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 <b \right\}. 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<w<t 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<t_2 \right\}

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<g_2<t_2. Consider the following set:

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

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<g_2<t_2<x_2+\frac{1}{k}<g_2+\frac{1}{k}

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<t_1 \text{ and }y_2<t_2 \right\}

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<g_1<t_1 and x_2<g_2<t_2. Consider the following set:

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

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<g_1<t_1<x_1+\frac{1}{k}<g_1+\frac{1}{k}

    x_2<g_2<t_2<x_2+\frac{1}{k}<g_2+\frac{1}{k}

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}

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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}