If $S$ is non-stationary in $[k]^{\omega}$ is there a choice-function on $S$ with bounded fibers? Fodor's Lemma : When $k$ is a regular uncountable cardinal, and $T$ is a stationary subset of $k$, any regressive $f:T\to k$  has a  fiber which is stationary in $k$. Corollary: $T$ is stationary in $k$ iff every regressive $f:T\to k$ has an unbounded fiber. (If $C$ is c.l.u.b. in $k$ and $C\cap T=\phi$ let $f(t)=\sup (t\cap C). $  The fibers of $f$ are bounded subsets of $k$.)
The theorem that is also called Fodor's Lemmma: When $k$ is a regular uncountable cardinal, and $S$ is stationary in $[k]^{\omega}$, any choice-function $g:S\to k$ has a fiber which is stationary in $[k]^{\omega}$.
Q: For regular uncountable cardinal $k$ : If $S$ is any non-stationary subset of $[k]^{\omega}$ does there exist a choice-function $f:S\to k$ such that no fiber of $f$ is unbounded in $[k]^{\omega}$?
I thought a proof of "yes" to this would be simple, but I got nowhere. I feel I am  missing something obvious.
 A: A set $C$ subset $[\kappa]^{\aleph_{0}} $ is club if there is a function $f \colon \kappa^{<\omega} \to \kappa$ such that C is the set of countable $a \subseteq \kappa$ closed under $f$. Let us say that $C$ is a unary club if there is a function
$g \colon \kappa \to [\kappa]^{\aleph_{0}}$ such that $C$ is the set of $b \in [\kappa]^{\aleph_{0}}$ such that
$b$ contains $g(\alpha)$ for all $\alpha \in b$. Your question amounts to asking whether every club contains a unary club, since we could take a club $C$ disjoint from your given nonstationary set $S$, and if $C' \subseteq C$ were a unary club, this would give us a function as desired on the compliment of $S'$ (the other direction is similar).
I claim that there is a club $C \subseteq [\omega_{2}]^{\aleph_{0}}$ which does not contain a unary club (this should show that the same holds for every uncountable $\kappa$). Fix  bijections $f_{\alpha} \colon \omega_{1} \to \alpha$ for all $\alpha \in [\omega_{1} , \omega_{2})$. Let $C$ be the set of countable $b \subseteq \omega_{2}$ such that $f_{\alpha}(\beta) \in b$ for all $\alpha \in b, \beta \in b \cap \omega_{1} $.
Now fix $g \colon\omega_{2} \to [\omega_{2}]^{\aleph_{0}}$. Since increasing the $g(\alpha)$'s thins the induced unary club, we may assume that $g(\alpha) = g(\beta)$ for all $\alpha < \omega_{2}$ and $\beta \in g(\alpha)$.
Fix cofinal $A \subseteq \omega_{2}$ and $\beta_{0} < \omega_1$ such that
$g(\alpha) \cap\omega_{1} \subseteq \beta_0$ for all $\alpha \in A$ and
$B = \omega_{2} \setminus \bigcup_{\alpha \in A} g(\alpha)$ has cardinality $\aleph_{2}$.
Fix a $\gamma \in B \setminus \bigcup_{\alpha <\omega_{1} }g(\alpha)$. Then $\gamma = f_{\alpha}(\delta)$ for some $\delta \in A \setminus (\gamma + 1)$ and $\delta < \omega_1$. Then $g(\gamma) \cup g(\delta)$ is in the unary club according to $g$, but not in $C$.
Note that if $D$ is the set of countable subsets of an uncountable cardinal $\kappa$ closed (in both directions) under a
fixed bijection between $\kappa$ and $\kappa^{<\omega}$, then the sets $C \cap D$ for $C$ club, and $C' \cap D$ for $C'$ a unary club do coincide.
