# Network Weight of Topological Spaces – II

This is a continuation of the discussion on network. In the previous post, I showed that the network weight (the minimum cardinality of a network) coincides with the weight for both metrizable spaces and locally compact spaces. In another post, I showed that this is true for compact spaces. I now show that this is also true for the class of Moore spaces. First, some definitions. A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$ and each open set $U \subset X$ with $x \in U$, there is some $n$ such that any open set in $\mathcal{D}_n$ containing the point $x$ is contained in $U$. A developable space is one that has a development. A Moore space is a regular developable space.

For a collection of $\mathcal{G}$ of subsets of a space $X$ and for $x \in X$, define $st(x,\mathcal{G})=\bigcup\lbrace{U \in \mathcal{G}:x \in U}\rbrace$. An equivalent way of defining a development: A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$, $\lbrace{st(x,\mathcal{G}_n):n \in \omega}\rbrace$ is a local base at $x$. For a basic introduction to Moore space and the Moore space conjecture, there are numerous places to look in the literature ([1] being one of them).

Theorem. If $X$ is a Moore space, then $nw(X)=w(X)$.

Proof. Since $nw(X) \leq w(X)$ always holds, we only need to show $w(X) \leq nw(X)$. To this end, we exhibit a base $\mathcal{B}$ with $\vert \mathcal{B} \lvert \leq nw(X)$. Let $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ be a development for $X$. Let $\mathcal{N}$ be a network with cardinality $nw(X)$.

For each $N \in \mathcal{N}$, choose open set $O(n,N) \in \mathcal{D}_n$ such that $N \subset O(n,N)$. Let $\mathcal{B}_n=\lbrace{O(n,N):N \in \mathcal{N}}\rbrace$ and $\mathcal{B}=\bigcup_{n<\omega}\mathcal{B}_n$. Note that $\lvert \mathcal{B} \lvert \leq nw(X)$. Because $\mathcal{N}$ is a network, each $\mathcal{B}_n$ is a cover of $X$. To see this, let $x \in X$. Choose some $V \in \mathcal{D}_n$ such that $x \in V$. There is some $N \in \mathcal{N}$ such that $x \in N \subset V$. Then $x \in O(n,N)$. For each $n$, $\mathcal{B}_n \subset \mathcal{D}_n$. The sequence $\lbrace{\mathcal{B}_n}\rbrace$ works like a development. We have just shown that $\mathcal{B}$ is a base for $X$.

Corollary. The example of Butterfly space is not a Moore space.

The example of the Butterfly (or Bow-tie) space is defined in this previous post. This space has a countable network and the weight of this space is continuum. Thus this space cannot be a Moore space.

Reference
[1] Steen, L. A. & Seebach, J. A. [1995] Counterexamples in Topology, Dover Books.