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The standard formulation of the Diamond Principle $\Diamond$ is as follows:

There exists a sequence $\langle f_\alpha:\ \alpha<\omega_1\rangle$ of functions $f_\alpha:\alpha\to2$ such that for all functions $f:\omega_1\to2$ the set $\{\alpha<\omega_1:\ f|_\alpha=f_\alpha\}$ is stationary in $\omega_1$.

Given any set $E\subseteq\omega_1$ by the Restricted Diamond Principle $\Diamond(E)$ we mean:

There exists a sequence $\langle f_\alpha:\ \alpha\in E\rangle$ of functions $f_\alpha:\alpha\to2$ such that for all functions $f:\omega_1\to2$ the set $\{\alpha\in E:\ f|_\alpha=f_\alpha\}$ is stationary in $\omega_1$.

Now, assume $\Diamond$ holds. Let $E$ and $F$ be such subsets of $\omega_1$ that $\Diamond(E)$ and $\Diamond(F)$ hold. How can it be proved that $\Diamond(E\cap F)$ also holds?

I need this to prove that the set $I=\{E\subseteq\omega_1: \neg\Diamond(E)\}$ is a normal ideal (i.e. an ideal closed under diagonal union and containing all countable subsets of $\omega_1$).

My second question also concerns the Diamond Principle. This principle can be stated in two similar, but equivalent to it, forms:

There exists a sequence $\langle f_\alpha:\ \alpha<\omega_1\rangle$ of functions $f_\alpha:\alpha\to\alpha$ such that for all functions $f:\omega_1\to\omega_1$ the set $\{\alpha<\omega_1:\ f|_\alpha=f_\alpha\}$ is stationary in $\omega_1$.

and:

There exists a sequence $\langle f_\alpha:\ \alpha<\omega_1\rangle$ of functions $f_\alpha:\alpha\to\alpha$ such that for all functions $f:\omega_1\to\omega_1$ there is $\alpha>0$ such that $f|_\alpha=f_\alpha$.

The latter easily follows from the former one (stationary sets are always nonempty). How can it be proved that existence of one $\alpha>0$ implies the existence of stationary many $\alpha$'s? I don't see even how it can imply the existence of just another one $\alpha$.

I appreciate any hints. TIA.

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2 Answers 2

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You cannot prove that $\diamondsuit(E\cap F)$ follows from $\diamondsuit(E)$ and $\diamondsuit(F)$, even assuming that $E\cap F$ is stationary. (Of course, if $E\cap F$ is not stationary, then $\diamondsuit(E\cap F)$ fails.)

The reason is that it is consistent to have disjoint stationary sets $A,B,C$ with $\diamondsuit(A)$, $\diamondsuit(B)$ and $\lnot\diamondsuit(C)$. But then you can take $E=A\cup C$, $F=B\cup C$.

That the set you call $I$ is a normal ideal is a result of Shelah (in his famous "Whitehead groups may be not free, even assuming CH. I", Israel J. Math. 28 (1977), no. 3, 193–204). The result holds for any $\kappa$ such that $\diamondsuit_\kappa$ holds, not just for $\kappa=\omega_1$.

The argument (for $\omega_1$) is as follows: (I use sets rather than characteristic functions, but the translation should be straightforward.) First, since $\diamondsuit$ holds, $\omega_1\notin I$. Obviously, if $A\in I$ and $B\subseteq A$, then $A\in I$. To prove that $I$ is normal (which also gives us that it is closed under finite unions, though this of course can be seen directly), suppose $\diamondsuit(E)$ holds, and $f:E\to\omega_1$ is regressive. We need to see that for some $\alpha$, $\diamondsuit(f^{-1}(\{\alpha\}))$ holds.

Start by noting that there is a sequence $(S_\alpha\mid\alpha<\omega_1)$ with $S_\alpha\subseteq\alpha\times\alpha$ and such that for any $X\subseteq\omega_1\times\omega_1$, we have that $\{\alpha\in E\mid X\cap(\alpha\times\alpha)=S_\alpha\}$ is stationary.

(This is easy: Fix a bijection $b:\omega_1\times\omega_1\to\omega_1$ and use it to "pull back" your original diamond sequence. Kunen has several variations of this idea.)

Ok, now, for $\alpha,\beta<\omega_1$, let $$S^\beta_\alpha=\{\zeta<\alpha\mid(\zeta,\beta)\in S_\alpha\}.$$ Check that for some $\beta$, the sequence $(S^\beta_\alpha\mid\alpha<\omega_1)$ is a $\diamondsuit(f^{-1}(\{\beta\}))$ sequence.

For this, assume otherwise, so for each $\beta$ there is a set $X_\beta\subseteq\omega_1$ that is not guessed correctly stationarily often, so there is a club $C_\beta$ such that $X_\beta\cap\alpha\ne S^\beta_\alpha$ whenever $\alpha\in C_\beta\cap f^{-1}(\{\beta\})$. Let $$X=\bigcup_\beta X_\beta\times\{\beta\}$$ and let $C$ be the diagonal intersection of the $C_\beta$. Then it follows just unraveling definitions, that $X\cap(\alpha\times\alpha)\ne S_\alpha$ whenever $\alpha\in C\cap E$. This contradicts that the sequence of $S_\alpha$ was a diamond sequence on $E$, and we are done.


As for the second question, we can in fact show that the following implies diamond:

There is a sequence $(S_\alpha\mid\alpha<\omega_1)$ such that $S_\alpha$ is a countable subset of ${\mathcal P}(\alpha)$ for each $\alpha$, and whenever $X\subseteq\omega_1$, then $X\cap\alpha\in S_\alpha$ for some infinite $\alpha$.

There are several approaches to this fact. It is not complicated, and you may benefit from working through the details. I recommend you take a look at Devlin's "Constructibility" book, where this is stated as an Exercise (in III.3) and a generous hint is provided. Or you can look at the paper where it originally appeared, "Variations on $\diamondsuit$", by Devlin, The Journal of Symbolic Logic, Vol. 44, No. 1, (Mar., 1979), pp. 51-58.

Kanamori wrote years ago a small survey on diamond that was to be part of the second volume of his book on large cardinals. I imagine if you email him he will send you a copy, it was to be part of Chapter VII, "Higher combinatorics". For a truly excellent and up to date survey on diamond together with references, see "Jensen's diamond principle and its relatives", by Assaf Rinot. The paper appeared in "Set Theory and Its Applications", a volume published by the AMS that I co-edited, and can be obtained from Assaf's page.

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Assaf's work on squares and diamonds is very interesting. I recall several presentations he made in our seminar over the past year. (Also, thanks for posting an answer, now I can delete mine :)) –  Asaf Karagila Jun 28 '11 at 20:01
    
@Asaf: Oh, don't delete it, please. I think people may benefit from it and from the comments. –  Andres Caicedo Jun 28 '11 at 20:02
    
thank you for your answer. It is very helpful, but one issue is not clear for me. You have proved that for elements from $F$ (the dual to $I$) the Fodor's lemma works. What about subsets which are $F$-positive and not in $F$? I ask because I know such a lemma: a filter $F$ is normal iff for each $F$-positive subset $S$ and regressive function $f:S\to\omega_1$ there is $F$-positive subset of $S$ on which $f$ is constant. I don't see how one can deduce the thesis of this lemma from what you have posted. –  Damian Sobota Jun 28 '11 at 22:19
    
I think I considered sets $E$ that are not in the ideal (i.e., such that $\diamondsuit(E)$ holds) rather than just sets whose complement is in the ideal. These are precisely the positive sets. The argument shows that if $E$ is positive and $f:E\to\omega_1$ is regressive, then $f$ is constant on a positive set, i.e., there is some $\alpha$ such that $\diamondsuit(f^{-1}(\{\alpha\}))$ holds. (Let me know in case I am confused about something.) –  Andres Caicedo Jun 28 '11 at 22:30
    
A little off-topic after reading the last part of your comment: What happened to further volumes of Kanamori's book "The higher infinite"? Had he desisted from writing them? –  Charlie Jun 30 '11 at 22:54

I will post an answer for the first question, I will add the second question later.

An ideal is normal if it extends the nonstationary ideal, or if its dual filter extends the club filter.

Namely, it is enough to show that for every club set in $\omega_1$ diamond holds. (For convenience I will work with a sequence of sets $\langle A_\alpha\mid\alpha<\omega_1\rangle$ for the diamond sequence, not functions, it is the same thing)

Suppose $E\subseteq\omega_1$ is a club, and that $\diamondsuit$ holds. Suppose $A\subseteq\omega_1$, we have that $D_A=\{\alpha<\omega_1\mid A\cap\alpha=A_\alpha\}$ is stationary in $\omega_1$, then $D_A\cap E$ is stationary (a club and a stationary set meet at a stationary set), but all $\alpha\in D_A\cap E$ are from the diamond sequence restricted to $E$. Therefore $\diamondsuit(E)$ holds, as needed.

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You cannot prove that $\diamond(E\cap F)$ follows from $\diamond(E)$ and $\diamond(F)$, even assuming that $E\cap F$ is stationary. (Of course, if $E\cap F$ is not stationary, then $\diamond(E\cap F)$ fails.) The reason is that it is consistent to have disjoint stationary sets $A,B,C$ with $\diamond(A)$, $\diamond(B)$ and $\lnot\diamond(C)$. But then you can take $E=A\cup C$, $F=B\cup C$. –  Andres Caicedo Jun 28 '11 at 17:13
    
@Andres: I wrote that $E$ is a club, not stationary. If $\diamondsuit(E)$ holds for all clubs, then it holds for the club filter. –  Asaf Karagila Jun 28 '11 at 17:14
    
Oh, I'm not criticizing your answer, just pointing out to user12690 how it relates to what they are asking. –  Andres Caicedo Jun 28 '11 at 17:16
    
@Andres: Oh. I see your point. I will add this to my answer when I edit it to address the second question! Thanks! –  Asaf Karagila Jun 28 '11 at 17:18
    
@Andres: so is it false within ZFC+$\Diamond$ that $F=\{E\subseteq\omega_1:\ \Diamond(E)\}$ is a filter? –  Damian Sobota Jun 28 '11 at 18:44

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