How to prove $X$ is normal? Let $X=[0,\omega_1]\setminus \{\omega_1\}=[0,\omega_1)$. I known every linear order space is hereditarily normal. However, I would like to know the easier proof by which we can conclude that $X$ is normal.
Thanks for your help.
 A: Let $A,B$ be closed disjoint subsets of $[0,\omega_1)$. Let $A'=\{\alpha\in A:\alpha$ is successor$\}$ and $B'=\{\beta\in B:\beta$ is successor$\}$, then clearly both $A'$ and $B'$ are open and disjoint. Now if $\alpha\in A$ is limit we can find some $\xi_{\alpha}<\alpha$ such that $(\xi_{\alpha},\alpha]\cap B=\emptyset$, and for all limit $\beta \in B$ there is $\xi_{\beta}<\beta$ such that $(\xi_{\beta},\beta]\cap A=\emptyset$. Let $A'',B''$ be the set of limit ordinals of $A,B$ respectively then $A' \cup \bigcup_{\alpha\in A''}(\xi_{\alpha},\alpha]$ and $B' \cup \bigcup_{\beta\in B''}(\xi_{\beta},\beta]$ are disjoint open sets separating $A$ and $B$
A: Given two disjoint closed sets then one of them is bounded. The reason is that every two closed and unbounded sets have a nonempty intersection (and in fact unbounded intersection).
Let us name them $A$ and $B$. Say that $A$ is bounded below $\alpha$, then it is sufficient to separate $A$ and $B\cap\alpha$. We want to write an open set which contains $A$, first take all the successor ordinals appearing in it, and if $\beta$ is a limit ordinal appear in $A$ then $\beta$ is not a member of $B$ and in particular not a limit point of $B$, so there is some $\gamma<\beta$ such that $(\gamma,\beta+1)$ is disjoint from $B$.
Let $U$ be the union of all those intervals (and successor ordinals), then $U$ is open and disjoint of $B$, but I claim that its closure is also disjoint from $B$. To see this suppose that $\alpha_n$ is an increasing sequence from $U$, then either its limit in $A$ and therefore not in $B$; or its limit is not in $A$ and therefore the sequence is from new points added to $U$. Therefore every $\alpha_n$ comes from an interval around some $\beta_n$ which is a limit ordinal from $A$. Therefore $\beta_n$ is a sequence from $A$ and it is not hard to see that both sequences have the same limit, $\beta$, and since $A$ is closed $\beta\in A\subseteq U$.
So in fact $U$ is clopen, and its complement is an open set which contains $B$. Therefore we separated $A$ from $B$ by disjoint open sets, as wanted.
A: The first exercise of “Linearly ordered spaces I” from Ryszard Engelking’s “General Topology” [Ex. 1.7.4 in Russian version, 1986] contains a relatively simple proof sketch of normality of each linearly ordered space.
Another way to obtain an easy proof may be based on a next claim (which I heard and which may be true or not :-) ): every two closed unbounded subsets of $\omega_1$ have a non-empty intersection. 
