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.