Open neighborhoods in ordinals with the order topology

Question

While learning about ordinals, my teacher made some remarks about any ordinal α being equipped with the order topology, and some facts, one of which was basically that the finite ordinals and ω are discrete topological spaces, and no ordinal beyond that is discrete.

This intuitively made sense to me, since any ordinal β greater than ω must contain ω, and ω shouldn't be an isolated point. Any open ball in β centered around ω should contain co-finitely many points of ω, i.e. it should contain, for some large enough $n$, $\{n, n+1, n+2, ...\}$. How do I pick this $n$ to make this rigorous?

If, for instance, $β=ω+1$, what my friend and I thought to try was to create a bijection $0 ↦ 1$, $n ↦ 1/n$ (for $n≠0$), and $ω ↦ 0$. Then for any $ε ∈ ℝ_{>0}$, if we just pick $n$ to be $\lceil 1/ε \rceil +1$, we can visibly see that for this $n$, $1/n$ will be $B(0,ε)$. Then after undoing transformation, is it the case that this same $n$ will belong to $B(ω,ε)$ in $ω+1$, and that it will be the largest (or second largest) $n$ for which it and all its successors belong to $B(ω,ε)$?

Frankly, while the bijection above helped me to think about this differently, I'd like a more direct way to look at this problem, since I'd also like to think about the generalization of this problem to any limit ordinal (to show that the set of limit points of an ordinal is precisely the set of limit ordinals less than ).

Thank you in advance for any pointers and comments!

Answer 1

What you want follows almost immediately from the definition of the order topology.

If $\langle X,\le\rangle$ is a linear order, the order topology on $X$ has as a subbase the family of open rays, $$\mathscr{S}=\{(\leftarrow,x):x\in X\}\cup\{(x,\to):x\in X\}\;.$$ If $X$ has neither a largest nor a smallest element, this generates the base of open intervals, $\{(x,y):x,y\in X\text{ and }x<y\}$. If $x$ has a largest element $x_r$, basic open nbhds of $x_r$ have the form $(x,x_r]$ for $x<x_r$. Similarly, if $X$ has a smallest element $x_\ell$, basic open nbhds of $x_\ell$ have the form $[x_\ell,x)$ for $x>x_\ell$.

Suppose that $\alpha>\omega+1$, and $U$ is an open nbhd of $\omega$ in the order topology on $\alpha$. Then by definition of the order topology there are ordinals $\xi,\eta\in\alpha$ such that $\omega\in(\xi,\eta)\subseteq U$. This clearly means that $\xi<\omega<\eta$, so $\xi=n$ for some natural number $n\in\omega$, and $\eta\ge\omega+1$. Clearly, then,

$$\omega\in(n,\omega+1)\subseteq(\xi,\eta)\subseteq U\;.$$

Thus, every open nbhd of $\omega$ in $\alpha$ contains an open nbhd of the form $(n,\omega+1)=(n,\omega]$, and these nbhds therefore form a local base at the point $\omega$.

If $\alpha=\omega+1$, we have to modify this slightly. In that case $\omega$ is the largest element of the linear order, so there is no $\eta\in\alpha$ with $\omega<\eta$. However, in this case the basic open nbhds of $\omega$ in the order topology already have the form $(n,\omega]$ for some $n\in\omega$.

In particular, $\omega$ is not an isolated point: if $\{\omega\}$ were open, there would be an $n<\omega$ such that $(n,\omega]\subseteq\{\omega\}$, which implies that $(n,\omega)=\varnothing$. This, of course, is false, since $n+1\in(n,\omega)$.

Larger limit ordinals can be treated similarly.

Answer 2

Since the order topology is not induced by a metric, talking about balls is not quite the right thing to do.

Instead, the order topology is by definition generated by the open intervals $(\alpha,\gamma)=\{\beta\mid \alpha<\beta<\gamma\}$, so any neighborhood $N$ of $\omega$ will have to contain one of these intervals which contains $\omega$.

Now if $\omega\in(\alpha,\gamma)\subseteq N$, then we also have $(\alpha,\omega+1)\subseteq N$, so we can restrict our attention to intervals of this form. And $\alpha$ is finite because $\alpha<\omega$, so we can conclude

A set of ordinals is a neighborhood of $\omega$ exactly if it contains $\omega$ and contains all the natural numbers larger than $\alpha$, for some finite $\alpha$.

In particular $\{\omega\}$ is not open, because it it nonempty yet doesn't contain any open interval.