Is every linear ordered set normal in its order topology?

Solution 1:

Don't try to disprove it, because the theorem is true! (The proof uses the axiom of choice, but that's O.K. because the axiom of choice is true.) Here is a proof I found in my notes. It proves that a linearly ordered topological space is not only normal but completely (or hereditarily) normal, i.e., if $A,B$ are sets (not necessarily closed) such that $A\cap\bar B=B\cap\bar A=\emptyset$, then there are disjoint open sets $U,V$ such that $A\subseteq U$ and $B\subseteq V$.

Without loss of generality, we assume that no point of $A\cup B$ is an endpoint of $X$. For each $a\in A$, choose $p_a,q_a\in X$ satisfying the conditions:

1. $p_a\lt a\lt q_a$;

2. $(p_a,q_a)\cap B=\emptyset$;

3. $(a,q_a)=\emptyset$ or $q_a\in A$ or $q_a\notin B\wedge(a,q_a)\cap A=\emptyset$;

4. $(p_a,a)=\emptyset$ or $p_a\in A$ or $p_a\notin B\wedge(p_a,a)\cap A=\emptyset$.

You can verify [*] that such points $p_a,q_a$ exist for each $a\in A$. Now it is clear that the set $$U=\bigcup_{a\in A}(p_a,q_a)$$ is an open set containing $A$, and that $V=X\setminus\bar U$ is an open set disjoint from $U$. Now you have to prove that $B\subseteq V$.

[*] Consider any $a\in A$; we want $q_a\gt a$ with $(a,q_a)\cap B=\emptyset$ and satisfying condition 3. First choose $q\gt a$ with $(a,q)\cap B=\emptyset$. Now we consider three cases.
Case I. If $(a,q)=\emptyset$, let $q_a=q$.
Case II. If $(a,q)\cap A\ne\emptyset$, choose $q_a\in(a,q)\cap A$.
Case III. If $(a,q)\ne\emptyset=(a,q)\cap A$, choose $q_a\in(a,q)$.

P.S. I've been asked to explain why $B\subseteq V$. Consider a point $b\in B$; I have to show that $b\in V$, i.e., that $b\notin\overline U$. In other words, I have to find $c,d\in X$ such that $c\lt b\lt d$ and $(p_a,q_a)\cap(c,d)=\emptyset$ for all $a\in A$.

Let $A'=\{a\in A:a\lt b\}$ and $A''=\{a\in A:a\gt b\}$. I have to show that (i) there is $c\lt b$ such that $(p_a,q_a)\cap(c,b)=\emptyset$ for all $a\in A'$, and (ii) there is $d\gt b$ such that $(p_a,q_a)\cap(b,d)=\emptyset$ for all $a\in A''$. By symmetry, it will suffice to prove (i).

Since $b\notin\overline A$, there are $c_0,d_0\in X$ such that $c_0\lt b\lt d_0$ and $A\cap(c_0,d_0)=\emptyset$. Now, if $(p_a,q_a)\cap(c_0,b)=\emptyset$ for all $a\in A'$, then I can take $c=c_0$. On the other hand, if there is just one $a\in A'$ with $(p_a,q_a)\cap(c_0,b)\ne\emptyset$, and if $q_a\lt b$ for that $a$, then I can take $c=q_a$. I will show that no other case can arise.

Consider any $a\in A'$; we have $a\le c_0$ since $a\lt b$ and $A\cap(c_0,d_0)=\emptyset$, and we have $q_a\le b$ since $b\notin(p_a,q_a)$. Consider the three alternatives in Condition 3. First, if $(a,q_a)=\emptyset$, then $(p_a,q_a)\cap(c_0,b)=(p_a,a]\cap(c_0,b)=\emptyset$. Second, if $q_a\in A$, then $q_a\le c_0$ and so $(p_a,q_a)\cap(c_0,b)=\emptyset$. Therefore, if $(p_a,q_a)\cap(c_0,b)\ne\emptyset$, then we must have $q_a\notin B$ and $(a,q_a)\cap A=\emptyset$. Since $q_a\le b$ and $q_a\notin B$, we have $q_a\lt b$ as desired.

Finally, assume for a contradiction that there are two different points $a,a_1\in A'$ such that $(p_a,q_a)\cap(c_0,b)\ne\emptyset$ and $(p_{a_1},q_{a_1})\cap(c_0,b)\ne\emptyset$; without loss of generality, we may assume $a\lt a_1$. Then $a_1\ge q_a$ since $(a,q_a)\cap A=\emptyset$, and $a_1\le c_0$ since $A\cap(c_0,d_0)=\emptyset$. Thus $q_a\le a_1\le c_0$; but then $(p_a,q_a)\cap(c_0,b)=\emptyset$, a contradiction.