Sorgenfrey line is normal
This question has been answered before.
Rephrasing it:
Consider the Sorgenfrey topology on the line.
0)Let $A,B$ closed sets s.t.
$A\cap B =\emptyset$.
Need to find open disjoint sets $U,V$ s. t.
$A\subset U$ and $B\subset V$.
1)Since $A\cap B= \emptyset$ we have $A\subset B^c$ (complement in $\mathbb{R}$) and $B\subset A^c$ where $A^c, B^c$ are open.
2)For $a \in A$, since
$a$ is an interior point of $B^c$, we have
$a \in [a, a+\epsilon_a) \subset B^c$,
and similarly for $b\in B$
$b \in [b,b+\epsilon_b) \subset A^c$.
3)$A\subset U:=\bigcup_{a \in A} [a,a+\epsilon_a) \subset B^c$,
and $B\subset V:=\bigcup_{b \in B}[b, b+\epsilon_b) \subset A^c$.
4)Assume the open sets $U, V$ are not disjoint, i.e. $U \cap V \not=\emptyset$.
Then there are $a\in A$ and $b \in B$ s.t.
$x \in [a,a+\epsilon_a) \cap [b,b+\epsilon_b)$;
Wlog let $a<b$:
We have $b\le x$ and
$a<b \le x <a+\epsilon_a$, since $x \in [a, a+\epsilon_a)$,
it follows $b \in [a, a +\epsilon_a)$; a contradiction to $U \cap B=\emptyset$.
Kindly check and comment.
Solution 1:
I object to:
Wlog let $a<b$:
Since $b\le x$ it follows (why?, $b < a+\epsilon_a$ is unclear, even though that is probably meant)
$b \in [a, a +\epsilon a)$ (typo?) a contradiction to $U \cap B=\emptyset$.
Added after correction
It looks better now, the reasoning is complete.