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 $\delta < \gamma$, $\lim_{\xi \to \delta} f ( \xi) = f(\delta).$