I have a question which regards a "counterexample" to Fodor's lemma, which states that a regressive function on a stationary set of a regular, uncountable cardinal is constant on a stationary subset of that set.
I'm looking for a regressive function $f:A\longrightarrow A$ which is not constant on any uncountable subset. So even if $A$ is stationary, Fodor's lemma wouldn't work.
Given an arbitrary $A$, is there always such function?
We are given an uncountable set $A$ of reals, and want to define $f:A\to A$ regressive with the property that $f$ is not constant on any uncountable set. (We need some convention on how to define $f(a)$ if $a$ is the minimum of $A$; this is not really important and I'll leave it up to you what to do here.)
Let $\kappa=|A|$. We proceed by induction on $\kappa$. There are two cases.
1. Suppose ${\rm cof}(\kappa)$ is countable. We can write $A=\bigcup_n A_n$ where the $A_n$ are pairwise disjoint, each $A_n$ is uncountable and of size strictly less than $\kappa$.
By induction, there is a regressive function $f_n:A_n\to A_n$ as desired. Let $f=\bigcup_n f_n$. It may be that $f$ needs to be modified on $\{a\in A\mid\exists n\,(a=\min A_n)\}$ (for instance if this set is different from $\{\min A\}$), but this set is at most countable, so whichever way we fix this, this is not an issue.
2. Suppose now that ${\rm cof}(\kappa)$ is uncountable. Let $$B=\{x\in A\mid |A\cap(-\infty,x]|<\kappa\}.$$ Note that if $x\in B$ then $A\cap(-\infty,x]\subseteq B$. It follows that $B$ itself has size less than $\kappa$. To see this, let $t=\sup B$ in the extended reals (so $t=+\infty$ is possible). Pick an increasing sequence $t_0,t_1,\dots$ of elements of $B$ with limit $t$, and note that $$ \{t\}\cup B=\{t\}\cup\bigcup_n B\cap(-\infty,t_n]\subseteq\{t\}\cup\bigcup_n A\cap(-\infty,t_n], $$ and the last expression shows that $B$ is a countable union of sets of size strictly below $\kappa$. Since ${\rm cof}(\kappa)>\omega$, the claim follows.
First, define a regressive $f:B\to B$. If $B$ is countable, do this any way you want (ensuring that $f$ is regressive, of course); note that since $B$ is regressive, any point has countable preimage. If $B$ is uncountable, use the inductive hypothesis to define $f$.
Finally, let $C=A\setminus B$. Note that $|C|=\kappa$ and indeed $|C\cap (-\infty,x]|=\kappa$ for any $x\in C$. Well-order $C$ in order type $\kappa$, $C=\{x_\alpha\mid \alpha<\kappa\}$. Now define $f:C\to C$ recursively: Suppose that $\alpha<\kappa$ and we have defined $f(x_\xi)$ for all $\xi<\alpha$. Define $f(x_\alpha)=x_\beta$ where $\beta$ is least such that $x_\beta<x_\alpha$ and $x_\beta$ is not in $f(\{x_\xi\mid \xi<\alpha\})$. This is possible because $|C\cap(-\infty,x_\alpha)|=\kappa>|\alpha|$. Note that in fact $f$ is injective on $C$.
Note that the idea of the whole construction is very simple, and it is just what we have in the last paragraph. Perhaps it is simpler to think first of the case when $A$ has size $\aleph_1$. We well-order $A$ and try to define $f$ recursively. The problem is that, if we do not ensure that every $a\in A$ has uncountably many elements of $A$ below, our recursion may run out of elements to pick. So we need to first separate the elements of $A$ with only countable many elements below, the set $B$. When $|A|=\aleph_1$, $B$ is countable.
Once we have this case, we need to generalize from $\aleph_1$ to an arbitrary uncountable size. Now $B$ needs not be countable, and indeed we need to redefine $B$ not as the set of elements of $A$ with only countably many elements below, but as we did above, as the set of elements of $A$ with fewer than $|A|$ many elements below (otherwise, we may still run out of elements in our recursive definition of $f$). But this creates a new problem, namely, it is not clear that $|B|<|A|$. This forces us to divide the argument into two cases, according to the cofinality of $|A|$. The presentation above is just the result of reorganizing this sketch.
