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**Proof:**

Let $(X,\leq,T_<)$ be a well ordered set with the order topology.

Take H, K closed in X, $H\cap K = \emptyset $

First, assume that neither of these sets contains the least or greatest element of X.

Then,

$\forall h \in H \subseteq (X-K), \exists x_h \in H$ such that $(x_h,h] \subseteq (X-K)$.

$\forall k \in K \subseteq (X-H), \exists x_k \in K$ such that $(x_k,k] \subseteq (X-H)$.

Now consider the following U and V,

$U = \cup_{h \in H} (x_h,h] $ and $V = \cup_{k \in K} (x_k,k] $

**Why are these sets U and V open in the order topology?**I thought only sets of the form $(a,b)$,$[a_0,b)$, and $(a,b_0]$ were open. Where $a_0$ is the smallest element (if there is one) and $b_0$ is the largest element (if there is one).

Claim: $U \cap V = \emptyset$

BWOC, assume $\exists z \in U \cap V$.

$\implies \exists h \in H, k \in K, h \ne k$ such that $z \in (x_h,h]$ and $z \in (x_k,k]$

$\implies x_h<z \leq h$ and $x_k<z \leq k$

We know $h \ne k$, because H and K are disjoint. So there are two possibilities:

Case I: h<k

Then $x_k<z \leq h<k$

$\implies h \in (x_k,k]$

However, this is a contradiction because by construction, $(x_k,k] \cap H = \emptyset$.

Case II: k>h

Then $x_h<z \leq k<h$

$\implies k \in (x_h,h]$

However, this is a contradiction because by construction, $(x_h,h] \cap K = \emptyset$.

Then our claim is true.

$\implies (X,\leq,T_<)$ is normal.

**What happens if the closed sets do contain endpoints of X?**