I've encountered the following definition in Kunen, Levy, and other places: A function $\mathbf{F}:\mathbf{ON}\to\mathbf{ON}$ is continuous iff for every limit ordinal $\lambda$, we have $\mathbf{F}(\lambda)=\sup\{\mathbf{F}(\alpha):\alpha<\lambda\}$. I will say such $\mathbf{F}$ are ordinally continuous.
If we consider $\mathbf{ON}$ in the order topology, this definition of continuous coincides with topological continuity for non-decreasing $\mathbf{F}:\mathbf{ON}\to\mathbf{ON}$. If we remove that requirement, though, there are functions that are ordinally continuous, but not topologically continuous, and vice versa. For example, defining $$\mathbf{F}(\xi)=\begin{cases}0 & \text{if }\xi=2\!\cdot\! n\!+\!1\text{ for some }n<\omega\\\xi & \text{otherwise}\end{cases}\qquad\text{and}\qquad\mathbf{G}(\xi)=\begin{cases}\omega+1 & \text{if }\xi=0\\\omega & \text{if }0<\xi<\omega\\\xi & \text{otherwise,}\end{cases}$$ then $\mathbf{F}$ is ordinally but not topologically continuous, and $\mathbf{G}$ is topologically but not ordinally continuous.
Before I proceed to ask my question, let me clarify one thing. When I speak of a "topology" on $\mathbf{ON}$, I'm not speaking of something that formally exists in ZF(C)–as such a creature would be a class of (sometimes proper) classes. Instead, we'll describe "topologies" on $\mathbf{ON}$ indirectly as follows. We'll say that a class $\mathbf{B}$ of sets of ordinals is a basis class iff $$\forall U\!,V\!\!\in\!\mathbf{B}\;\forall\alpha\!\in\! U\cap V\;\exists W\!\!\in\!\mathbf{B}\;(\alpha\in W\subseteq U\cap V).$$ Given a basis class $\mathbf{B}$, we'll say that a subclass $\mathbf{M}$ of $\mathbf{ON}$ is "($\mathbf{B}$-)open" iff one of the following holds:
(i) $\mathbf{M}=\mathbf{ON}$
(ii) $\forall\alpha\!\in\!\mathbf{M}\;\exists U\!\in\!\mathbf{B}\;(\alpha\!\in\!U\!\subset\!\mathbf{M}).$
For further discussion of why I chose these particular definitions of basis class and openness, see this post.
Question: Is there a way to "topologize" $\mathbf{ON}$ such that the ordinally continuous functions $\mathbf{ON}\to\mathbf{ON}$ are precisely the topologically continuous functions $\mathbf{ON}\to\mathbf{ON}$? If so, what's an example? If not, how can one show that there is no way?
Remark: When considering $\mathbf{ON}$ in the order topology, limit ordinals and limit points are identical. It would, of course, be ideal to find a topology in which this still held and where topologically continuous and ordinally continuous functions are the same, but I would still be interested in any topology satisfying only the latter.
Current Goals: (A) I'd like to generalize Brian M. Scott's result from below to other limit ordinals. In other words, assuming that $\mathbf{ON}$ has been "topologized" in such a way that "ordinally continuous" and "topologically continuous" are identical, I'd like to determine for which limit ordinals $\lambda$ we can conclude that $[0,\lambda]$ is contained in all open classes containing $\{\lambda\}$. (Brian showed that this property holds for $\lambda=\omega$. Does this hold for all limit ordinals $\lambda$? Only when $\lambda$ is an aleph? Only when $\lambda$ is a regular aleph? Only when $\lambda=\omega$? Only when [fill in the blank appropriately]?
(B) I'd like to find a counterexample similar to $\mathbf{B}_1$ (from my answer below) satisfying the condition that limit points and limit ordinals are identical.
If you can help me with (A) or (B), but not yet answer my overarching question, let me know, and I'll make a new question for you to answer. (Heck, I'll even give you a portion of the bounty offered on this question, if it's a good answer.)
Best Answer
You define the class $\mathcal{C} := \{F: {\bf ON}\to {\bf ON}: \forall \mbox{ limit } \lambda \; F(\lambda) = \sup(F(\alpha) | \alpha < \lambda)\}$. And you ask: is there a topology $T$ on the ordinals such that $\mathcal{C}$ contains exactly the functions continuous in $T$?
In any topology, if $f$ and $g$ are continuous then so is their composition $g \circ f$. The following are in $\mathcal{C}$:
$$f: \alpha \mapsto \begin{cases} {2\alpha} \quad\mbox{if } \alpha < \omega ,\\ \alpha \quad\text{otherwise}; \end{cases}$$
$$g: \alpha \mapsto \begin{cases}0 \quad \mbox{if } \alpha < \omega \text{ and $\alpha$ is even} ,\\ \alpha \quad \text{otherwise}. \end{cases}$$ (In English, $f$ doubles finite numbers, $g$ annihilates finite even numbers, and everywhere else they're the identity map.)
However the composition $g \circ f (\alpha)$ takes value $0$ for finite $\alpha$, but value $\omega$ at $\alpha = \omega$. So it does not lie in $\mathcal{C}$. Hence $\mathcal{C}$ cannot contain exactly the continuous functions of any topology.