# The First Uncountable Ordinal

This is a basic discussion on the first uncountable ordinal $\omega_1$ and its immediate successor $\omega_1+1$. The goal here is to consider these two ordinals as topological spaces with the order topology. These are interesting examples often presented in beginning courses in topology. Both spaces are rich ground for counterexamples. I can still remember the moment when these two spaces were presented in my first graduate level topology course.

Without thinking in terms of ordinals, the set $\omega_1$ can be thought of as an uncountable well-ordered set such that every initial segment is countable. The set $\omega_1+1$ is then simply the set $\omega_1$ with an additional point that is greater than all points of $\omega_1$. However we think of these spaces, this is the notation I use in this post: $[0,\omega_1)=\omega_1$ and $[0,\omega_1]=\omega_1+1$.

Some basic properties of $[0,\omega_1]$. It is compact and not first countable. Note that there is no countable local base at the point $\omega_1$.

Some basic properties of $[0,\omega_1)$. The space $[0,\omega_1)$ is locally compact, first countable and not Lindelof. Note that the $\lbrace{[0,\alpha]:\alpha<\omega_1}\rbrace$ is an open cover that has no countable subcover. Even though $[0,\omega_1)$ is not compact, it has compact like properties. It is pseudocompact and countably compact.

Several of the facts I discuss in this post are proven by the Pressing Down Lemma. A subset $S$ of $[0,\omega_1)$ is called a stationary subset if $S$ has nonempty intersection with every closed and unbounded set in $[0,\omega_1)$. This lemma can be found in [Kunen].

Pressing Down Lemma. Let $S$ be a stationary subset of $[0,\omega_1)$. Let $f:S \rightarrow \omega_1$ such that for each $\gamma \in S$, $f(\gamma)<\gamma$, then for some $\alpha<\omega_1$, $f^{-1}\lbrace{\alpha}\rbrace$ is a stationary subset of $[0,\omega_1)$.

One of the facts discussed below concerns function space with the pointwise convergent topology. Let $X$ be a space and let $C(X)$ be the set of all real-valued continuous functions defined on $X$. Define $[x,V]=\lbrace{f\in C(X): f(x) \in V}\rbrace$ where $V$ is an open interval of $\mathbb{R}$. The topology generated by the subbase $\lbrace{[x,V]: x \in X, V}$ is an open interval of $\mathbb{R}\rbrace$ is called the pointwise convergent topology on $C(X)$ and is denoted by $C_p(X)$.

I now discuss the following facts in a little more details.

A. Both $[0,\omega_1)$ and $[0,\omega_1]$ are completely normal (i.e. hereditarily normal).
B. Every continuous function $f:[0,\omega_1)\rightarrow \mathbb{R}$ is eventually constant.
C. Both $[0,\omega_1)$ and $[0,\omega_1]$ are not perfectly normal.
D. $[0,\omega_1)$ is countably compact.
E. $[0,\omega_1)$ is pseudocompact.
F. $[0,\omega_1)$ is sequentially compact.
G. $[0,\omega_1)$ is not paracompact.
H. The Stone-Cech compactification of $[0,\omega_1)$ is $[0,\omega_1]$.
J. $C_p([0,\omega_1))$ is not separable.

Proof of A. Spaces with the order topology are completely normal. This fact is quoted in [Steen & Seebach]. Thus both $[0,\omega_1)$ and $[0,\omega_1]$ are completely normal (i.e. hereditarily normal).

Proof of B. Let $f:[0,\omega_1)\rightarrow \mathbb{R}$ be continuous. For each limit ordinal $\alpha<\omega_1$, choose $g_1(\alpha)<\alpha$ such that $f([g_1(\alpha),\alpha]) \subset (f(\alpha)-\frac{1}{2},f(\alpha)+\frac{1}{2})$. The function $g_1$ is a pressing down function and by the Pressing Down Lemma, there is a stationary set $S_1$ and there is an $\alpha_1$ such that for each $\beta \in S_1$, $g_1(\beta)=\alpha_1$.

Here’s the $n^{th}$ iteration where $n$ is an interger greater than $1$. For each $\alpha \in S_{n-1}$, choose $g_n(\alpha)<\alpha$ such that $f([g_n(\alpha),\alpha]) \subset (f(\alpha)-\frac{1}{2n},f(\alpha)+\frac{1}{2n})$. By the Pressing Down Lemma, there is some $\alpha_n$ and there is some stationary set $S_n$ such that for each $\beta \in S_n$, $g_n(\beta)=\alpha_n$. At this point, the function $f$ moves in a narrow band of width $\frac{1}{n}$ among all the points beyond $\alpha_n$. For all $\gamma>\alpha_n$, $\delta>\alpha_n$, $\vert f(\gamma)-f(\delta) \vert < \frac{1}{n}$. To see this, fix such $\gamma$ and $\delta$. Since $S_n$ is stationary, I can choose some $\beta \in S_n$ such that $\beta>\gamma$ and $\beta>\delta$. Both of these inequalities hold: $\vert f(\beta)-f(\gamma) \vert<\frac{1}{2n}$ and $\vert f(\beta)-f(\delta) \vert<\frac{1}{2n}$. This implies $\vert f(\gamma)-f(\delta) \vert < \frac{1}{n}$.

Let $\alpha$ be the least upper bound of all the $\alpha_n$. Based on the discussion in the $n^{th}$ iteration, it is now clear that for all $\gamma$ and $\delta$ in $[\alpha,\omega_1)$, $\vert f(\gamma)-f(\delta) \vert<\frac{1}{n}$ for all positive integer $n$. Thus the continuous function $f$ is constant on $[\alpha,\omega_1)$.

Proof of C. In $[0,\omega_1]$, the point $\omega_1$ is not a $G_\delta$ point. Let $\lbrace{[\alpha_n,\omega_1]}\rbrace_{n<\omega}$ be a sequence of open interval containing the point $\omega_1$. Then choose a countable ordinal $\alpha$ that is greater than all $\alpha_n$. Then $[\alpha,\omega_1]$ is contained in the intersection of $\cap_{n<\omega} [\alpha_n,\omega_1]$.

In $[0,\omega_1)$, the set $A=\lbrace{\alpha<\omega_1: \alpha}$ is a limit ordinal$\rbrace$ is a closed set that is not $G_\delta$. To this end, let $\lbrace{O_n}\rbrace_{n<\omega}$ be a sequence of open sets containing $A$. I claim that for some $\alpha<\omega_1$, $[\alpha,\omega_1)$ is contained in each $O_n$. Thus each $O_n$ would contain many successor ordinals and $A$, the set of all limit ordinals, cannot be a $G_\delta$ set in $[0,\omega_1)$.

First let me prove this claim. For each open set $O$ containing $A$, there is some $\alpha<\omega_1$ such that $[\alpha,\omega_1) \subset O$. For each limit ordinal $\gamma \in A$, there is a $g(\gamma)$ such that $[g(\gamma),\gamma] \subset O$ since $O$ is open. By the Pressing Down Lemma, there is some $\alpha<\omega_1$ and there is some stationary set $S$ such that all points in $S$ are mapped to $\alpha$ by $g$. Then it is easy to see that $[\alpha,\omega_1) \subset O$.

Now, go back to the sequence of open sets $\lbrace{O_n}\rbrace_{n<\omega}$, each of which contains $A$. For each $n$, let $\alpha_n$ be the ordinal as in the claim in the preceding paragraph. Let $\alpha$ be an ordinal that is greater than the least upper bound of all $\alpha_n$. It is clear that $[\alpha,\omega_1) \subset O_n$ for each $n$.

Thus both $[0,\omega_1)$ and $[0,\omega_1]$ are examples of completely normal spaces that are not perfectly normal.

Proof of D. A space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. Let $\lbrace{O_n}\rbrace_{n<\omega}$ be an open cover of $[0,\omega_1)$. Suppose it has no finite subcover. Then for each $n$, there is some $\alpha_n$ such that it does not belong to $O_i$ for all $i \leq n$. Let $\alpha$ be the least upper bound of all $\alpha_n$. Then $\alpha$ must belong to some $O_m$. This would mean that $O_m$ contains many $\alpha_n$ for $n>m$, which is a contradiction.

Proof of E. A space $X$ is pseudocompact if every real-valued continuous function defined on $X$ is bounded. Fact B, is a stronger statement of this fact.

Proof of F. A space $X$ is sequentially compact if every sequence of points in $X$ has a convergent subsequence. This is true for $[0,\omega_1)$ since the least upper bound of a countably infinite sequence is the limit of a subsequence.

Proof of G. Consider the open cover $V=\lbrace{[0,\alpha]}\rbrace_{\alpha<\omega_1}$. It can be shown that this does not have a locally finite open refinement. Let $\lbrace{U_z:z \in H}\rbrace$ be an open refinement of $V$. For each limit ordinal $\alpha<\omega_1$, it is in some $U_z$. So choose $g(\alpha)<\alpha$ such that $[g(\alpha),\alpha] \subset U_z$. Once again, this is a pressing down function. So there is some $\alpha$ and there is a stationary set $S$ such that $g(\gamma)=\alpha$ for all $\gamma \in S$. This means that the point $\alpha$ is in $U_z$ for uncountably many $z$. Thus any open cover has no locally finite open refinement (in fact, no locally countable open refinement).

Proof of H. Every continuous function defined on $[0,\omega_1)$ is bounded and is eventually constant (from point B) and thus can be extended to $[0,\omega_1]$. We can simply define $f(\omega_1)$ to be the eventual constant value. A subspace $W$ of a space $Y$ is $C^*$-embedded in $Y$ if every bounded continuous real-valued function on $W$ can be extended to $Y$. According to theorem 19.12 in [Willard], if $Y$ is a compactification of $X$ and if $X$ is $C^*$-embedded in $Y$, then $Y$ is the Stone-Cech compactification of $X$. Thus $[0,\omega_1)$ is $C^*$-embedded in $[0,\omega_1]$ and $[0,\omega_1]$ is the Stone-Cech compactification of $[0,\omega_1)$. Here the Stone-Cech compactification agrees with the one-point compactification.

Proof of I. I now show that no countable subspace of $C_p([0,\omega_1))$ is dense in $C_p([0,\omega_1))$. Let $\lbrace{f_n}\rbrace_{n<\omega}$ be a countable family of continuous functions. For each $n$, let $\alpha_n$ be such that $f_n$ is constant on $[\alpha_n,\omega_1)$. Let $\alpha$ be the least upper bound of all $\alpha_n$. Choose two distinct countable ordinal $\gamma>\alpha$ and $\delta>\alpha$. Then $[\gamma, (0,\frac{1}{4})] \cap [\delta,(\frac{1}{2},1)]$ is an open set in the function space that does not contain any $f_n$.

Reference
[Kunen]
Kunen, K., [1980] Set Theory, An Introduction to Independence Proofs, First Edition, North-Holland, New York

[Steen & Seebach]
Steen, L. A. and Seebach, J. A., [1995], Counterexamples in Topology, Dover Edition

[Willard]
Willard, S., [1970] General Topology, Addison-Wesley Publishing Company, Inc.

## 3 thoughts on “The First Uncountable Ordinal”

1. How can I prove that the topological countable product of (\omega_1 + 1) (with the order topology) is the union of (\omega_1) closed nowhere dense sets?