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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
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@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

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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.

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Ok, so how do we define $g$ in general? –  Martin Brandenburg Sep 25 '10 at 9:59
    
Here is one way: 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? The first step would be to prove this. Then, for the A of the exercise, take lambda=sup(A) (note lambda is indeed a cardinal). To do the first step, consider a club disjoint from A, and note that elements of A sit between consecutive elements of the club. This should give you a head start. –  Andres Caicedo Sep 25 '10 at 16:19
    
Hm no I don't know this lemma. And why is your $\lambda$ regular? –  Martin Brandenburg Sep 27 '10 at 10:29
    
It needs not be. That's part of the point. –  Andres Caicedo Sep 27 '10 at 16:20

I'm no specialist, but see how this goes. When you say "the subset A∩λ of λ is not stationary", stationary in what? Presumably not stationary in λ. So there is a closed and unbounded subset of λ that is disjoint from A. Pick any element of this subset greater than sup{g(α):α<λ}

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Non-stationary doesn't mean bounded. For example, the set of all successor ordinals in $\lambda$ is unbounded, but doesn't intersect with the club of $\lambda \cap Lim$ (where $Lim$ is the class of limit ordinals). –  Asaf Karagila Sep 25 '10 at 2:16

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