## Set theory seminar – Marion Scheepers: Coding strategies (I)

September 25, 2010

This semester, the seminar started with a series of talks by Marion. The first talk happened on September 14.

We consider two games relative to a (proper) ideal $J\subset{\mathcal P}(S)$ for some set $S$. The ideal $J$ is not assumed to be $\sigma$-complete; we denote by $\left< J\right>$ its $\sigma$-closure, i.e., the collection of countable unions of elements of $J$. Note that $\left< J\right>$ is a $\sigma$-ideal iff $\left< J\right>$ is an ideal iff $S\notin\left< J\right>$.

The two games we concentrate on are the Random Game on $J$, $RG(J)$, and the Weakly Monotonic game on $J$, $WMG(J)$.

In both games, players I and II alternate for $\omega$ many innings, with I moving first, moving as follows:

$\begin{array}{cccccc} I&O_0\in\left< J\right>&&O_1\in\left< J_2\right>&&\cdots\\ II&&T_0\in J&&T_1\in J \end{array}$

In $RG(J)$ we do not require that the $O_i$ relate to one another in any particular manner (thus “random”), while in $WMG(J)$ we require that $O_1\subseteq O_2\subseteq\dots$ (thus “weakly”, since we allow equality to occur).

In both games, player II wins iff $\bigcup_n T_n\supseteq\bigcup_n O_n$. Obviously, II has a (perfect information) winning strategy, with $=$ rather than the weaker $\supseteq$.

However, we are interested in an apparently very restrictive kind of strategy, and so we will give some leeway to player II by allowing its moves to over-spill if needed. The strategies for II we want to consider we call coding strategies. In these strategies, II only has access to player I’s latest move, and to its own most recent move. So, if $F$ is a coding strategy, and II follows it in a run of the game, then we have that for every $n$,

$T_n=F(T_{n-1},O_n)$,

with $T_{-1}=\emptyset$.

The underlying goal is to understand under which circumstances player II has a winning coding strategy in $WMG(J)$. Obviously, this is the case if II has a winning coding strategy in $RG(J)$.

Theorem 1. For an ideal $J\subset{\mathcal P}(S)$, the following are equivalent:

1. II has a winning coding strategy in $RG(J)$.
2. ${\rm cf}(\left< J\right>,{\subset})\le|J|$.

Corollary. ${\sf GCH}$ implies that for any ideal $J\subset{\mathcal P}(S)$, II has a winning strategy in $WMG(J)$.

We can reformulate our goal as asking how much one can weaken ${\sf GCH}$ in the corollary.

Let’s denote by ${\sf wSCH}$, the weak singular cardinals hypothesis, the statement that if $\kappa$ is singular strong limit of uncountable cofinality, then for no cardinal $\lambda$ of countable cofinality, we have $\kappa<\lambda<2^\kappa$.

By work of Gitik and Mitchell, we know that the negation of ${\sf wSCH}$ is equiconsistent with the existence of a $\kappa$ of Mitchell order $o(\kappa)=\kappa^{+\omega}+\omega_1$.

Theorem 2. The following are equivalent:

1. ${\sf wSCH}$.
2. For each ideal $J$ on a singular strong limit $\kappa$ of uncountable cofinality, II has a winning strategy in $RG(J)$.

We now begin the proof of Theorem 1.

$(1.\Rightarrow2.)$ Suppose II has a winning coding strategy $F$ in $RG(J)$. We want to show that ${\rm cf}(\left< J\right>,{\subset})\le|J|$. For this, we will define a map $f:J\to\left< J\right>$ with $\subset$-cofinal range, as follows: Given $X\in J$, let $T_0=X$ and $T_{n+1}=F(T_n,\emptyset)$ for all $n$. Now set

$f(X)=\bigcup_n T_n$.

To see that $f$ is cofinal, given $O\in\left< J\right>$, let $X=F(\emptyset,O)$, so that the $T_n$ are II’s responses using $F$ in a run of the game where player I first plays $O$ and then plays $\emptyset$ in all its subsequent moves. Since $F$ is winning, we must have $f(X)\supseteq O$.

## 580 -Cardinal arithmetic (10)

March 9, 2009

Let me begin with a couple of comments that may help clarify some of the results from last lecture.

First, I want to show a different proof of Lemma 21.2, that I think is cleaner than the argument I gave before. (The argument from last lecture, however, will be useful below, in the proof of Kunen’s theorem.)

Lemma 1 If ${\kappa}$ is measurable, ${{\mathcal U}}$ is a ${\kappa}$-complete nonprincipal ultrafilter over ${\kappa,}$ and ${j_{\mathcal U}:V\rightarrow M}$ is the corresponding ultrapower embedding, then ${{}^\kappa M\subset M.}$

Proof: Recall that if ${\pi}$ is Mostowski’s collapsing function and ${[\cdot]}$ denotes classes in ${V^\kappa/{\mathcal U},}$ then ${M=\{\pi([f]):f\in{}^\kappa V\}.}$ To ease notation, write ${\langle f\rangle}$ for ${\pi([f]).}$

Let ${h:\kappa\rightarrow M.}$ Pick ${f:\kappa\rightarrow V}$ such that for all ${\alpha<\kappa,}$ ${h(\alpha)=\langle f(\alpha)\rangle.}$

Lemma 2 With notation as above, ${\langle f\rangle=j_{\mathcal U}(f)(\langle{\rm id}\rangle)}$ for any ${f:\kappa\rightarrow V.}$

Proof: For a set ${X}$ let ${c_X:\kappa\rightarrow V}$ denote the function constantly equal to ${X.}$ Since ${\pi}$ is an isomorphism, ${\mbox{\L o\'s}}$‘s lemma gives us that the required equality holds iff

$\displaystyle \{\alpha<\kappa : f(\alpha)=((c_f)(\alpha))({\rm id}(\alpha))\}\in{\mathcal U},$

but this last set is just ${\{\alpha<\kappa:f(\alpha)=f(\alpha)\}=\kappa.}$ $\Box$

From the nice representation just showed, we conclude that ${\langle f(\alpha)\rangle=j_{\mathcal U}(f(\alpha))(\langle{\rm id}\rangle)}$ for all ${\alpha<\kappa.}$ But for any such ${\alpha,}$ ${j_{\mathcal U}(f(\alpha))=j_{\mathcal U}(f)(\alpha)}$ because ${{\rm cp}(j_{\mathcal U})=\kappa}$ by Lemma 21 from last lecture. Hence, ${h=(j_{\mathcal U}(f)(\alpha)(\langle{\rm id}\rangle):\alpha<\kappa),}$ which is obviously in ${M,}$ being definable from ${j_{\mathcal U}(f),}$ ${\langle{\rm id}\rangle,}$ and ${\kappa.}$ $\Box$

The following was shown in the proof of Lemma 20, but it deserves to be isolated.

Lemma 3 If ${{\mathcal U}}$ is a normal nonprincipal ${\kappa}$-complete ultrafilter over the measurable cardinal ${\kappa,}$ then ${{\mathcal U}=\{X\subseteq\kappa:\kappa\in i_{\mathcal U}(X)\},}$ i.e., we get back ${{\mathcal U}}$ when we compute the normal measure derived from the embedding induced by ${{\mathcal U}.}$ ${\Box}$

Finally, the construction in Lemma 10 and preceeding remarks is a particular case of a much more general result.

Definition 4 Given ${f:I\rightarrow J}$ and an ultrafilter ${{\mathcal D}}$ over ${I,}$ the projection ${f_*({\mathcal D})}$ of ${{\mathcal D}}$ over ${J}$ is the set of ${X\subseteq J}$ such that ${f^{-1}(X)\in{\mathcal D}.}$

Clearly, ${f_*({\mathcal D})}$ is an ultrafilter over ${J.}$

Notice that if ${\kappa={\rm add}({\mathcal D}),}$ ${(X_\alpha:\alpha<\kappa)}$ is a partition of ${I}$ into sets not in ${{\mathcal D},}$ and ${f:I\rightarrow\kappa}$ is given by ${f(x)=}$ the unique ${\alpha}$ such that ${x\in X_\alpha,}$ then ${f_*({\mathcal D})}$ is a ${\kappa}$-complete nonprincipal ultrafilter over ${\kappa.}$ (Of course, ${\kappa=\omega}$ is possible.)

For a different example, let ${{\mathcal U}}$ be a ${\kappa}$-complete nonprincipal ultrafilter over the measurable cardinal ${\kappa,}$ and let ${f:\kappa\rightarrow\kappa}$ represent the identity in the ultrapower by ${{\mathcal U},}$ ${\langle f\rangle=\kappa.}$ Then ${f_*({\mathcal U})}$ is the normal ultrafilter over ${\kappa}$ derived from the embedding induced by ${{\mathcal U}.}$

Definition 5 Given ultrafilters ${{\mathcal U}}$ and ${{\mathcal V}}$ (not necessarily over the same set), say that ${{\mathcal U}}$ is Rudin-Keisler below ${{\mathcal V},}$ in symbols, ${{\mathcal U}\le_{RK}{\mathcal V},}$ iff there are sets ${S\in{\mathcal U},}$ ${T\in{\mathcal V},}$ and a function ${f:T\rightarrow S}$ such that ${{\mathcal U}\upharpoonright S=f_*({\mathcal V}\upharpoonright T).}$

Theorem 6 Let ${{\mathcal U}}$ be an ultrafilter over a set ${X}$ and ${{\mathcal V}}$ an ultrafilter over a set ${Y.}$ Suppose that ${{\mathcal U}\le_{RK}{\mathcal V}.}$ Then there is an elementary embedding ${j:V^X/{\mathcal U}\rightarrow V^Y/{\mathcal V}}$ such that ${j\circ i_{\mathcal U}=i_{\mathcal V}.}$

Proof: Fix ${T\in{\mathcal U}}$ and ${S\in{\mathcal V}}$ for which there is a map ${f:S\rightarrow T}$ such that ${{\mathcal U}\upharpoonright T=f_*({\mathcal V}\upharpoonright S).}$ Clearly, ${V^X/{\mathcal U}\cong V^T/({\mathcal U}\upharpoonright T)}$ as witnessed by the map ${[f]_{\mathcal U}\mapsto[f\upharpoonright T]_{{\mathcal U}\upharpoonright T},}$ and similarly ${V^Y/{\mathcal V}\cong V^S/({\mathcal V}\upharpoonright S),}$ so it suffices to assume that ${S=Y}$ and ${T=X.}$

Given ${h:X\rightarrow V,}$ let ${h_*:Y\rightarrow V}$ be given by ${h_*=h\circ f.}$ Then ${j([h]_{\mathcal U})=[h_*]_{\mathcal V}}$ is well-defined, elementary, and ${j\circ i_{\mathcal U}=i_{\mathcal V}.}$

In effect, ${h=_{\mathcal U}h'}$ iff ${\{x\in X:h(x)=h'(x)\}\in{\mathcal U}}$ iff ${\{y\in Y:h\circ f(y)=h'\circ f(y)\}\in{\mathcal V}}$ iff ${h_*=_{\mathcal V}h'_*,}$ where the second equivalence holds by assumption, and it follows that ${j}$ is well-defined.

If ${c_B^A}$ denotes the function with domain ${A}$ and constantly equal to ${B,}$ then for any ${x,}$ ${j\circ i_{\mathcal U}(x)=j([c^X_x]_{\mathcal U})=[(c^X_x)_*]_{\mathcal V}=[c^Y_x]_{\mathcal V}=i_{\mathcal V}(x)}$ since ${(c^X_x)_*=c^Y_x}$ by definition of the map ${h\mapsto h_*.}$ This shows that ${j\circ i_{\mathcal U}=i_{\mathcal V}.}$

Elementarity is a straightforward modification of the proof of Lemma 10 from last lecture. $\Box$

One can show that Theorem 6 “very nearly” characterizes the Rudin-Keisler ordering, see for example Proposition 0.3.2 in Jussi Ketonen, Strong compactness and other cardinal sins, Annals of Mathematical Logic 5 (1972), 47–76.