ordinal space

Let $\alpha$ be an ordinal. The set ordered by $\le$ is a well-ordered 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. Another way of saying that $W(\alpha )$ is well-ordered is that for any non-empyt subset $S$ of $W(\alpha )$, $\bigwedge S$ exists. Clearly, $0\in W(\alpha )$ is its least element. If in addition , $W(\alpha )$ is also atomic, with $1$ as the sole atom.

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 one-point compactification of ${\mathrm{\Omega }}_0$.

Title	ordinal space
Canonical name	OrdinalSpace
Date of creation	2013-03-22 17:10:56
Last modified on	2013-03-22 17:10:56
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	8
Author	CWoo (3771)
Entry type	Definition
Classification	msc 54F05