One way to find collectionwise normal spaces

Collectionwise normality is a property that is weaker than paracompactness and stronger than normality (see the implications below). Normal spaces need not be collectionwise normal. Bing’s Example G is an example of a normal and not collectionwise normal space (see the blog post “Bing’s Example G”). We discuss one instance when normal spaces are collectionwise normal, giving a way to obtain collectionwise normal spaces that are not paracompact.

    \text{ }
    \text{paracompact} \Longrightarrow \text{collectionwise normal} \Longrightarrow \text{normal}

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Collectionwise Normal Spaces

A normal space is one in which any two disjoint closed sets can be separated by disjoint open sets. By induction, in a normal space any finite number of disjoint closed sets can be separated by disjoint open sets. Of course, the inductive reasoning cannot be carried over to the case of infinitely many disjoint closed sets. In the real line with the usual topology, the singleton sets \left\{x \right\}, where x is rational, are disjoint closed sets that cannot be simultaneously separated by disjoint open sets. In order to separate an infinite collection of disjoint closed sets, it makes sense to restrict on the type of collections of closed sets. A space X is collectionwise normal if every discrete collection of closed subsets of X can be separated by pairwise disjoint open subsets of X. The following is a more specific definition.

    Definition
    A space X is collectionwise normal if for every discrete collection \mathcal{A} of closed subsets of X, there exists a pairwise disjoint collection \mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\} of open subsets of X such that A \subset U_A for each A \in \mathcal{A}.

For more details about the definitions of collectionwise normality, see “Definitions of Collectionwise Normal Spaces”. The implications displayed above are repeated below. None of the arrows is reversible.

    \text{ }
    \text{paracompact} \Longrightarrow \text{collectionwise normal} \Longrightarrow \text{normal}

As indicated above, Bing’s Example G is an example of a normal and not collectionwise normal space (see the blog post “Bing’s Example G”). The propositions in the next section can be used to obtain collectionwise normal spaces that are not paracompact.

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When Normal implies Collectionwise Normal

Being able to simultaneously separate any discrete collection of closed sets is stronger than the property of merely being able to separate finite collection of disjoint closed sets. It turns out that the stronger property of collectionwise normality is required only for separating uncountable discrete collections of closed sets. As the following lemma shows, normality is sufficient to separate any countable discrete collection of closed sets.

Lemma 1
Let X be a normal space. Then for every discrete collection \left\{C_1,C_2,C_3,\cdots \right\} of closed subsets of X, there exists a pairwise disjoint collection \left\{O_1,O_2,O_3,\cdots \right\} of open subsets of X such that C_i \subset O_i for each i.

Proof of Lemma 1
Let \left\{C_1,C_2,C_3,\cdots \right\} be a discrete collection of closed subsets of X. For each i, choose disjoint open sets U_i and V_i such that C_i \subset U_i and \cup \left\{C_j: j \ne i \right\} \subset V_i. Let O_1=U_1. For each i>1, let O_i=U_i \cap V_1 \cap \cdots \cap V_{i-1}. It follows that O_i \cap O_j = \varnothing for all i \ne j. It is also clear that for each i, C_i \subset O_i. \blacksquare

We have the following propositions.

Proposition 2
Let X be a normal space. If all discrete collections of closed subsets of X are at most countable, then X is collectionwise normal.

Proposition 3
Let X be a normal space. If all closed and discrete subsets of X are at most countable (such a space is said to have countable extent), then X is collectionwise normal.

Proposition 4
Any normal and countably compact space is collectionwise normal.

Proposition 2 follows from Lemma 1. As noted in Proposition 3, any space in which all closed and discrete subsets are countable is said to have countable extent. It is easy to verify that X has countable extent if and only if all discrete collections of closed subsets of X are at most countable. If X is a countably compact space, then every infinite subset of X has a limit point. Thus Proposition 4 follows from the fact that any countably compact space has countable extent.

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Paracompact Spaces

One way to find a collectionwise normal space that is not paracompact is to find a non-paracompact space that satisfies Propositions 3, 4 or 5. For example, \omega_1, the space of all countable ordinals with the order topology, is not paracompact. Since \omega_1 is normal and countably compact, it is collectionwise normal by Proposition 4. For a basic discussion of \omega_1 as a topological space, see “The First Uncountable Ordinal”.

As the following theorem shows, paracompact spaces are collectionwise normal. Thus the class of collectionwise normal spaces includes all metric spaces and paracompact spaces.

Theorem 5
If a space X is paracompact, then X is collectionwise normal.

Proof of Theorem 5
Let X be a paracompact space. Let \mathcal{A} be a discrete collection of closed subsets of X. For each x \in A, let O_x be open such that x \in O_x and O_x meets at most one element of \mathcal{A}. Let \mathcal{O}=\left\{O_x: x \in X \right\}. By the paracompactness of X, \mathcal{O} has a locally finite open refinement \mathcal{V}=\left\{V_x: x \in X \right\} such that V_x \subset O_x for each x \in X.

For each A \in \mathcal{A}, let W_A=\cup \left\{\overline{V}: V \in \mathcal{V} \text{ and } \overline{V} \cap A=\varnothing \right\} and let U_A=X-W_A. Each W_A is a closed set since \mathcal{V} is locally finite. Thus each U_A is open. Furthermore, for each A \in \mathcal{A}, A \subset U_A. It is easily checked that \left\{U_A: A \in \mathcal{A} \right\} is pairwise disjoint. \blacksquare

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Reference

  1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
  2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

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Definitions of Collectionwise Normal Spaces

The notion of collectionwise normal spaces is a property stronger than normal spaces. There are several ways of defining the notion of collectionwise normality. We show that they are equivalent.

We only consider spaces in which any set with only one point is considered a closed set (i.e. T_1 spaces). Let X be a space. Let \mathcal{A} be a collection of subsets of X. We say \mathcal{A} is pairwise disjoint if A \cap B = \varnothing for all A,B \in \mathcal{A} where A \ne B. We say \mathcal{A} is a discrete collection of subsets of X if for each x \in X, there is an open neighborhood O with x \in O such that O meets at most one element of \mathcal{A}. When \mathcal{A} is such a collection, we also say that \mathcal{A} is discrete in X (or discrete if X is understood).

    Definition 1
    A space X is said to be collectionwise normal if for every discrete collection \mathcal{A} of closed subsets of X, there is a collection \mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\} of open subsets of X such that \mathcal{U} is pairwise disjoint and A \subset U_A for each A \in \mathcal{A}.

In other words, every discrete collection of closed sets can be separated by pairwise disjoint open sets. Clearly any collectionwise normal space is normal. When discrete collection of closed sets in the definition of “collectionwise normal” is replaced by discrete collection of singleton sets, the space is said to be collectionwise Hausdorff. Clearly any collectionwise normal space is collectionwise Hausdorff.

Some authors define a collectionwise normal space as one in which every discrete collection of sets can be separated by pairwise disjoint open sets. We have the following definition.

    Definition 2
    A space X is said to be collectionwise normal if for every discrete collection \mathcal{A} of subsets of X, there is a collection \mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\} of open subsets of X such that \mathcal{U} is pairwise disjoint and A \subset U_A for each A \in \mathcal{A}.

It is clear that Definition 2 implies Definition 1. For any discrete collection \mathcal{A} of subsets, \mathcal{A}^*=\left\{\overline{A}: A \in \mathcal{A} \right\} is also a discrete collection. Thus Definition 1 implies Definition 2. The following is another way of defining collectionwise normal.

    Definition 3
    A space X is said to be collectionwise normal if for every discrete collection \mathcal{A} of closed subsets of X, there is a collection \mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\} of open subsets of X such that \mathcal{U} is a discrete collection and A \subset U_A for each A \in \mathcal{A}.

Since discrete collection is pairwise disjoint collection of sets, Definition 3 implies Definition 1. We show that Definition 1 implies Definition 3. Let X be a collectionwise normal space according to Definition 1. Let \mathcal{A} be a discrete collection of closed subsets of X. Let \mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\} be as in Definition 1. Let H=X-\cup \mathcal{U} and K=\cup \mathcal{A}. Note that H and K are disjoint closed subsets of X.

Suppose H=\varnothing. Then the sets in \mathcal{U} are both open and closed. Thus \mathcal{U} is the desired discrete collection of open sets separating sets in \mathcal{A}. So assume that H \ne \varnothing. Since X is a normal space, we have O_H and O_K, disjoint open subsets of X, such that H \subset O_H and K \subset O_K. For each A \in \mathcal{A}, let V_A=U_A \cap O_K. Let \mathcal{V}=\left\{V_A: A \in \mathcal{A} \right\}.

We claim that \mathcal{V} is a discrete collection of open sets separating sets in \mathcal{A}. Let x \in X. Suppose x \notin H. We have x \in U_A for some A \in \mathcal{A}. Then U_A meets only one set of \mathcal{A}, namely V_A. So assume x \in H. Then x \in O_H and O_H does not meet any V_A. Thus \mathcal{A} is a discrete collection of open sets. It is clear that A \subset V_A for each A \in \mathcal{A}.

Thus all three statements defining the notion of collectionwise normal spaces are equivalent. It is clear that any space satisfying any one of these collectionwise normal definitions is a normal space. The notion of collectionwise normality is stronger than normality. Bing’s Example G is a normal space that is not collectionwise normal (see “Bing’s Example G”).

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

A subspace of Bing’s example G

Bing’s Example G is the first example of a topological space that is normal but not collectionwise normal (see [1]). Example G was an influential example from an influential paper. The Example G and its subspaces had been extensively studied. In addition to being normal and not collectionwise normal, Example G is not perfectly normal and not metacompact. See the previous post “Bing’s Example G” for a basic discussion of Example G. In this post we focus on one subspace of Example G examined by Michael in [3]. This subspace is normal, not collectionwise normal and not perfectly normal just like Example G. However it is metacompact. In [3], Michael proved that any metacompact collectionwise normal space is paracompact (metacompact was called pointwise paracompact in that paper). This subspace of Example G demonstrates that collectionwise normality in Michael’s theorem cannot be replaced by normality.

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Bing’s Example G

For a more detailed discussion of Bing’s Example G in this blog, see the blog post “Bing’s Example G”. For the sake of completeness, we repeat the definition of Example G. Let P be any uncountable set. Let Q be the set of all subsets of P. Let F=2^Q be the set of all functions f: Q \rightarrow 2=\left\{0,1 \right\}. Obviously 2^Q is simply the Cartesian product of \lvert Q \lvert many copies of the two-point discrete space \left\{0,1 \right\}, i.e., \prod \limits_{q \in Q} \left\{0,1 \right\}. For each p \in P, define the function f_p: Q \rightarrow 2 by the following:

    \forall q \in Q, f_p(q)=1 if p \in q and f_p(q)=0 if p \notin q

Let F_P=\left\{f_p: p \in P \right\}. Let \tau be the set of all open subsets of 2^Q in the product topology. The following is another topology on 2^Q:

    \tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}

Bing’s Example G is the set F=2^Q with the topology \tau^*. In other words, each x \in F-F_P is made an isolated point and points in F_P retain the usual product open sets.

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Michael’s Subspace of Example G

For each f \in F, let supp(f) be the support of f, i.e., supp(f)=\left\{q \in Q:f(q) \ne 0 \right\}. Michael in [3] considered the following subspace of F.

    M=F_P \cup \left\{f \in F: supp(f) \text{ is finite} \right\}

Michael in [3] used the letter G to denote the space M. We choose another letter to distinguish it from Example G. The subspace M consists of all points f_p \in F_P and all other f \in F such that f(q)=1 for only finitely many q \in Q. The space M is normal and not collectionwise Hausdorff (hence not collectionwise normal and not paracompact). By eliminating points f \in F that have values of 1 for infinitely many q \in Q, we obtain a subspace that is metacompact. We discuss the following points:

  • The space M is normal.
  • The space M is not collectionwise Hausdorff and hence not collectionwise normal.
  • The space M is metacompact.
  • The space M is not perfectly normal.

The space M is normal since the space F that is Example G is hereditarily normal (see the section called Bing’s Example G is Completely Normal in the post “Bing’s Example G”).

To show that the space M is not collectionwise Hausdorff, it is helpful to first look at M as a subspace of the product space 2^Q. The product space 2^Q has the countable chain condition (CCC) since it is a product of separable spaces. Note that M is dense in the product space 2^Q. Thus M as a subspace of the product space has the CCC.

In the space M, the set F_P is still a closed and discrete set. In the space M, open sets containing points of F_P are the same as product open sets in 2^Q relative to the set M. Since M has CCC (as a subspace of the product space 2^Q), M cannot have uncountably many pairwise disjoint open sets containing points of F_P (in either the product topology or the Example G subspace topology). It follows that M is not collectionwise Hausdorff. If it were, there would be uncountably many pairwise disjoint product open sets separating points in F_P, which is not possible.

To see that M is metacompact, let \mathcal{U} be an open cover of M. For each p \in P, choose U_p \in \mathcal{U} such that f_p \in U_p. For each p \in P, let W_p=\left\{f \in M: f(\left\{p \right\})=1 \right\}. Let \mathcal{V} be the following:

    \mathcal{V}=\left\{U_p \cap W_p: p \in P \right\} \cup \left\{\left\{x \right\}: x \in M-F_P \right\}

Note that \mathcal{V} is a point-finite open refinement of \mathcal{U}. Each U_p \cap W_p contains only one point of F_P, namely f_p. On the other hand, each f \in M with finite support can belong to at most finitely many U_p \cap W_p.

The space M is not perfectly normal. This point is alluded to in [3] by Michael and elsewhere in the literature, e.g. in Bing’s paper (see [1]) and in Engelking’s general topology text (see 5.53 on page 338 of [2]). In fact Michael indicated that one can obtain a perfectly normal example with the aforementioned properties using Example H defined in [1] instead of using the subspace M defined here in this post.

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Reference

  1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
  2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  3. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
  4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012