# How can I prove that the Sorgenfrey line is a Lindelöf space?

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.

• Does this help? Dec 12, 2013 at 10:44

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 $b=\sup C_a.$

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

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 $$b > a$$.

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 $$x, then $$q_x. 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 $$\mathbb{R}_l$$.