0

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!

Martin Sleziak
  • 56,060
Michela
  • 359

2 Answers2

1

(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.

DanielWainfleet
  • 59,529
0

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

CopyPasteIt
  • 11,870