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I've been reviewing exams, and I came across this problem that I am having trouble with. Can anyone help me out?

Show that every paracompact space with the Suslin property is Lindelöf.

A topological space has the Suslin property if there is no uncountable family of pairwise disjoint non-empty open subsets of $X$.

Thanks in advance for any help!

1 Answers 1

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Let $X$ be a paracompact space with the Suslin property, and let $\mathscr{U}$ be an open cover of $X$. Since $X$ is paracompact, $\mathscr{U}$ has a $\sigma$-discrete open refinement, i.e., an open refinement $\mathscr{R}=\bigcup_{n\in\Bbb N}\mathscr{R}_n$ such that each $\mathscr{R}_n$ is discrete. In particular, each $\mathscr{R}_n$ is a pairwise disjoint collections of open sets, so each $\mathscr{R}_n$ is countable. But then $\mathscr{R}$ is itself countable, and $X$ is Lindelöf. If you insist on getting a countable subcover of $\mathscr{U}$ instead of a countable open refinement, for each $R\in\mathscr{R}$ choose a $U_R\in\mathscr{U}$ such that $R\subseteq U_R$; then $\{U_R:R\in\mathscr{R}\}$ is a countable subcover of $\mathscr{U}$.

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    It is a Good job.2013-05-20