# 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).

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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.

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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.

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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.

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

# 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 $K$. $\blacksquare$

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Reference

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

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

# Equivalent conditions for hereditarily Lindelof spaces

A topological space $X$ is Lindelof if every open cover $X$ has a countable subcollection that also is a cover of $X$. A topological space $X$ is hereditarily Lindelof if every subspace of $X$, with respect to the subspace topology, is a Lindelof space. In this post, we prove a theorem that gives two equivalent conditions for the hereditarily Lindelof property. We consider the following theorem.

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

1. The space $X$ is a hereditarily Lindelof space.
2. Every open subspace of $X$ is Lindelof.
3. For every uncountable subspace $Y$ of $X$, there exists a point $y \in Y$ such that every open subset of $X$ containing $y$ contains uncountably many points of $Y$.

This is an excellent exercise for the hereditarily Lindelof property and for transfinite induction (for one of the directions). The equivalence $1 \longleftrightarrow 3$ is the exercise 3.12.7(d) on page 224 of [1]. The equivalence of the 3 conditions of Theorem 1 is mentioned on page 182 (chapter d-8) of [2].

Proof of Theorem 1
The direction $1 \longrightarrow 2$ is immediate. The direction $2 \longrightarrow 3$ is straightforward.

$3 \longrightarrow 1$
We show $\text{not } 1 \longrightarrow \text{not } 3$. Suppose $T$ is a non-Lindelof subspace of $X$. Let $\mathcal{U}$ be an open cover of $T$ such that no countable subcollection of $\mathcal{U}$ can cover $T$. By a transfinite inductive process, choose a set of points $\left\{t_\alpha \in T: \alpha < \omega_1 \right\}$ and a collection of open sets $\left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\}$ such that for each $\alpha < \omega_1$, $t_\alpha \in U_\alpha$ and $t_\alpha \notin \cup \left\{U_\beta: \beta<\alpha \right\}$. The inductive process is possible since no countable subcollection of $\mathcal{U}$ can cover $T$. Now let $Y=\left\{t_\alpha: \alpha<\omega_1 \right\}$. Note that each $U_\alpha$ can at most contain countably many points of $Y$, namely the points in $\left\{t_\beta: \beta \le \alpha \right\}$.

For each $\alpha$, let $V_\alpha$ be an open subset of $X$ such that $U_\alpha=V_\alpha \cap Y$. We can now conclude: for every point $t_\alpha$ of $Y$, there exists an open set $V_\alpha$ containing $t_\alpha$ such that $V_\alpha$ contains only countably many points of $Y$. This is the negation of condition 3. $\blacksquare$

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Remarks

Condition 3 indicates that every uncountable set has a certain special type of limit points. Let $p \in X$. We say $p$ is a limit point of the set $Y \subset X$ if every open set containing $p$ contains a point of $Y$ different from $p$. Being a limit point of $Y$, we only know that each open set containing $p$ contain infinitely many points of $Y$ (assuming a $T_1$ space). Thus the limit points indicated in condition 3 are a special type of limit points. According to the terminology of [1], if $p$ is a limit point of $Y$ satisfying condition 3, then $p$ is said to be a condensation point of $Y$. According to Theorem 1, existence of condensation point in every uncountable set is a strong topological property (being equivalent to the hereditarily property). It is easy to see that of condition 3 holds, all but countably many points of any uncountable set $Y$ is a condensation point of $Y$.

In some situations, we may not need the full strength of condition 3. In such situations, the following corollary may be sufficient.

Corollary 2
If the space $X$ is hereditarily Lindelof, then every uncountable subspace $Y$ of $X$ contains one of its limit points.

As noted earlier, if every uncountable set contains one of its limits, then all but countably many points of any uncountable set are limit points. To contrast the hereditarily Lindelof property with the Lindelof property, consider the following theorem.

Theorem 3
If the space $X$ is Lindelof, then every uncountable subspace $Y$ of $X$ has a limit point.

The condition “every uncountable subspace $Y$ of $X$ has a limit point” has another name. When a space satisfies this condition, it is said to have countable extent. The ideas in Corollary 2 and Theorem 3 are also discussed in this previous post.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.

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

# An example of a normal but not Lindelof Cp(X)

In this post, we discuss an example of a function space $C_p(X)$ that is normal and not Lindelof (as indicated in the title). Interestingly, much more can be said about this function space. In this post, we show that there exists a space $X$ such that

• $C_p(X)$ is collectionwise normal and not paracompact,
• $C_p(X)$ is not Lindelof but contains a dense Lindelof subspace,
• $C_p(X)$ is not first countable but is a Frechet space,
• As a corollary of the previous point, $C_p(X)$ cannot contain a copy of the compact space $\omega_1+1$,
• $C_p(X)$ is homeomorphic to $C_p(X)^\omega$,
• $C_p(X)$ is not hereditarily normal,
• $C_p(X)$ is not metacompact.

A short and quick description of the space $X$ is that $X$ is the one-point Lindelofication of an uncountable discrete space. As shown below, the function space $C_p(X)$ is intimately related to a $\Sigma$-product of copies of real lines. The results listed above are merely an introduction to this wonderful example and are derived by examining the $\Sigma$-products of copies of real lines. Deep results about $\Sigma$-product of real lines abound in the literature. The references listed at the end are a small sample. Example 3.2 in [2] is another interesting illustration of this example.

We now define the domain space $X=L_\tau$. In the discussion that follows, the Greek letter $\tau$ is always an uncountable cardinal number. Let $D_\tau$ be a set with cardinality $\tau$. Let $p$ be a point not in $D_\tau$. Let $L_\tau=D_\tau \cup \left\{p \right\}$. Consider the following topology on $L_\tau$:

• Each point in $D_\tau$ an isolated point, and
• open neighborhoods at the point $p$ are of the form $L_\tau-K$ where $K \subset D_\tau$ is countable.

It is clear that $L_\tau$ is a Lindelof space. The Lindelof space $L_\tau$ is sometimes called the one-point Lindelofication of the discrete space $D_\tau$ since it is a Lindelof space that is obtained by adding one point to a discrete space.

Consider the function space $C_p(L_\tau)$. See this post for general information on the pointwise convergence topology of $C_p(Y)$ for any completely regular space $Y$.

All the facts about $C_p(X)=C_p(L_\tau)$ mentioned at the beginning follow from the fact that $C_p(L_\tau)$ is homeomorphic to the $\Sigma$-product of $\tau$ many copies of the real lines. Specifically, $C_p(L_\tau)$ is homeomorphic to the following subspace of the product space $\mathbb{R}^\tau$.

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

Thus understanding the function space $C_p(L_\tau)$ is a matter of understanding a $\Sigma$-product of copies of the real lines. First, we establish the homeomorphism and then discuss the properties of $C_p(L_\tau)$ indicated above.

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The Homeomorphism

For each $f \in C_p(L_\tau)$, it is easily seen that there is a countable set $C \subset D_\tau$ such that $f(p)=f(y)$ for all $y \in D_\tau-C$. Let $W_0=\left\{f \in C_p(L_\tau): f(p)=0 \right\}$. Then each $f \in W_0$ has non-zero values only on a countable subset of $D_\tau$. Naturally, $W_0$ and $\Sigma_{\alpha<\tau}\mathbb{R}$ are homeomorphic.

We claim that $C_p(L_\tau)$ is homeomorphic to $W_0 \times \mathbb{R}$. For each $f \in C_p(L_\tau)$, define $h(f)=(f-f(p),f(p))$. Here, $f-f(p)$ is the function $g \in C_p(L_\tau)$ such that $g(x)=f(x)-f(p)$ for all $x \in L_\tau$. Clearly $h(f)$ is well-defined and $h(f) \in W_0 \times \mathbb{R}$. It can be readily verified that $h$ is a one-to-one map from $C_p(L_\tau)$ onto $W_0 \times \mathbb{R}$. It is not difficult to verify that both $h$ and $h^{-1}$ are continuous.

We use the notation $X_1 \cong X_2$ to mean that the spaces $X_1$ and $X_2$ are homeomorphic. Then we have:

$C_p(L_\tau) \ \cong \ W_0 \times \mathbb{R} \ \cong \ (\Sigma_{\alpha<\tau}\mathbb{R}) \times \mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$

Thus $C_p(L_\tau) \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$. This completes the proof that $C_p(L_\tau)$ is topologically the $\Sigma$-product of $\tau$ many copies of the real lines.

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Looking at the $\Sigma$-Product

Understanding the function space $C_p(L_\tau)$ is now reduced to the problem of understanding a $\Sigma$-product of copies of the real lines. Most of the facts about $\Sigma$-products that we need have already been proved in previous blog posts.

In this previous post, it is established that the $\Sigma$-product of separable metric spaces is collectionwise normal. Thus $C_p(L_\tau)$ is collectionwise normal. The $\Sigma$-product of spaces, each of which has at least two points, always contains a closed copy of $\omega_1$ with the ordered topology (see the lemma in this previous post). Thus $C_p(L_\tau)$ contains a closed copy of $\omega_1$ and hence can never be paracompact (and thus not Lindelof).

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Consider the following subspace of the $\Sigma$-product $\Sigma_{\alpha<\tau}\mathbb{R}$:

$\sigma_\tau=\left\{ x \in \Sigma_{\alpha<\tau}\mathbb{R}: x_\alpha \ne 0 \text{ for at most finitely many } \alpha<\tau \right\}$

In this previous post, it is shown that $\sigma_\tau$ is a Lindelof space. Though $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is not Lindelof, it has a dense Lindelof subspace, namely $\sigma_\tau$.

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A space $Y$ is first countable if there exists a countable local base at each point $y \in Y$. A space $Y$ is a Frechet space (or is Frechet-Urysohn) if for each $y \in Y$, if $y \in \overline{A}$ where $A \subset Y$, then there exists a sequence $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $A$ such that the sequence converges to $y$. Clearly, any first countable space is a Frechet space. The converse is not true (see Example 1 in this previous post).

For any uncountable cardinal number $\tau$, the product $\mathbb{R}^\tau$ is not first countable. In fact, any dense subspace of $\mathbb{R}^\tau$ is not first countable. In particular, the $\Sigma$-product $\Sigma_{\alpha<\tau}\mathbb{R}$ is not first countable. In this previous post, it is shown that the $\Sigma$-product of first countable spaces is a Frechet space. Thus $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is a Frechet space.

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As a corollary of the previous point, $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ cannot contain a homeomorphic copy of any space that is not Frechet. In particular, it cannot contain a copy of any compact space that is not Frechet. For example, the compact space $\omega_1+1$ is not embeddable in $C_p(L_\tau)$. The interest in compact subspaces of $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is that any compact space that is topologically embeddable in a $\Sigma$-product of real lines is said to be Corson compact. Thus any Corson compact space is a Frechet space.

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It can be readily verified that

$\Sigma_{\alpha<\tau}\mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \cdots \ \text{(countably many times)}$

Thus $C_p(L_\tau) \cong C_p(L_\tau)^\omega$. In particular, $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$ due to the following observation:

$C_p(L_\tau) \times C_p(L_\tau) \cong C_p(L_\tau)^\omega \times C_p(L_\tau)^\omega \cong C_p(L_\tau)^\omega \cong C_p(L_\tau)$

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As a result of the peculiar fact that $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$, it can be concluded that $C_p(L_\tau)$, though normal, is not hereditarily normal. This follows from an application of Katetov’s theorem. The theorem states that if $Y_1 \times Y_2$ is hereditarily normal, then either $Y_1$ is perfectly normal or every countably infinite subset of $Y_2$ is closed and discrete (see this previous post). The function space $C_p(L _\tau)$ is not perfectly normal since it contains a closed copy of $\omega_1$. On the other hand, there are plenty of countably infinite subsets of $C_p(L _\tau)$ that are not closed and discrete. As a Frechet space, $C_p(L _\tau)$ has many convergent sequences. Each such sequence without the limit is a countably infinite set that is not closed and discrete. As an example, let $\left\{x_1,x_2,x_3,\cdots \right\}$ be an infinite subset of $D_\tau$ and consider the following:

$C=\left\{f_n: n=1,2,3,\cdots \right\}$

where $f_n$ is such that $f_n(x_n)=n$ and $f_n(x)=0$ for each $x \in L_\tau$ with $x \ne x_n$. Note that $C$ is not closed and not discrete since the points in $C$ converge to $g \in \overline{C}$ where $g$ is the zero-function. Thus $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$ is not hereditarily normal.

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It is well known that collectionwise normal metacompact space is paracompact (see Theorem 5.3.3 in [4] where metacompact is referred to as weakly paracompact). Since $C_p(L_\tau)$ is collectionwise normal and not paracompact, $C_p(L_\tau)$ can never be metacompact.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Bella, A., Masami, S., Tight points of Pixley-Roy hyperspaces, Topology Appl., 160, 2061-2068, 2013.
3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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