Network Weight of Topological Spaces – I

In the previous post, I discussed the notion of network of a topological space. It was noted that for any space X, the network weight (the least cardinality of a network for X) is always \leq the weight (the least cardinality of a base for X). When X is compact, the network weight and weight would coincide. Are there other classes of spaces for which network weight = weight? I would like to discuss two other classes of spaces where network weight and weight coincide, namely metrizable spaces and locally compact spaces. The following two theorems are proved. For a basic discussion on network, see the previous post.

Theorem 1. If X is metrizable, then nw(X)=w(X).

Proof. For the case of w(X)=\omega, we have nw(X)=\omega. Now consider the case that w(X) is an uncountable cardinal. Based on the Bing-Nagata-Smirnov metrization theorem, any metrizable space has a \sigma-discrete base. Let \mathcal{B}=\bigcup_{n<\omega} \mathcal{B}_n be a \sigma-discrete base for the metrizable space X. Now let \mathcal{K}=\lvert \mathcal{B} \lvert. Because each \mathcal{B}_n is a discrete collection of open sets, any network woulld have cardinality at least as big as \lvert \mathcal{B}_n \lvert for each n. If \mathcal{K}=\lvert \mathcal{B}_n \lvert for some n, then \mathcal{K} \leq nw(X). If \mathcal{K} is the least upper bound of \lvert \mathcal{B}_n \lvert, then \mathcal{K} \leq nw(X). Both cases imply w(X) \leq nw(X). Since nw(X) \leq w(X) always hold,  we have w(X)=nw(X).

Theorem 2. If X is a locally compact space, then nw(X)=w(X).

Proof. Let X be locally compact where nw(X)=\mathcal{K}. The idea is that we can obtain a base for X of cardinality \leq \mathcal{K} (i.e. w(X) \leq nw(X)). Let \mathcal{N} be a network whose cardinality is \mathcal{K}. Here’s a sketch of the proof. Each point in X has an open neighborhood whose closure is compact. For the compact closure of such open neighborhood, the weight would coincide with the network weight. Thus we can find a base of size \leq \mathcal{K} within such open neighborhood. Because \lvert \mathcal{N} \lvert=\mathcal{K}, we only need to consider \mathcal{K} many such open neighborhoods with compact closure. Thus we can obtain a base for X of cardinality \leq \mathcal{K}. To make this sketch more precise, consider the following three claims.

Claim 1
The collection of all N \in \mathcal{N}, where \overline{N} is compact, is a cover of the space X.

Claim 2
For every compact set A \subset X, there is an open set U such that A \subset U and \overline{U} is compact.

Claim 3
If U \subset X is open with \overline{U} compact, then we can obtain a base \mathcal{B}_U for the open subspace U with \vert \mathcal{B}_U \vert \leq \mathcal{K}.

For each N \in \mathcal{N} in Claim 1, we can select an open U (as in Claim 2)such that \overline{N} \subset U and \overline{U} is compact. Let \mathcal{B} be the union of all the \mathcal{B}_U in Claim 3 over all such U. There are \leq \mathcal{K} many N in Claim 1. Thus \vert \mathcal{B} \lvert \leq \mathcal{K}. Note that \mathcal{B} would form a base for the whole space X.

Both Claim 1 and Claim 2 are direct consequence of locally compactness. To see Claim 3, let U be open such that \overline{U} is compact. We have nw(\overline{U}) \leq nw(X)=\mathcal{K} (the network weight of a subspace cannot exceed the original network weight). By the result in the previous post, we have w(\overline{U})=nw(\overline{U}). We now have w(U) \leq w(\overline{U}) (the weight of a subspace cannot exceed the weight of the space containing it). So the weight of any open subspace with compact closure cannot exceed \mathcal{K}.

Corollary. For both metrizable spaces and locally compact spaces X, w(X) \leq \lvert X \lvert.


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