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[SOLVED] The Order Topology on a Well Ordered Set is Normal

joypav

Active member
Mar 21, 2017
151
I am supposed to be presenting to my topology class the proof that the order topology on a well ordered set is normal. I'd love some help! If you could make sure the proof is correct and clarify some things I would really appreciate it!

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?
 

joypav

Active member
Mar 21, 2017
151
I believe I figured it out. But I will answer myself for completeness! And you can tell me if I'm wrong...

FIRST QUESTION: 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).

By $(x_h,h]$ we are referring to (or this is the same as) the set $(x_h,h')$ where
$h'=min \left\{ m\in H : m>h \right\} $
Meaning, h' is the "next" element after h. We know this h' exists because X is well ordered, so every subset of X has a minimal element. Do the same to find a k'.

Then,

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

So U and V are unions of open sets $\implies$ U and V are open.

SECOND QUESTION: What happens if the closed sets do contain endpoints of X?

Say, WLOG $minX = h_0 \in H$.
We would simply like to union the singleton $h_0$ to our open set U. So re-write U in the following manner,

$U = \cup_{h \in H} (x_h,h] \cup \left\{ h_0 \right\} = \cup_{h \in H} (x_h,h') \cup [h_0,h_0') $

$ [h_0,h_0') $ is an open set in the order topology, since $minX = h_0 $.
Where, similarly as before, $h_0'$ is the "next" element after $h_0$.