# 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 $X$, $w(X) \leq \lvert X \lvert$.