# Let $X$ be a Moore space and $e(X)=\omega$. Is it metrizable?

Let $X$ be a Moore space and $e(X)=\omega$. Is it metrizable?

What I’ve tried: I list these facts:

1 A space $X$ is a Moore space iff $X$ is a $\sigma$-space and a $p$-space.

2 If $X$ is a $p$-space, then $nw(X)=w(X)$.

3 A space $X$ is a $\sigma$-space if $X$ has a $\sigma$-discrete network.

Since if $X$ is a $\sigma$-space and $e(X)=\omega$, we know that $nw(X)=\omega$, see the proof. So by 2, we can conclude that $w(X)=nw(X)=\omega$, and hence it is metrizable.

However I’m not sure. Thanks for any help.

#### Solutions Collecting From Web of "Let $X$ be a Moore space and $e(X)=\omega$. Is it metrizable?"

Here are the references.

A Moore space is a regular developable space. [Gru, 1.3] A space $X$ is a $\sigma$-space if $X$ has a $\sigma$-discrete network. [Gru, 4.3] Every Moore space $X$ is a $\sigma$-space. [Gru, 4.5] Moreover, a space $X$ is a Moore space iff $X$ is a $\sigma$-space and a $p$-space (or a $w\Delta$-space). [Gru, 4.7.(i)] If $X$ is a $\sigma$-space and $e(X)=\omega$, we know that $nw(X)=\omega$, see the proof. If $X$ is a $p$-space, then $nw(X)=w(X)$. [Gru, 4.2] By Nagata-Smirnov Theorem [Eng, 4.4.7], a topological space $X$ is metrizable iff $X$ is regular and has a $\sigma$-locally finite base. In particular, a regular second countable space is metrizble.

References

[Eng] Ryszard Engelking. General Topology (Russian version, 1986).

[Gru] Gary Gruenhage, Generalized Metric Spaces, in K. Kunen and J. E. Vaughan, Handbook of set theoretic topology, Elsevier, 1984.