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I was reading about developments and came across the statement: Every regular Lindelof space with a development has a countable base. I'm having trouble showing this, and I was wondering if anyone could help?

By the way, a development for a space $X$ is a sequence of open covers $\gamma_n$ such that for every $x \in X$, the sequence $\{St(x, \gamma_n)\}^{\infty}_{n=1}$ is a base at $x$.

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up vote 2 down vote accepted

Since $X$ is Lindelöf, each ‘layer’ $\gamma_n$ of the development has a countable subcover $\beta_n$. Let $x\in X$ be arbitrary, and let $U$ be any open set containing $x$. There is some $n$ such that $\operatorname{St}(x,\gamma_n)\subseteq U$. Since $\beta_n$ is still a cover of $X$, there is some $B\in\beta_n\subseteq\gamma_n$ such that $x\in B\subseteq\operatorname{St}(x,\gamma_n)\subseteq U$. Thus, $$\beta=\bigcup_{n\in\Bbb N}\beta_n$$ is a base for $X$. Since each $\beta_n$ is countable, so is $\beta$, and $X$ is second countable.

Added: Note that regularity wasn’t actually needed.

A developable $T_3$-space is known as a Moore space; Moore spaces have quite a lot in common with metrizable spaces, all of which are of course Moore spaces themselves. This result is an example of that: it’s well-known that a metrizable space is Lindelöf iff it is second countable, and that separability is equivalent to each of these. This last equivalence does not extend to Moore spaces: the tangent disk space is a separable, completely regular Moore space that is neither Lindelöf nor second countable. (It’s also not normal.)

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