# linear ordered topological spaces are $T_4$- guided proof.

I would like to prove the following exercise, for which my topology book give a hint.

"Let $X$ be a linear ordered topological space, than $X$ is $T_4$".

A space is $T_4$ if for all $A,B$ closed disjoint set, there exist a neighborhood of $A$ and one of $B$ which are disjoint.

The hint goes as follows: consider $A,B$ disjoint closed subset of $X$ and let $$A^*=\bigcup \{[a,b]\mid a,b\in A, [a,b]\cap B=\emptyset\}$$ $$B^*=\bigcup \{[c,d]\mid c,d\in B, [c,d]\cap A=\emptyset\}$$ These two set are disjoint (this I can easily prove). Then, partion $A^*, B^*$ into their convex component and intersect them component-weise with open sets.

A set $C$ is convex if $\alpha, \beta\in C$ then $[\alpha,\beta]\subset C$.

I don't understand where all of this will take me and hence how to get there. Can anyone help me to understand better? Thanks a lot!

• If you show that $A^*$ and $B^*$ are open, you are done, right? – RKD Jul 17 '17 at 18:09
• @Ravi I thought that too, but overall I don't understand the instruction... – Michela Jul 17 '17 at 19:13
• Check out this proof and the one following for inspiration. They contain some of the ideas (convex components etc.) – Henno Brandsma Jul 17 '17 at 19:29
• @HennoBrandsma. Proving it is normal is a terrible, tedious proof. Proving it is monotonically normal seemed much quicker besides being a stronger property. – William Elliot Jul 18 '17 at 2:43
• Maybe it's of no use to you, if you have to follow the hint you were given, but there is a reasonably straightforward proof in the answer to this question. – bof Jul 18 '17 at 4:39

## 2 Answers

(I). A modification of this approach:

Notation: $In[x,x']=[x,x']\cup [x',x].$ That is, $In[x,x']$ is the closed interval with end-points $x,x'.$

Define an equivalence relation $E$ on $A$ where $aEa'$ iff $In[a,a']\cap B=\phi.$ Let $A'$ be the set of the convex hulls of the $E$-equivalence classes. (The convex hull of $Y$ is $\cup \{In[y,y']:y,y'\in Y\}.$)

Show that for each $C\in A'$ there is a convex open set $C'$ such that $C\subset \overline {C'}\subset X$ \ $B.$

Show that for each $b\in B$ there is a convex open set $J(b)$ such that $J(b)\cap (\cup \{\overline {C'}: C\in A'\})=\phi.$

Then $A\subset \cup \{C': C\in A'\}$ and $B\subset \cup \{J(b):b\in B\},$ while the open sets $\cup \{C':C \in A'\},\;\cup \{J(b):b\in B\}$ are disjoint.

(II). A variant of this is to note that when $A$ is closed in $X$ then $X$ \ $A=\cup F$ where $F$ is a family of pairwise-disjoint maximal convex open sets. (Maximal in the sense that if $f\in F$ and if $g$ is a convex open set with $f\subset g \subset X$ \ $A$ then $g=f.$)

Show that for each $f\in F$ such that $B\cap f\ne \phi$ there exists a convex open set $C(f)$ such that $B\cap f \subset C(f)\subset \overline {C(f)}\subset f.$

So $B\subset C=\cup \{C(f): f\in F \land B\cap f \ne \phi\}.$ And show that $A\cap \overline C=\phi,$ so $A\subset X$ \ $\overline C.$

BTW. To obtain $F$ let $\equiv$ be the equivalence relation on $X$ \ $A$ where $x\equiv x'$ iff In[x,x']\cap A=\phi. Then $F$ is the set of $\equiv$ equivalence classes.

BTW. Linear spaces are hereditarily collection-wise normal. I found this useful for some other problems. General Topology by R. Engelking has a lot of material on linear spaces in the exercises and problems.

In what follows we tackle the problem in small bite size pieces, and being the director, we can abruptly cut the action, setting up for the next take.

Take 1

As a warm-up, we will show that if $A = [a_0, a_1]$ and $B=[b_0,b_1]$ are disjoint closed intervals, then they can be separated by open sets.

We can assume that $a_1 \lt b_0$.

If there are no points between $a_1$ and $b_0$, we can separate with two open rays as follows,
$A \subset (-\infty,b_0)$
$B \subset (a_1,+\infty)$

If there is point $\gamma$ between $a_1$ and $b_0$, we can separate with two open rays as follows,
$A \subset (-\infty,\gamma)$
$B \subset (\gamma,+\infty)$

Take 2

$X = \Bbb Q$
$A^* = \Bbb Z$
$B^* = \Bbb Z+\frac{1}{2}$

By the way $A^*$ and $B^*$ are defined, if $a \in A^*$, then if there is anything at all in $A^* \cup B^*$ to the right of $a$, it has to be a $B^*$ singleton component (this also goes when looking to the left). So take care of business by creating an open set about $a$ that could not possibly intersect with the open sets you will be creating about components in $B^*$.

In our 'model' for the problem, every singleton component has stuff from the other set on either side, and by setting up the notation/indexing you can get

$A^* \subset \cup \,U_a = P$
$B^* \subset \cup \,V_b = Q$

with both the open sets $P$ and $Q$ being disjoint.

These ideas can be reformulated and then applied to hint/technique proposed in the more general question.

Take 3

In the general setup now, with a slight staging twist:

Suppose for some $b_0 \in X$,
$[b_0,+\infty) \cap A = \emptyset$

Using DanielWainfleet's In[] notation, set

$B^{+\infty}_\text{Chunk} = \cup In[b_0,b]$, where $b \in B$ and $In[b_0,b] \cap A = \emptyset$.

If this set does not have a minimum, then the union of the open rays $(b, +\infty)$ with $b \in B^{+\infty}_\text{Chunk}$ contains it.

If $A$ is nonempty, select an $a_0 \in A$ such that
$a_0$ is a lower bound of $B^{+\infty}_\text{Chunk}$
AND
For every $b \in B$, if $b \notin B^{+\infty}_\text{Chunk}$ then $b \lt a_0$

Set
$A^{ndx(?)}_\text{Chunk} = \cup In[a_0,a]$, where $a \in A$ and $In[a_0,a] \cap B = \emptyset$.

Exercise: Show that the two chunks sets can be separated by open sets in $X$.
If there is point $\gamma$ to the right of chunk $A$ and to the left of chunk $B$ - check.
So assume there is no such $\gamma$.
If $A$ chunk has a max and $B$ chunk has a min - check.
If no max for $A$ chunk and no min for chunk - check.
Other two cases - contradiction - check.

You can see what is going on now. Finding the open sets to separate the sets $A$ and $B$ is a finite 'localization' problem, but you need the axiom of choice to take care of all business at once, intersecting a finite number of open rays at at a time as necessary. To dot the i's and cross the t's will require careful indexing with the set theoretic proof.

I plan to supply a complete answer in a separate post here, but it will take some time to put together.

END OF TAKEs