Mfrhtyj

How can I prove that the Sorgenfrey line is a Lindelöf space? Now, Sorgenfrey line is $mathbb{R}$ with the basis of ${[a,b) mid a,binmathbb{R}, a<b}$, 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.

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 $Xsubseteq R$ 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 $ainmathbb R$, and let $C_a={x: xge a,text{ and the interval }[a,x]text{ is countably covered}}$. 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 $mathbb R=bigcup_{ninmathbb N}[-n,n]$.

P.S. I have been asked to explain why assuming that $sup C_a=binmathbb R$ leads to a contradiction. Let $b_n=b-frac{b-a}{2^n}$ for $n=1,2,3,dots,$ so that $alt b_nlt b$ and $b_nto b.$ Thus for each $n$ there is a countable collection $mathcal S_nsubseteqmathcal U$ such that $[a,b_n]$ is covered by $mathcal S_n,$ and the half-open interval $[a,b)$ is covered by the countable collection $bigcup_{ninmathbb N}mathcal S_n.$ Moreover, since $mathcal U$ covers $mathbb R,$ there is some $Uinmathcal U$ such that $bin U.$ Since $U$ is Sorgenfrey-open, there is some neighborhood $[b,b+varepsilon)$ of $b$ (with $varepsilongt0$) such that $[b,b+varepsilon)subseteq U.$ Then $[a,b+varepsilon)$ is covered by ${U}cupbigcup_{ninmathbb N}mathcal S_n,$ whence $b+fracvarepsilon2in C_a,$ contradicting our assumption that $b=sup C_a.$

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 $b > a$.