# ordinal space

Let $\alpha$ be an ordinal   . The set $W(\alpha):=\{\beta\mid\beta<\alpha\}$ ordered by $\leq$ 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 $W(\alpha)\neq\varnothing$, or $0<\alpha$.

Before examining some basic topological structures of $W(\alpha)$, let us look at some of its order structures  .

1. 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. 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 $1<\alpha$, $W(\alpha)$ is also atomic, with $1$ as the sole atom.

3. 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 $b\leq a<\alpha$, therefore $b\in W(\alpha)$ as well.

4. 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 $z<\alpha$. This means that $z+1\leq\alpha$. If $z+1<\alpha$, then $z+1\in W(\alpha)$ so that $z+1\leq\bigvee W(\alpha)=z$, 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 $0<\alpha$, a typical open interval may be written $(x,y)$, where $0\leq x\leq y<\alpha$. 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 $a_{i}. 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 $a_{i} 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. 1.

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

2. 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. 3.

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

4. 4.
5. 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 $\Omega_{0}$. $\Omega_{0}$ is not a compact space.

• $W(\omega_{1}+1)$, or $\Omega$. $\Omega$ is compact, and, in fact, a one-point compactification of $\Omega_{0}$.

## References

• 1 S. Willard, General Topology, Addison-Wesley, Publishing Company, 1970.
Title ordinal space OrdinalSpace 2013-03-22 17:10:56 2013-03-22 17:10:56 CWoo (3771) CWoo (3771) 8 CWoo (3771) Definition msc 54F05