ordinal space
Let $\alpha $ be an ordinal^{}. The set $$ ordered by $\le $ is a wellordered set. $W(\alpha )$ becomes a topological space^{} if we equip $W(\alpha )$ with the interval topology. An ordinal space $X$ is a topological space such that $X=W(\alpha )$ (with the interval topology) for some ordinal $\alpha $. In this entry, we will always assume that $W(\alpha )\ne \mathrm{\varnothing}$, or $$.
Before examining some basic topological structures of $W(\alpha )$, let us look at some of its order structures^{}.

1.
First, it is easy to see that $W(\alpha )=\uparrow y\cup W(y)$, for any $y\in W(\alpha )$. Here, $\uparrow y$ is the upper set of $y$.
 2.

3.
Next, $W(\alpha )$ is bounded complete. If $S\subseteq W(\alpha )$ is bounded from above by $a\in W(\alpha )$, then $b=\bigvee S$ is an ordinal such that $$, therefore $b\in W(\alpha )$ as well.

4.
Finally, we note that $W(\alpha )$ is a complete lattice^{} iff $\alpha $ is not a limit ordinal^{}. If $W(\alpha )$ is complete^{}, then $z=\bigvee W(\alpha )\in W(\alpha )$. So $$. This means that $z+1\le \alpha $. If $$, then $z+1\in W(\alpha )$ so that $z+1\le \bigvee W(\alpha )=z$, a contradiction^{}. As a result, $z+1=\alpha $. On the other hand, if $\alpha =z+1$, then $z=\bigvee W(\alpha )\in W(\alpha )$, so that $W(\alpha )$ is complete.
In any ordinal space $W(\alpha )$ where $$, a typical open interval may be written $(x,y)$, where $$. If $y$ is not a limit ordinal, we can also write $(x,y)=[x+1,z]$ where $z+1=y$. This means that $(x,y)$ is a clopen set if $y$ is not a limit ordinal. In particular, if $y$ is not a limit ordinal, then $\{y\}=(z,y+1)$ is clopen, where $z+1=y$, so that $y$ is an isolated point. For example, any finite ordinal is an isolated point in $W(\alpha )$.
Conversely, an isolated point can not be a limit ordinal. If $y$ is isolated, then $\{y\}$ is open. Write $\{y\}$ as the union of open intervals $({a}_{i},{b}_{i})$. So $$. Since $y+1$ covers $y$, each ${b}_{i}$ must be $y+1$ or $({a}_{i},{b}_{i})$ would contain more than a point. If $y$ is a limit ordinal, then $$ so that, again, $({a}_{i},{b}_{i})$ would contain more than just $y$. Therefore, $y$ can not be a limit ordinal and all ${a}_{i}$ must be the same. Therefore $({a}_{i},{b}_{i})=(z,y+1)$, where $z$ is the predecessor of $y$: $z+1=y$.
Several basic properties of an ordinal space are:

1.
Isolated points in $W(\alpha )$ are exactly those points that are limit ordinals (just a summary of the last two paragraphs).

2.
$W(y)$ is open in $W(\alpha )$ for any $y\in W(\alpha )$. $W(y)$ is closed iff $y$ is not a limit ordinal.

3.
For any $y\in W(\alpha )$, the collection^{} of intervals of the form $(a,y]$ (where $$) forms a neighborhood base of $y$.

4.
$W(\alpha )$ is a normal space^{} for any $\alpha $;

5.
$W(\alpha )$ is compact^{} iff $\alpha $ is not a limit ordinal.
Some interesting ordinal spaces are

•
$W(\omega )$, which is homeomorphic to the set of natural numbers $\mathbb{N}$.

•
$W({\omega}_{1})$, where ${\omega}_{1}$ is the first uncountable ordinal. $W({\omega}_{1})$ is often written ${\mathrm{\Omega}}_{0}$. ${\mathrm{\Omega}}_{0}$ is not a compact space.

•
$W({\omega}_{1}+1)$, or $\mathrm{\Omega}$. $\mathrm{\Omega}$ is compact, and, in fact, a onepoint compactification of ${\mathrm{\Omega}}_{0}$.
References
 1 S. Willard, General Topology, AddisonWesley, Publishing Company, 1970.
Title  ordinal space 

Canonical name  OrdinalSpace 
Date of creation  20130322 17:10:56 
Last modified on  20130322 17:10:56 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  8 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 54F05 