An inconsistent version of Jensen's principle $\diamondsuit$

Question

In page 93 of Kunen's book, exercise 55 says:

Show that the following version of $\diamondsuit$ is inconsistent: There are $A_\alpha \subset \alpha$ for $\alpha < \omega_1$, such that for all stationary $A\subset \omega_1$, $\exists \alpha \in A (A \cap \alpha=A_\alpha)$.

I know that there is a version of this principle taking closed unbounded sets, but why we can't use stationary sets? Why is this version of Jensen's principle $\diamondsuit$ inconsistent?

Answer 1

By assumption (since $\omega_1$ itself is stationary), the set

$S=\{\alpha<\omega_1:A_\alpha=\alpha\}$

is stationary.

Clearly $S$ cannot equal all of $\omega_1$, so choose $\beta<\omega_1$ with

$S\cap\beta\neq\beta$.

The set $S\setminus\beta$ is stationary, and so there is an $\alpha\in S\setminus\beta$ with

$S\cap\alpha=A_\alpha$.

By choice of $S$, we have $S\cap\alpha=\alpha$, but $\beta\leq\alpha$ and we have a contradiction.

Edit: Answered too fast -- I see that I misread the principle (it doesn't a priori guarantee a stationary set of successful guesses). Will fix up if I have time.

Answer 2

I see that I completely forgot to post my own revised solution. Since it’s completely different from jolubo’s, it may still be of some interest.

There seems to be a small error in the problem statement: this version of $\diamondsuit$ is trivially true if you take every $A_\alpha$ to be empty. I’ll assume that at least one $A_\alpha\ne\varnothing$; if $S=\{\alpha\in\omega_1:A_\alpha\ne\varnothing\}$, this implies that $S$ is stationary. To see this, suppose that $C$ is a cub disjoint from $S$. Fix $\alpha_0\in S$ arbitrarily, and let $A=\{\alpha_0\}\cup(K\setminus\alpha_0)$. Clearly $A$ is stationary, but $A\cap\alpha_0=\varnothing\ne A_{\alpha_0}$, and if $\alpha_0<\alpha\in A$, then $A\cap\alpha\ne\varnothing=A_\alpha$. Thus, no such $C$ can exist.

Revised: Assume that such a sequence exists. Recursively define a function $\varphi:\omega_1\to 2$ by setting $\varphi(0)=1$, and $\varphi(\eta)=1$ iff $\{\xi<\eta:\varphi(\xi)=1\}\ne A_\eta$ for $0<\eta<\omega_1$. Let $$X=\{\xi\in\omega_1:\varphi(\xi)=1\}\;;$$ by construction $X\cap\xi\ne A_\xi$ for $\xi\in X$, so $X$ is non-stationary, and $Y=\omega_1\setminus X$ is stationary (indeed, contains a cub). Thus, $Y\cap\eta=A_\eta$ for some $\eta\in Y$. But $\eta\in Y$ implies that $\varphi(\eta)=0$, so $$A_\eta=\{\xi<\eta:\varphi(\xi)=1\}=X\cap\eta=\eta\setminus Y\;.$$ This contradiction establishes the inconsistency of this version of $\diamondsuit$.

Answer 3

Funnily enough, such a sequence does actually exist. Just let $A_\alpha = \emptyset$ for every $\alpha$, then for any stationary set, the least element makes the right guess.

So we need to assume some kind of nontriviality, say, $A_\alpha \neq \emptyset$ for stationarily many $\alpha$.

Now then, by Fodor's Lemma, the least element of $A_\alpha$ is constant for stationarily many $\alpha$. Say, $\min{A_\alpha} = \beta$ when $\alpha \in S$, and $S$ is stationary.

But $S \setminus \{\beta\}$ is stationary as well, yet cannot be guessed by $A_\alpha$ when $\alpha \in S$.