Is the function of the limit = the limit of the function when I'm talking about continuous functions of ordinals? 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  $\delta < \gamma$, $$\lim_{\xi \to \delta} f ( \xi) = f(\delta).$$
 A: 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.

A: 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.
