Proving that $L-\{x\}$ is not connected.

Question

Problem: Let L be a linearly ordered set and give it the order topology. Suppose $x$ is an element of L which is not maximal or minimal. Prove the $L-\{x\}$ is not connected.

So I have to show that there exists some separation of $L-\{x\}$. But isn't $(-\infty,x)\cup (x,\infty)=$$L-\{x\}$ or in the case where L has a largest and or smallest element $a_0$ and $b_0$ respectively, the separation would be $(a_0,x)\cup (x,b_0)=$$L-\{x\}$. These sets are open in the order topology, and open rays are open too, hence it is a separation right?

Answer 1

It's almost correct. To be almost pedantic:

$L \setminus \{x\} = L_x \cup R_x$, where $L_x = \{y \in L: y < x\}$ and $R_x = \{y \in L: y > x\}$. These sets are subbasic open (open rays) in the order topology, so open sets of $L$ in the order topology.

And if $z \in L\setminus\{x\}$ then either $z > x$ or $z < x$ according to the axioms of a linear order. This proves the union.

The sets are also disjoint: otherwise we'd have $p < x < p$, when $p\in L_x \cap R_x$ which cannot be.

Both are non-empty as $x$ is not a minimal element: there is at least one $p < x$< so $L_x \neq \emptyset$, and as it is not maximal, at least one $q > x$, so $q \in R_x \neq \emptyset$. Hence it is a separation.