any disjoint closed subsets $H$ and $K$ of $Y$, there exist disjoint $G_\delta$ subsets $V_H$ and $V_K$ of $Y$ such that $H \subset V_H$ and $K \subset V_K$. Clearly any normal space is subnormal.

A space $Y$ is pseudonormal if for any disjoint closed subsets $H$ and $K$ of $Y$ (one of which is countable), there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$. The space $\omega_1 \times (\omega_1+1)$ is pseudonormal (see this previous post). The Sorgenfrey plane is an example of a subnormal space that is not pseudonormal (see here). Thus the two weak forms of normality (pseudonormal and subnormal) are not equivalent.

The same two disjoint closed sets that prove the non-normality of $\omega_1 \times (\omega_1+1)$ are also used for proving non-subnormality. The two closed sets are:

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

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

The key tool, as in the proof for non-normality, is the Pressing Down Lemma ([1]). The lemma has been used in a few places in this blog, especially for proving facts about $\omega_1$ (e.g. this previous post on the first uncountable ordinal). Lemma 1 below is a lemma that is derived from the Pressing Down Lemma.

Pressing Down Lemma
Let $S$ be a stationary subset of $\omega_1$. Let $f:S \rightarrow \omega_1$ be a pressing down function, i.e., $f$ satisfies: $\forall \ \alpha \in S, f(\alpha)<\alpha$. Then there exists $\alpha<\omega_1$ such that $f^{-1}(\alpha)$ is a stationary set.

Lemma 1
Let $L=\left\{(\alpha,\alpha) \in \omega_1 \times \omega_1: \alpha \text{ is a limit ordinal} \right\}$. Suppose that $L \subset \bigcap_{n=1}^\infty O_n$ where each $O_n$ is an open subset of $\omega_1 \times \omega_1$. Then $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty O_n$ for some $\gamma<\omega_1$.

Proof of Lemma 1
For each $n$ and for each $\alpha<\omega_1$ where $\alpha$ is a limit, choose $g_n(\alpha)<\alpha$ such that $[g_n(\alpha),\alpha] \times [g_n(\alpha),\alpha] \subset O_n$. The function $g_n$ can be chosen since $O_n$ is open in the product $\omega_1 \times \omega_1$. By the Pressing Down Lemma, for each $n$, there exists $\gamma_n < \omega_1$ and there exists a stationary set $S_n \subset \omega_1$ such that $g_n(\alpha)=\gamma_n$ for all $\alpha \in S_n$. It follows that $[\gamma_n,\omega_1) \times [\gamma_n,\omega_1) \subset O_n$ for each $n$. Choose $\gamma<\omega_1$ such that $\gamma_n<\gamma$ for all $n$. Then $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset O_n$ for each $n$. $\blacksquare$

Theorem 2
The product space $\omega_1 \times (\omega_1+1)$ is not subnormal.

Proof of Theorem 2
Let $H$ and $K$ be defined as above. Suppose $H \subset \bigcap_{n=1}^\infty U_n$ and $K \subset \bigcap_{n=1}^\infty V_n$ where each $U_n$ and each $V_n$ are open in $\omega_1 \times (\omega_1+1)$. Without loss of generality, we can assume that $U_n \cap (\omega_1 \times \left\{\omega_1 \right\})=\varnothing$, i.e., $U_n$ is open in $\omega_1 \times \omega_1$ for each $n$. By Lemma 1, $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty U_n$ for some $\gamma<\omega_1$.

Choose $\beta>\gamma$ such that $\beta$ is a successor ordinal. Note that $(\beta,\omega_1) \in \bigcap_{n=1}^\infty V_n$. For each $n$, there exists some $\delta_n<\omega_1$ such that $\left\{\beta \right\} \times [\delta_n,\omega_1] \subset V_n$. Choose $\delta<\omega_1$ such that $\delta >\delta_n$ for all $n$ and that $\delta >\gamma$. Note that $\left\{\beta \right\} \times [\delta,\omega_1) \subset \bigcap_{n=1}^\infty V_n$. It follows that $\left\{\beta \right\} \times [\delta,\omega_1) \subset [\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty U_n$. Thus there are no disjoint $G_\delta$ sets separating $H$ and $K$. $\blacksquare$

____________________________________________________________________

Reference

1. Kunen, K., Set Theory, An Introduction to Independence Proofs, First Edition, North-Holland, New York, 1980.

____________________________________________________________________

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