# 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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$\copyright \ \ 2012$