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How can I prove that the Sorgenfrey line is a Lindelöf space? Now, Sorgenfrey line is $\mathbb{R}$ with the basis of $\{[a,b) \mid a,b\in\mathbb{R}, a<b\}$, and in general, a topological space is called a "Lindelöf space" iff every open cover has a countable subcover. Please show me an elegant proof.

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  • $\begingroup$ Does this help? $\endgroup$
    – user642796
    Dec 12, 2013 at 10:44

3 Answers 3

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Here's a simple direct proof, which works just as well for the Sorgenfrey topology as for the usual topology of the line.

Let $\mathcal U$ be a collection of Sorgenfrey-open sets that covers $\mathbb R$. Let's say that a set $X\subseteq R$ is countably covered if $X$ is covered by countably many members of $\mathcal U$. We want to show that $\mathbb R$ is countably covered.

Consider any $a\in\mathbb R$, and let $C_a=\{x: x\ge a,\text{ and the interval }[a,x]\text{ is countably covered}\}$. It's easy to see that $\sup C_a=\infty$; assuming the contrary leads to a contradiction. Hence every finite interval $[a,b]$ is countably covered, and so is $\mathbb R=\bigcup_{n\in\mathbb N}[-n,n]$.

P.S. I have been asked to explain why assuming that $\sup C_a=b\in\mathbb R$ leads to a contradiction. Let $b_n=b-\frac{b-a}{2^n}$ for $n=1,2,3,\dots,$ so that $a\lt b_n\lt b$ and $b_n\to b.$ Thus for each $n$ there is a countable collection $\mathcal S_n\subseteq\mathcal U$ such that $[a,b_n]$ is covered by $\mathcal S_n,$ and the half-open interval $[a,b)$ is covered by the countable collection $\bigcup_{n\in\mathbb N}\mathcal S_n.$ Moreover, since $\mathcal U$ covers $\mathbb R,$ there is some $U\in\mathcal U$ such that $b\in U.$ Since $U$ is Sorgenfrey-open, there is some neighborhood $[b,b+\varepsilon)$ of $b$ (with $\varepsilon\gt0$) such that $[b,b+\varepsilon)\subseteq U.$ Then $[a,b+\varepsilon)$ is covered by $\{U\}\cup\bigcup_{n\in\mathbb N}\mathcal S_n,$ whence $b+\frac\varepsilon2\in C_a,$ contradicting our assumption that $b=\sup C_a.$

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  • $\begingroup$ Wow! I got it!! Thanks a lot!!! $\endgroup$
    – user115322
    Jan 10, 2014 at 14:29
  • $\begingroup$ Why does $\sup C_a \in \mathbb{R}$ lead to a contradiction? Thanks :) $\endgroup$
    – user370967
    May 10, 2018 at 17:03
  • $\begingroup$ Thanks it is clear now ! $\endgroup$
    – user370967
    May 11, 2018 at 4:30
  • $\begingroup$ Nice proof! Reminds me of the Henie-Borel lemma's proof in mathematical analysis $\endgroup$
    – Harmonia
    Apr 18 at 6:03
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The proof given by bof is correct, but incomplete. Here is a completion.

When bof assumed ad absurdum that $\sup{C_a} = b \in \mathbb{R}$, he implicitely assumed that $b > a$, which a priori does not need to be true (as $b$ could be equal to $a$). For the sake of completeness, first remark that $C_a$ is non-empty, as the interval $[a, a]$ is trivially countably covered, so $a \in C_a$.

Now let $A$ be a Sorgenfrey-open set that covers $a$. Then, because $A$ is open, there must exist a neighbourhood $[a, a+\epsilon[$ in $A$ for some $\epsilon > 0$. Now observe that the interval $[a, a+\frac{\epsilon}{2}]$ is covered by $A$, so $a + \frac{\epsilon}{2}$ will certainly be a member of $C_a$. It follows that $b = \sup{C_a} \geq a + \frac{\epsilon}{2} > a$, which justifies bofs assumption that $b > a$.

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Let $\mathscr{U}=\{[a_\alpha,b_\alpha)\}_{\alpha\in J}$ be a covering of $\mathbb{R}$ by basis elements for the lower limit topology, and $C=\bigcup\limits_{\alpha\in J}(a_\alpha,b_\alpha)$ be a subset of $\mathbb{R}$. For a point $x\in \mathbb{R}-C$, we know that $x$ belongs to no open interval $(a_n,b_n)$. Therefore $x=a_\beta$ for some index $\beta$. Choose such a $\beta$ and $q_x\in(a_\beta,b_\beta)\cap\mathbb{Q}$. Since $(a_\beta,b_\beta)\subset C$, $(a_\beta,q_x)=(x,q_x)$. So if $x,y\in \mathbb{R}-C$ with $x<y$, then $q_x<q_y$. Thus the map $$\begin{aligned} \varphi:\mathbb{R}-C&\longrightarrow\mathbb{Q},\\ x&\longmapsto q_x \end{aligned}$$ is injective, and then $\mathbb{R}-C$ is countable.

Choosing for each element of $\mathbb{R}-C$ an element of $\mathscr{U}$ containing it, we obtain a countable subcollection $\mathscr{U}'$ of $\mathscr{U}$ that covers $\mathbb{R}-C$. Topologize $C$ as a subspace of $\mathbb{R}$ satisfying the second countability axiom, and then $C$ is covered by $(a_\alpha,b_\alpha)$, which are open in $\mathbb{R}$ and hence open in $C$. Then some countable subcollection covers $C$. Suppose this subcollection consists of the elements $(a_\alpha,b_\alpha)$ for $\alpha=\alpha_1,\alpha_2,\cdots$. Then the collection $$\mathscr{U}''=\{[a_\alpha,b_\alpha)\mid\alpha=\alpha_1,\alpha_2,\cdots\}$$ is a countable subcollection of $\mathscr{U}$ that covers the set $C$, and $\mathscr{U}'\cup\mathscr{U}''$ is a countable subcollection of $\mathscr{U}$ that covers $\mathbb{R}_l$.

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