Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x$.

share|improve this question

1 Answer 1

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.