$a,b \in X$ and $a<b$, then there is a neighbourhood $U$ of $a$ and neighbourhod of $V$ of $b$ such that $x<y$ whenever $x\in U$ and $y \in V$

Question

Let ($X,\le$) be a linear order set, and let $\tau$ denote the order topology on $X$. Prove that for $a,b \in X$ and $a<b$, then there is a neighbourhood $U$ of $a$ and neighbourhod of $V$ of $b$ such that $x<y$ whenever $x\in U$ and $y \in V$

$U=\{x\in X:x\leq a \}$ and $V=\{x\in X:x\ge b\}$. Am I correct?

Answer 1

bof's comment has solved the problem. In the case where there is a point c between a and b, let $U=\{x\in X: x<c\}$ and $V=\{x\in X: x>c\}$. U and V satisfy the requirement; otherwise, $U=\{x\in X: x<b\}$ and $V=\{x\in X: x>a\}$ will do the trick.