Conflicting definitions of “continuity” of ordinal-valued functions on the ordinals

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.)

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What would be a counterxample? –  Andres Caicedo Apr 12 '12 at 3:41
@AndresCaicedo A function that is topologically continuous, but not continuous in the sense defined above, would be $\mathbf{G}(\xi)=\begin{cases}\omega+1 & \xi=0\\ \omega & 0<\xi<\omega\\ \xi & \mathrm{otherwise}.\end{cases}$ –  Cameron Buie Apr 12 '12 at 3:50
@AndresCaicedo A function that is continuous in the sense defined above, but not topologically continuous , would be $\mathbf{F}(\xi)=\begin{cases}0 & \xi=2\cdot n+1\: \mathrm{for\: some}\: n<\omega\\ \xi & \mathrm{otherwise}.\end{cases}$ –  Cameron Buie Apr 12 '12 at 3:51
Is 'sup' really the right thing? Are you sure you don't want something more like $\mathop{\mathrm{inf}}_{\beta < \lambda} \mathop{\mathrm{sup}}_{\beta < \alpha < \lambda} F(\alpha)$? –  Hurkyl May 21 '12 at 8:58
The definition you mention only applies when the union is a limit ordinal: my definition would also require, for example: $$f(0 \cup 2) = f(0) \cup f(2)$$ and thus $f(2) \geq f(0)$. The definition you give doesn't require this. –  Hurkyl Nov 22 '12 at 3:01

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.

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Your conclusion rests on the assumption that $g\circ f$ is continuous in your arbitrary topology. This would of course be true if $f,g$ are continuous, but this need not be the case. Perhaps I'm missing something? –  Cameron Buie Nov 23 '12 at 18:41
In any topology, the composition of continuous functions is continuous. If $U$ is open, so is $f^{-1}(U)$ (as $f$ continuous), and hence so is $g^{-1}(f^{-1}(U))$ (as $g$ is continuous. But this is exactly the preimage $(g \circ f)^{-1}(U)$. So as the inverse image of any open $U$ is open, $g \circ f$ is continuous. –  HTFB Nov 23 '12 at 21:45
I know the definition of continuity, and why compositions of continuous functions are continuous. My point is that none of $f,g,g\circ f$ need be continuous in an arbitrary topology on the ordinals. Can you prove that $f,g$ are necessarily continuous? If not, your answer doesn't help much, though it may further narrow the field of possibilities. –  Cameron Buie Nov 23 '12 at 22:15
The $f$ and $g$ of my example are in $\mathcal{C}$, or in your terminology ordinally continuous. If they're not both continuous in a topology $T$, then that $T$ is not a candidate for the topology you are looking for. If they are continuous in $T$, then so is their composition. But that composition is not in $\mathcal{C}$, so again $T$ is not a candidate. So by exclusion no such $T$ exists. –  HTFB Nov 23 '12 at 23:30
Ah! I see now. Excellent work, and thanks for your help! –  Cameron Buie Nov 23 '12 at 23:32

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.)

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I think you meant $n\notin V$ where it says $n\notin\omega$? Very nice proof; any insight into how you came up with it? –  joriki Apr 12 '12 at 10:19
@joriki: I did indeed; thanks for catching it. Not really; after looking at Cameron’s examples, I decided that there probably wasn’t such a topology and concentrated on what would have to happen at $\omega$ if there were one. –  Brian M. Scott Apr 12 '12 at 10:26
@BrianM.Scott: I'm actually considering the case where our open subclasses may be proper, but this gives me some ideas how I might approach proving that no such topology can exist at all. Thanks! –  Cameron Buie Apr 12 '12 at 17:50
@Cameron: I conjecture that you’re right, but I was getting too sleepy to think straight after I got that far! –  Brian M. Scott Apr 12 '12 at 19:34
I've finally had a moment to sit down and think about this. Working in ZF, where proper classes don't formally exist, it is problematic to define a topology on a proper class in the traditional sense (since you'll be dealing with a collection of classes, some of which are proper). It is more convenient to describe a class of sets to serve as a basis. Given that particular restriction, your approach does more than simply give me ideas, it completely rules out the possibility of the existence of any topology describable in this way--simply take $V$ to be a basis neighborhood of $\omega$. Thanks! –  Cameron Buie Apr 17 '12 at 3:18

Consider the classes $$\mathbf{B}_0=\bigl\{\{0\},\{\omega\}\bigr\}\cup\bigl\{\{\alpha,\omega\}:0<\alpha<\omega\bigr\}\cup\bigl\{\{0,\alpha\}:\alpha>\omega\bigr\}$$ and $$\mathbf{B}_1=\bigl\{\{0\}\bigr\}\cup\bigl\{\{\alpha+1\}:\alpha>0\bigr\}\cup\bigl\{[0,\omega\cdot 2\cdot\beta):\beta>0\bigr\}.$$ Both classes have the nice property that for each $\alpha$, there is a $\subseteq$-least set $V$ in the class such that $\alpha\in V$, which is sufficient to show that they are basis classes as defined above.

In the "topology" induced by $\mathbf{B}_0$, we see that $0,\omega$ are (the only) isolated points, so with $\mathbf{F}$ as in $(1)$ of Brian's answer, we see that $\mathbf{F}^{-1}\bigl[\{0\}\bigr]=\omega$ is not open, since (for example) it contains no nbhd of $1$ (as $\omega\notin\omega$), but $\{0\}$ is open, so $\mathbf{F}$ is not topologically continuous, even though $\omega$ is isolated.

In the "topology" induced by $\mathbf{B}_1$, we see that the isolated points are precisely $0$ and the successor ordinals other than $1$--so it nearly has the property "limit point iff limit ordinal"--and the $\subseteq$-least nbhd of $\omega$ is $[0,\omega\cdot 2)$, so every nbhd of $\omega$ contains $[0,\omega]$. But $[0,\omega\cdot 2)$ is also the $\subseteq$-least nbhd of $1$, so again, $\mathbf{F}^{-1}\bigl[\{0\}\bigr]=\omega$ is not open, and for the same reason, even though $\{0\}$ is open, so $\mathbf{F}$ is not continuous. However, the identity function $\mathbf{ON}\to\mathbf{ON}$ is continuous--necessarily, since it is a homeomorphism, as a bijective open map that is its own inverse.

From Brian's answer, we can certainly conclude that necessary conditions for a topology of the type I seek are that either (1) $\omega$ is isolated or (2) $\omega$ is a limit point and every nbhd of $\omega$ contains $[0,\omega]$. That substantially narrows my options, so again, I want to thank Brian M. Scott for his answer. Neither of these topologies satisfies "topologically continuous iff ordinally continuous". In fact, in both cases, neither continuity implies the other. Defining $$\mathbf{F}_0(\xi)=\begin{cases}\omega & \text{if }\xi=0\\0 & \text{if }0<\xi<\omega\\\xi & \text{otherwise}\end{cases}\qquad\text{and}\qquad\mathbf{G}_0(\xi)=\begin{cases}0 & \text{if }0<\xi\le\omega\\\omega & \text{otherwise,}\end{cases}$$ we see that, in the "topology" induced by $\mathbf{B}_0$, $\mathbf{F}_0$ is ordinally but not topologically continuous, and $\mathbf{G}_0$ is topologically but not ordinally continuous. Defining $$\mathbf{F}_1(\xi)=\begin{cases}\omega\cdot 2 & \text{if }\xi=0\text{ or }\xi=\omega\\0 & \text{if }0<\xi<\omega\\\xi & \text{otherwise}\end{cases}\qquad\text{and}\qquad\mathbf{G}_1(\xi)=\begin{cases}\omega\cdot 2 & \text{if }\xi=1\text{ or }\xi=\omega\\\omega\cdot 2+\xi & \text{if }\xi\ge\omega\cdot 2\\0 & \text{otherwise,}\end{cases}$$ we see that, in the "topology" induced by $\mathbf{B}_1$, $\mathbf{F}_1$ is ordinally but not topologically continuous, and $\mathbf{G}_1$ is topologically but not ordinally continuous. If you can't see why one or more of the $4$ preceding claims are true, let me know, and I'll give justification(s).

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