Oracle forcing part II

Putting oracle forcing into context

$\sigma$-closed
$\Downarrow$
uberoracle-proper $\Rightarrow$ strong $\bar{M}$-proper $\Longrightarrow$
$\Rightarrow$ $\bar{M}$-proper
proper
$\Uparrow$ $\Uparrow$ $\Uparrow$ $\Uparrow$
uberoracle-cc $\Rightarrow$ strong $\bar{M}$-cc $\Rightarrow$ $\bar{M}$-cc $\Rightarrow$ ccc
$\Uparrow$
$\sigma$-centred

Note: All arrows are strict. The only arrow which could possibly be added is $\sigma$-centred $\Rightarrow \bar{M}$-proper; this is unknown. The picture is supposed to show that strong oracle-proper implies proper, but oracle-proper does not imply proper.

Proofs of these implications and counterexamples to show strictness are given below (at least the ones that are not obvious, well-known or mentioned in the previous post), working our way from left to right in the picture.

Claim Any $\sigma$-closed forcing is $\bar{M}$-proper for any oracle $\bar{M}$.

Proof Starting with a condition $p$ and $\delta \in \omega_1$, basically use the closure of the forcing to extend $p$ to a $(\delta, M_\delta)$-generic condition.
$\Box$

To see that uberoracle-cc implies uberoracle-proper and that strong oracle-cc implies strong oracle-proper, one simply uses the fact that if a forcing is $\bar{M}$-cc for a specific oracle $\bar{M}$ then it is $\bar{M}$-proper for the same oracle (see Abraham’s notes for a proof of this latter fact).

The following claim (proof due to Martin is forthcoming) implies both that the notion of uberoracle-proper is strictly stronger than oracle-proper and that uberoracle-cc is stronger than oracle-cc.

Claim:

  1. For any oracle $\bar{M}$ there is a Suslin tree $T$ such that $T$ is $\bar{M}$-cc.
  2. For any Suslin tree $T$ and any oracle $\bar{M}$ there is $\bar{M}^\prime \trianglerighteq \bar{M}$ such that $T$ is not $\bar{M}^\prime$-proper.

The next claim can be found in Shelah 100.

Claim For any strong oracle $\bar{M}$, if $P$ is strong $\bar{M}$-proper and $|P| \leq \aleph_1$ then $P$ is proper.

Proof Assume $P = \omega_1$. Fix $p \in P$. Let $N$ be a countable elementary submodel of some large $H(\xi)$ such that $\bar{M}, P \in N$. Let $\delta^* = N \cap \omega_1$.

Let $S$ be the set of all $\delta < \omega_1$ such that there exists $q \leq p$ where $q$ is $(\delta, M_\delta)$-generic which is club as $P$ is $\bar{M}$-proper and $\bar{M}$ is a strong oracle. Therefore we may assume that that $\delta^* \in S$.

Now let $E \in N$ such that $E \subseteq P$ and dense in $P$. The set $C$ of $\delta \in \omega_1$ such that $E \cap P \upharpoonright \delta$ is dense in $P\upharpoonright \delta$ is club and $C \in N$ so $\delta^* \in C$. Also $E \cap P \upharpoonright \delta^* \in N$. We want to see that $E \cap P\upharpoonright \delta^* \in M_{\delta^*}$.

The set
$$\{N : N \cap [\omega_1]^{\aleph_0} \subseteq M_{(N \cap \omega_1)}\}$$
is club in $[H(\xi)]^{\aleph_0}$.
Why? Given $N_i : i < \omega$ in this set, we have
$$(\bigcup N_i) \cap [\omega_1]^{\aleph_0} \subseteq \bigcup_i M_{(N_i \cap \omega_1)} \subseteq M_{(\bigcup N_i \cap \omega_1)}$$
so it is closed. Thus the $N$ fixed above is a member of this set, which gives
$$E \cap P\upharpoonright \delta^* \in M_{\delta^*}.$$

Now by $\bar{M}$-properness we have $E \cap P\upharpoonright \delta^* = E \cap N$ is predense in $P$ below $q$.

$\Box$

This next one though, proved by Martin, comes as somewhat a surprise and makes us think twice about bothering at all with the weak definition of oracle-proper.

Claim: There is a $P$ which is $\bar{M}$-proper and $|P| = \aleph_1$ but $P$ is not proper.

Proof Let $S$ be a stationary, co-stationary subset of $\omega_1$ and let $P$ be the forcing which collapses $\omega_1 \setminus S$. That is, conditions are continuous functions $f : \alpha \rightarrow \alpha$ where $\alpha$ is a successor ordinal and $f(\beta) \in S$ for all $\beta < \alpha$. This forcing is not proper, as models $N$ such that $N \cap \omega_1 \not\in S$ do not have $(N,P)$-generic conditions.

Let $\bar{M}$ be defined as $\{M_\delta : \delta \in S\}$. Given $\delta \in S$ and $p \in M_\delta$ we will find $q \leq p$ which is $(\delta, M_\delta)$-generic. Denote by $P \upharpoonright \delta = \{f \in P : f \subset \delta \times \delta\}$. Enumerate by $\langle A_n : n < \omega\rangle$ the set of $A \subseteq P\upharpoonright \delta$ such that $D \in M_\delta$ and are antichains in $P\upharpoonright \delta$. We may extend $p$ in $\omega$-steps such that each $p_n$ forces that the generic intersects $A_n$ at a point $a_n$. Then $q = \bigcup p_n \cup \{(\delta, \delta)\}$ is a condition in $P$ and forces that for all $n < \omega$ the $P$-generic filter meets $A_n$ at $a_n$ (i.e. is non-empty).
$\Box$

Finally we show that the oracle-cc to oracle-proper implications are strict.

Claim: There is a forcing which is uberoracle-proper but neither $\omega$-proper nor ccc.

Proof Let conditions in $P$ be finite partial functions $p : \omega_1 \rightarrow \omega_1$ which are weakly increasing. This is proper, but not Axiom A, see Jech Ch. 31 exercises.

To see that this forcing is oracle proper for any oracle, let $p \in P$ and $\delta$ be such that $P \upharpoonright \delta = \{p \upharpoonright (\delta \times \delta) : p \in P\}$ (happens on a club). Then for $M_\delta$ we let $q = p \cup \{\delta, \delta\}$ which is $(\delta, M_\delta)$-generic.
$\Box$

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