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This is also from Kunen, Set Theory, ch. II:

Let $A$ be a set of infinite cardinals such that for each $\lambda$ regular $A\cap\lambda$ is not stationary in $\lambda$. Show that there is an injective function $g$ with domain $A$ such that $\forall\alpha\in A(g(\alpha)<\alpha)$.

I tried using $\lambda=\sup A^+$, took a closed unbounded set $C$ disjoint from $A\cap\lambda=A$ and defined $g(\alpha)=\sup(C\cap\alpha)$ if $\sup(C\cap\alpha)>g(\eta)\forall\eta<\alpha$ and $g(\alpha)=\eta$ if $\sup(C\cap\alpha)=g(\eta),\eta<\alpha$, but then realized it could happen that $\alpha>\eta=\sup(C\cap\beta)=g(\beta),\eta<\beta<\alpha$, so g wouldn't be injective.

Help, please!

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Eran wrote: In the second definition above, just take another sup (on all those $\eta$'s) that satisfy the condition. Since the set of these $\eta$'s cannot be unbounded in $\alpha$ (otherwise we would get unbounded values of $(C\cap\alpha)$ and $\alpha$ would go into $C$) we are done. –  Asaf Karagila May 26 '12 at 22:24
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1 Answer

Let $\lambda=(\sup A)^+$, as you chose it, and $C$ a club in $\lambda$ which is disjoint from $A$.

Claim: Suppose $\alpha$ is a limit ordinal and $C$ is a club in some cardinal $\lambda>\alpha$. If $\sup(C\cap\alpha)=\alpha$ then $\alpha\in C$.

Proof. $C$ is closed therefore if $D\subseteq C$ is bounded (below $\lambda$) then $\sup D\in C$. In particular $\sup(C\cap\alpha)$ is bounded. $\square$

Since $A$ is a set of cardinals, every $\alpha\in A$ is a limit ordinal. The above claim holds for the chosen $\lambda$ and $C$ from the first line. That is, $\sup(C\cap\alpha)<\alpha$ for all $\alpha\in A$.

So now we can take $g(\alpha)=\sup(C\cap\alpha)$. Indeed restricted to $A$, $g$ has to be regressive, by the aforementioned argument.

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thanks for the clarification on that point. Sure we get a regressive function. but how to make it injective? that's where i'm actually stucked. –  el barto May 26 '12 at 1:20
    
@elbarto: I haven't seen that requirement. I'll give it some thought. –  Asaf Karagila May 26 '12 at 9:33
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