Conflicting definitions of "continuity" of ordinal-valued functions on the ordinals

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.

It can’t be done if you require the elements of the topology (other than $\mathbf{ON}$, of course) to be sets.

If $\omega$ is an isolated point, the function

$$\mathbf{F}(\xi)=\begin{cases} 0,&\text{if }\xi<\omega\\ \xi&\text{if }\xi\ge\omega \end{cases}\tag{1}$$

is topologically continuous but not ordinally continuous, so assume that $\omega$ is not an isolated point. Suppose that $\omega$ has an open nbhd $V$ disjoint from an infinite $A\subseteq\omega$, where $V$ is a set. Let $\lambda$ be a limit ordinal greater than any element of $V$, and let $\{a_n:n\in\omega\}$ be an increasing enumeration of $A$. Then

$$\mathbf{F}(\xi)=\begin{cases} a_\xi,&\text{if }\xi\in \omega\\\ \omega,&\text{if }\xi=\omega\\ \lambda+\xi,&\text{if }\xi>\omega \end{cases}$$

is ordinally continuous but not topologically continuous: $\mathbf{F}^{-1}[V]=\{\omega\}$.

Now suppose that $\omega\setminus V$ is finite for every nbhd $V$ of $\omega$, but there is are an $n\in\omega$ and a nbhd $V$ of $\omega$ such that $n\notin V$. Let $\lambda$ be as before. Then

$$\mathbf{F}(\xi)=\begin{cases} n,&\text{if }\xi\in\omega\text{ is even}\\ \xi,&\text{if }\xi\in \omega\text{ is odd}\\ \omega,&\text{if }\xi=\omega\\ \lambda+\xi,&\text{if }\xi>\omega \end{cases}$$

is ordinally but not topologically continuous.

The only remaining possibility is that every nbhd of $\omega$ contains $[0,\omega]$. In that case $(1)$ is not ordinally continuous, but if it’s not topologically continuous, then neither is the identity map.

Added: As Cameron shows with the clever examples in his answer, my first and fourth assertions are false. (I suspect that I was unconsciously thinking only of $T_1$ topologies, though even that may not be sufficient to salvage them.) To ensure the topological continuity of $(1)$, I should have assumed not just that the point $\omega$ is isolated point, but that the set $\omega$ is clopen. Then if $0\notin V$, $\mathbf{F}^{-1}[V]=V\setminus\omega$, and if $0\in V$, $\mathbf{F}^{-1}[V]=V\cup\omega$, both of which are open if $V$ is.

(I’ll probably have more later.)