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Let $f: \bf{Ord} \to \bf{Ord}$ be a continuous, weakly increasing function, and let $\langle \alpha_{\xi} \mid \xi < \gamma \rangle$ be an increasing sequence of ordinals. Is it true that $$\displaystyle f\left(\lim_{\xi \to \gamma}\alpha_{\xi}\right) = \lim_{\xi \rightarrow \gamma}f(\alpha_{\xi})?$$ If it is true how do I show it?

The reason I am asking is because I have a continuous function $f: \lambda \to \kappa$ where $\kappa$ is a singular cardinal and $\operatorname{cf}\kappa = \lambda$, which enumerates a club $C$ in $\kappa$. Given a set $A \subseteq \kappa$, I want to show that $f^{-1}(A)$ is stationary in $\lambda$ if and only if $A \cap C$ is stationary in $\kappa$ if and only if $A$ is stationary in $\kappa$.

The part I am stuck on is $f^{-1}(A)$ stationary in $\lambda \implies A$ stationary in $\kappa$.

My idea was that given a club $D \subseteq \kappa$, then $f^{-1}(D)$ would be a club in $\lambda$. Showing closure of $f^{-1}(D)$ would follow from the above statement, I think: that is, given an increasing sequence $\langle \tau_{\nu} \mid \nu < \gamma \rangle$ with $\gamma < \lambda$, then $\displaystyle f\left(\lim_{\nu \to \gamma}\tau_{\nu}\right) = \lim_{\nu \rightarrow \gamma}f(\tau_{\nu})$, which implies $\displaystyle\lim_{\nu \to \gamma}\tau_{\nu} \in f^{-1}(A \cap C)$. Is there another way to see this if the above is not true?

Thanks for any help.

EDIT: The definition I am using is that $f$ continuous $\iff$ given any limit ordinal $\delta < \gamma$, $$\lim_{\xi \to \delta} f ( \xi) = f(\delta).$$

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Man, we're really lacking seriousness when choosing names for things in mathematics. A "club"? Please. –  Patrick Da Silva Feb 27 '12 at 18:48
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You mean continuous and (weakly) increasing, right? This already will show half of the desired equality. –  g.castro Feb 27 '12 at 18:56
    
@Patric: And simplicial "stuff"? Is that better? At least club is a short for $\mathbf{CL}\text{osed and }\mathbf{U}\text{n}\mathbf{B}\text{ounded}$. –  Asaf Karagila Feb 27 '12 at 18:57
    
@Asaf : All I'm saying is that sometimes you give your math texts to someone from the outside and they can laugh at you for the choice of words. –  Patrick Da Silva Feb 27 '12 at 19:04
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It may be helpful to note that continuity of these functions can be seen topologically: if you endow every ordinal with the order topology, then (set-theoretic) continuous functions correspond exactly to (topologically) continuous functions. (Actually -- looking at what you want to show -- being (set-theoretically) closed is also the same as being (topologically) closed.) –  Arthur Fischer Feb 27 '12 at 20:07

2 Answers 2

up vote 3 down vote accepted

As we generally only speak of limits of non-decreasing sequences of ordinals, we may assume that the function $f$ is itself non-decreasing.

Let $\alpha = \lim_{\xi \to \gamma} \alpha_\gamma$. If $f ( \alpha ) \neq \lim_{\xi \to \gamma} f ( \alpha_\xi )$ there are two possibilities:

  • If $f ( \alpha ) < \lim_{\xi \to \gamma} f ( \alpha_\xi )$ then there must be a $\beta < \gamma$ such that $f ( \alpha ) < f ( \alpha_\xi )$ for all $\beta \leq \xi < \gamma$. This clearly contradicts that $f$ is non-decreasing.
  • If $f ( \alpha ) > \lim_{\xi \to \gamma} f ( \alpha_\xi )$, then by continuity we have that $\lim_{\xi \to \gamma} f ( \alpha_\xi ) < f ( \alpha ) = \lim_{\beta \to \alpha} f ( \beta )$, and therefore there must be a $\delta < \alpha$ such that $\lim_{\xi \to \gamma} f ( \alpha_\xi ) < f ( \beta )$ for all $\delta \leq \beta < \alpha$. But as $\alpha = \lim_{\xi \to \gamma} \alpha_\xi$ there is a $\xi < \gamma$ such that $\delta < \alpha_\xi$. This again contradicts that $f$ is non-decreasing.
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Typically, continuous functions are defined with respect to the underlying topology so that what you said is already true, for appropriate extensions of the idea of "limit." In this case, the definition of limit is precisely what it ought to be (though the traditional definition of "sequence" is replaced with "function from an ordinal," rather than "function from $\omega$").

That said, the property you want is true, and in fact it's the definition of continuity given on Wikipedia http://en.wikipedia.org/wiki/Continuous_function_(set_theory), noting that the limit of a sequence is defined to be the $\liminf$ and $\limsup$ where they agree, and is undefined otherwise.

If you have a different definition of continuous in mind, please post it, and we will demonstrate the equivalence.

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Richard, while you wrote true things, you should beware of making such claims. The fact a certain name is used need not imply of any connection to other contexts. For example we call a continuous function a normal function. Does that mean anything about the function now? Does it have anything to do with other "normal" stuff in mathematics? –  Asaf Karagila Feb 27 '12 at 20:29
    
@Richard I have included my definition in the question now –  Paul Slevin Feb 27 '12 at 21:08
    
@asaf I did say typically –  Richard Rast Feb 28 '12 at 17:16

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