Linearly ordered X is regular

Question

Prove that every linearly ordered space X is regular.

Can anyone please help me with this proof?

I started with letting $x$ belong to $x$ and take a nbhd $U=(a,b)$ of $x$ and then taking $A=(a,x)$, $B=(x,b)$ and then i found just the case when A and B not the empty set. But then I got stuck.

Thanks for any help.

There are four cases depending upon (a,x) and (x,b) being empty or not. — William Elliot, Apr 11 '17 at 19:45
http://math.stackexchange.com/q/1189654/4280 has the Munkres proof you're trying for. — Henno Brandsma, Apr 12 '17 at 04:31
The answer to this question proves that every linearly ordered space (or subspace of a linearly ordered space) is normal (and therefore regular). — bof, Apr 12 '17 at 23:04

score 0 · Answer 1 · answered Apr 12 '17 at 22:17

Def'n: $[x,\to)=\{y:y\geq x\}.\quad$ $(x,\to)=\{y:y>x\}.\quad$ $ (\leftarrow,x]=\{y:y\leq x\}.\quad$ $ (\leftarrow,x)=\{y:y<x\}.$

Let $x\in X$ and $x\not \in A=\bar A\subset X.$

If the closed set $[x,\to)$ is open, let $L=[x,\to)$ and $L^*=(\leftarrow,x)$. If $[x,\to)$ is not open then take $y<x$ such that $A\cap [y,x]=\phi,$ and let $L=(y,\to)$ and $L^*=(\leftarrow,y).$

If the closed set $(\leftarrow,x]$ is open, let $U=(\leftarrow,x]$ and $U^*=(x,\to).$ If $(\leftarrow,x]$ is not open then take $z>x$ such that $A\cap [x,z]=\phi,$ and let $U=(\leftarrow,z)$ and $U^*=(z,\to).$

Let $C=L\cap U$ and $D=L^*\cup U^*.$ In all cases, $C$ and $D$ are open and disjoint, with $x\in C$ and $A\subset D. $

Linear spaces are normal spaces. In fact they are hereditarily collectionwise-normal. — DanielWainfleet, Apr 13 '17 at 09:25

Linearly ordered X is regular

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