How to understand closed subsets of limit ordinal?

Question

On page 20, Constructibility, K.J.Devlin,

(Let $\alpha$ is a limit ordinal)A set $A \subseteq \alpha$ is closed, iff $\bigcup A \cap \gamma \in A$ for all $\gamma < \alpha$. Equivalently, if we define a limit point of $A$ to be any limit ordinal $γ$ such that ,$A \cap γ$ is unbounded in $γ$($\sup{A \cap γ} = γ$), then $A$ will be closed in $α$ iff it contains all its limit points below $α$.

This definition is different from what I expect. It doesn't mention what is an open set, or what is the enveloping space. It would be more illuminating if it explicitly define a topology on the whole space, i.e. the proper class of all ordinals, and then show $A$ to be a closed set relative to $\alpha$ under conditions discribed. Could we do that?

Answer 1

Like Martin said, every linearly ordered set $(X,<)$ has a natural order topology, which is the smallest topology that contains all sets of the form $L_x = \{ y \in X: y < x \}$ and $U_x = \{ y \in X: y > x \}$, the lower and upper sets.

Note that this means that all open intervals $(a,b)$, with $a,b \in X$ are in the order topology as well, as $(a,b) = L_b \cap U_a $. In fact, it's not too hard to check the open intervals together with the upper and lower sets for a base for the order topology. So if we give an ordinal $\alpha$ the order topology, all successor ordinals $\beta = \gamma+1$ are isolated points, as $\{\beta\} = (\gamma, \beta+1)$, where we also use that every ordinal has a successor. The same holds for $\{0\} = L_1$.

So the only non-isolated points are the limit ordinals. And if $\alpha$ is a limit, in some open set $O$, and then some open interval $(\beta,\gamma)$ containing $\alpha$ exists such that $(\beta, \gamma) \subset O$. As $\alpha < \gamma, \alpha+1 \le \gamma$, so $(\beta, \alpha] = (\beta, \alpha+1) \subset (\beta, \gamma)$. So $\alpha$ is an interior point of $O$ iff there exists some $\beta < \alpha$ such that $(\beta, \alpha] \subset O$. Or, sets of the form $(\beta, \alpha]$ where $\beta < \alpha$ form a local base at a limit ordinal $\alpha$.

From these remarks we can see that the definition of closed as used in set theory, for subsets of ordinals, is just the same as being closed in the order topology on that same ordinal.

Answer 2

Yes, you can do this. But notice that the class of ordinal numbers is not a set, thus cannot be endowed with a topology within the chosen universe of set theory. But every ordinal number is an ordered set, and thus carries the order topology. Then one can easily show that $A \subseteq \alpha$ is closed with respect to this topology iff it is closed in the sense of the definition you mentioned.