# set of infinite cardinals admits an injective regressive function

Let $A$ be a set of infinite cardinals. Assume that for every regular $\lambda$, the subset $A \cap \lambda$ of $\lambda$ is not stationary. Then I want to prove that there is an injective function $g$ on $A$ which satisfies $g(\alpha) < \alpha$ for all $\alpha \in A$.

This is an exercise in Kunen's set theory. I dont' understand if $g$ should be a map from $A \to A$. But then the claim would be false if $A = \{\kappa\}$. But if $g$ may have arbitrary ordinal values, then the claim is somehow trivial, also without the assumptions. So I don't know what this exercise about.

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What does "stationary" mean? Why is a singleton not stationary? –  Arturo Magidin Sep 24 '10 at 16:35
Thanks. Unfortunately, it does not help me answer your question. –  Arturo Magidin Sep 24 '10 at 17:27
@Arturo: if you look at the measure which gives all sets containing a closed and unbounded set a measure of 1, and all sets which are complements of such a measure of 0, then stationary sets are those of positive measure. –  Asaf Karagila Sep 24 '10 at 22:47

The exercise is asking for a map from $A$ to the ordinals, not to itself. Why do you think the exercise is trivial? If $\lambda$ is regular and $A\cap\lambda$ is non-stationary (in $\lambda$) then there is a club subset of $\lambda$ disjoint from $A$. You can then pick $g(\lambda)$ in this club. You need to ensure, however, that this element is not $g(\tau)$ for some prior $\tau$, and this requires some argument: If, as suggested in another answer, you are to pick something larger than $\sup(g(\alpha)\mid\alpha\in A\cap\lambda)$, you need to argue that this supremum is strictly below $\lambda$. But this is not automatic (or true in general), as $A$ may contain all cardinals up to and including some inaccessible. Also, $A$ may contain singular cardinals, and you need to define $g$ on them as well.
Ok, so how do we define $g$ in general? –  Martin Brandenburg Sep 25 '10 at 9:59
Hm no I don't know this lemma. And why is your $\lambda$ regular? –  Martin Brandenburg Sep 27 '10 at 10:29
@AndresCaicedo You wrote "Do you know how to prove that a subset A of a cardinal lambda is non-stationary iff there is an injective regressive function on A?" But this cannot be true for silly reasons like $A = \omega_1 \subseteq \lambda = \omega_2$. I think the best one can do is the following: For any subset $A$ of a regular cardinal $\lambda$, $A$ is non stationary iff there is a regressive map $h:A \to \lambda$ which is not constant on an unbounded subset of $\lambda$. –  hot_queen Jul 7 at 13:58