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I think this is quite awkward, but I decided to ask for help anyway.

This is the problem:

Show that every perfectly normal Lindelöf space has the Souslin property.

I've tried to slove this problem by using the fact that every open subset of a perfectly normal space is also Lindelöf, so the two open neighborhoods which seperates two closed subsets from each other (normality) have countable open cover. Now in worst case all of the open sets which these two countable open cover are obtained from, are disjoint, which is still countable and the wanted property holds. But I know there's something wrong with this and I don't know what to do.

Any help would be greatly appreciated.

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Note that if $X$ is perfectly normal and Lindelöf, every open subspace $Y$ of $X$ is also Lindelöf, because these are $F_\sigma$ in $X$ and an $F_\sigma$ in a Lindelöf space is still Lindelöf. This is enough for our purposes: if $\mathcal{U}$ then is a pairwise disjoint family of non-empty open sets in $X$, the union $O=\bigcup \mathcal U$ is open and Lindelöf, and the $\mathcal{U}$ form a disjoint open cover of $O$, so in fact must be its own countable subcover (we cannot omit any member). So $\mathcal{U}$ is at most countable and $X$ is ccc.

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  • $\begingroup$ For me $X$ is ccc is synonymous with $X$ has has the Souslin property, i.e. every family of non-empty, open, pairwise disjoint sets is at most countable. $\endgroup$ – Henno Brandsma Jan 24 at 19:09
  • $\begingroup$ Thank you very very much! $\endgroup$ – Mahan Mehravard Jan 24 at 19:13
  • $\begingroup$ @MahanMehravard You're welcome. You were on the right track with the start that open subsets are Lindelöf, just take that idea a bit further.. $\endgroup$ – Henno Brandsma Jan 24 at 19:14
  • $\begingroup$ I'm now understanding it! Thank you again! $\endgroup$ – Mahan Mehravard Jan 24 at 19:23
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Let $X$ be perfectly normal and Lindelöf, and suppose that $\mathscr{U}$ is an uncountable family of pairwise disjoint, non-empty open subsets of $X$. Let $F=X\setminus\bigcup\mathscr{U}$, and let $G$ be any open set containing $F$. Then $\{G\}\cup\mathscr{U}$ is an open cover of $X$, so there is a countable $\mathscr{U}(G)\subseteq\mathscr{U}$ such that $\{G\}\cup\mathscr{U}(G)$ covers $X$. It follows that $U\subseteq G$ for each $U\in\mathscr{U}\setminus\mathscr{U}(G)$.

Now $F$ is closed, so there are open sets $G_n$ for $n\in\Bbb N$ such that $F=\bigcap_{n\in\Bbb N}G_n$. Each family $\mathscr{U}(G_n)$ is countable, so there is a $U\in\mathscr{U}\setminus\bigcup_{n\in\Bbb N}\mathscr{U}(G_n)$. Then $U\subseteq G_n$ for each $n\in\Bbb N$, so $U\subseteq F$, which is impossible. Thus, $X$ has no uncountable family of pairwise disjoint, non-empty open sets and is therefore ccc (i.e., has the Suslin property).

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  • $\begingroup$ I don't know how to say thank you! $\endgroup$ – Mahan Mehravard Jan 24 at 19:13
  • $\begingroup$ @MahanMehravard: You’re very welcome. $\endgroup$ – Brian M. Scott Jan 24 at 19:14

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