# How to show the Sorgenfrey line is hereditarily lindelöf?

How to show the Sorgenfrey line is hereditarily lindelöf? I know the Sorgenfrey line is lindelöf and hence it is closed hereditarily. The definition of Sorgenfrey line see here. Thanks for any help.

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Just to be clear, you are trying to show that every subset of the Sorgenfrey line is lindelöf? –  MJD Mar 6 '13 at 6:46
Yes, it is. Thanks for help. –  Paul Mar 6 '13 at 6:53

It suffices to show that for any family $\mathcal{U}$ of open sets in the Sorgenfrey line there is a countable $\mathcal{U}_0 \subseteq \mathcal{U}$ such that $\bigcup \mathcal{U}_0 = \mathcal{U}$.

For each $U \in \mathcal{U}$ consider the $\mathrm{Int}_{\mathbb R} ( U )$, where the interior is taken with respect to the usual metric topology on $\mathbb R$. As $\mathbb{R}$ is second-countable (and hence is itself hereditarily Lindelöf) there is a countable $\{ U_i : i \in \mathbb{N} \} \subseteq \mathcal{U}$ such that $$\bigcup_{i \in \mathbb{N}} \mathrm{Int}_{\mathbb R} ( U_i ) = \bigcup \{ \mathrm{Int}_{\mathbb R} ( U ) : U \in \mathcal{U} \}.$$

Note that if $x \in \bigcup \{ U : U \in \mathcal{U} \} \setminus \bigcup_{i \in \mathbb N} U_i$ then in particular $x \notin \mathrm{Int}_{\mathbb R} ( U )$ for all $U \in \mathcal{U}$, and so let us consider $A = \bigcup \{ U : U \in \mathcal{U} \} \setminus \bigcup \{ \mathrm{Int}_{\mathbb R} ( U ) : U \in \mathcal{U} \}$. I claim that $A$ is countable. For each $x \in A$ there is a $\epsilon_x > 0$ such that $[ x , x+ \epsilon_x ) \subseteq U$ for some $U \in \mathcal{U}$, and note that $( x , x+ \epsilon_x ) \subseteq \mathrm{Int}_{\mathbb R} ( U )$, so $( x , x+ \epsilon_x ) \cap A = \emptyset$. It follows that $\{ ( x , x+\epsilon_x ) : x \in A \}$ is a pairwise disjoint family of open sets in the metric topology on $\mathbb{R}$ and is therefore countable.

For each $x \in A$ pick $U_x \in \mathcal{U}$ containing $x$. Then $\mathcal{U}_0 = \{ U_i : i \in \mathbb N \} \cup \{ U_x : x \in A \}$ is a countable subfamily of $\mathcal{U}$ with the same union.

This is just to justify the first sentence/paragraph.

($\Rightarrow$) If $X$ is hereditarily Lindelöf, then any family $\mathcal{U}$ of open subsets of $X$ is a cover of $A = \bigcup \mathcal{U}$ by open subsets of $X$, and so there is a countable $\mathcal{U}_0 \subseteq \mathcal{U}$ such that $A \subseteq \bigcup \mathcal{U}_0 \subseteq \bigcup \mathcal{U}$.

($\Leftarrow$) If $X$ is not hereditarily Lindelöf, then there is a subset $A \subseteq X$ and a cover $\mathcal{U}$ of $A$ by open subsets of $X$ such that $A \not\subseteq \bigcup \mathcal{U}_0$ for any countable $\mathcal{U}_0 \subseteq \mathcal{U}$. Clearly, $\bigcup \mathcal{U}_0 \neq \bigcup \mathcal{U}$ for all countable $\mathcal{U}_0 \subseteq \mathcal{U}$.

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Why we only need to show that...(the first line)? –  Paul Mar 6 '13 at 8:24
@John: A justification for this has been added to the answer. –  Arthur Fischer Mar 6 '13 at 8:45

Consider any set $S\subseteq\mathbb R$, and let $\mathcal U$ be a collection of Sorgenfrey-open sets which covers $S$. Let's say that a set of reals is countably covered if it's covered by countably many members of $\mathcal U$. I have to show that $S$ is countably covered.

For $x,y\in\mathbb R$, define the relation $x\equiv y$ to mean that the set $\{s\in S:x\le s\le y\text{ or }y\le s\le x\}$ is countably covered.

It's easy to see that $\equiv$ is an equivalence relation on $\mathbb R$ and that the equivalence classes are convex subsets of $\mathbb R$, i.e., intervals or singletons. Moreover, using the fact that a convex subset of $\mathbb R$ is a countable union of closed intervals, we can see that the intersection of $S$ with each equivalence class is countably covered. Since each point of $S$ is in a nondegenerate equivalence class (an interval, not a singleton), and since there are only countably many such classes, it follows that $S$ is countably covered.

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