3
$\begingroup$

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!

  • 0
    [Eran](http://math.stackexchange.com/users/18508/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.*2012-05-26

1 Answers 1

1

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.

  • 0
    @elbarto: I haven't seen that re$q$uirement. I'll give it some thought.2012-05-26