This is Shelah's simpler notion of 'completeness' for not adding reals via forcing. Here $P$ and $Q$ are forcing notions.

Let $S_{\aleph_0}(\kappa)$ be the set of all countable subsets of a cardinal $\kappa$. A family of subsets of $S_{\aleph_0}(\kappa)$, $\varepsilon$, is said to be nontrivial iff there is a cardinal $\mu$, for all $\lambda > \mu$, there is a countable $N \preccurlyeq H(\lambda)$ such that $\varepsilon \in N$ and $N \cap \kappa \in A$ for every $A \in \varepsilon \cap N$. We call such an $N$ suitable for $\varepsilon$. By a theorem, we can always assume $S_{\aleph_0}(\kappa) \in \varepsilon$.

$p := \{ p_n : n < \omega \}$ is a generic sequence for $(N, P)$ iff for each $n$, $p_n \in P \cap N$, $p_{n+1} \geq$ (extends) $p_n$, and $p$ intersects all dense open sets $\chi \in \mathcal{P}(P) \cap N$. We say the pair $(N, P)$ is complete iff every generic sequence for $(N, P)$ has an upper bound in $P$.

We can now define $\varepsilon$-completeness. $P$ is $\varepsilon$-complete for some nontrivial subset $\varepsilon$ of $S_{\aleph_0}(\kappa)$, iff there is a cardinal $\mu$, for all $\lambda > \mu$, for every $N \preccurlyeq H(\lambda)$, $N$ suitable for $\varepsilon$ such that $P \in N$, the pair $(N, P)$ is complete.

Shelah mentions in passing, in his book, that if $Q$ is not $\varepsilon$-complete whereas $P$ is, then forcing with $P$ will not make $Q$ $\varepsilon$-complete in the generic extension. Why is this so? Proving the same statement with the word 'proper' in place of '$\varepsilon$-complete' is easy, but even with some modifications, I am stuck.

My reference for this is Shelah's 'Proper and Improper Forcing', Chapter 5, Section 1. http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pl/1235419814

Much thanks in advance!


Let $\lambda$ be so that $\varepsilon$, $Q$, $P \in H(\lambda)$. We shall show that for any $p \in P$, there is $p^{*} \in P$, $p \leq p^{*}$ with $p^{*} \Vdash$ "if $\lambda$ is a cardinal, then $Q$ is not $\varepsilon$-complete".

Define $NCom(\lambda, A) :=$ "there is a countable $N \preccurlyeq H(\lambda)$, such that $N$ is suitable for $\varepsilon$, $Q$, $A \in N$ (thus $A \subset N$ if $A$ is countable) and the pair $(N, Q)$ is not complete."

We prove that if $A$ is a countable subset of $H(\lambda)$, then $NCom(\lambda, A)$. Assume otherwise. By $Q$ being not $\varepsilon$-complete, we have $\mu > \lambda$ and countable $M \preccurlyeq H(\mu)$, $M$ suitable for $\varepsilon$, $Q \in M$ with the pair $(M, Q)$ not complete. Then there are $\lambda'$, $A' \in M$ so that $Q$, $A'$, $\varepsilon \in H(\lambda')$ and $M \models \neg NCom(\lambda', A')$. But $M \cap H(\lambda')$ witnesses $NCom(\lambda', A')$, a contradiction.

Let $N$ witness $NCom(\lambda, \{P, p\})$, and $\chi := \langle q_n : n < \omega \rangle$ be a generic sequence for $(N, Q)$ with no upper bound in $Q$. Let $\langle \pi_n : n < \omega \rangle$ be a list of all $Q$-names in $N$, and $\langle y_n : n < \omega \rangle$ be a list of dense open subsets of $P$ in $N$.

We defined $\langle p_n : n < \omega \rangle$ by induction, satisfying the following conditions,

  1. $p_0 = p$
  2. $p_n \in N$
  3. $p_{n+1} \in y_n$
  4. $p_{n+1} \Vdash$ "$\pi_n$ is not a dense open subset of $Q$" or $p_{n+1} \Vdash \pi_n \cap \chi \neq 0$.

Let $p_n$ be defined. If $p_n \not \Vdash$ "$\pi_n$ is a dense open subset of $Q$", choose $p' \in N$, $p' \geq p_n$ with $p' \Vdash$ "$\pi_n$ is not a dense open subset of $Q$". Otherwise, $X := \{q \in Q : \exists p \geq p_n (p \Vdash q \in \pi_n) \}$ is a dense open subset of $Q$, so there is $m < \omega$ such that $q_m \in X$. Choose $p' \in N$ such that $p' \geq p_n$, $p' \Vdash q_m \in \pi_n$. Now let $p_{n+1} \in N$, $p_{n+1} \geq p'$ with $p_{n+1} \in y_n$. This completes the inductive step.

Clearly $\langle p_n : n < \omega \rangle$ is a generic sequence for $(N, P)$. Since $P$ is $\varepsilon$-complete, let $p^{*}$ be an upper bound of the sequence. Now $p^{*} \Vdash$ "$N[G]$ is suitable for $\varepsilon$ and $\chi$ is a generic sequence for $(N[G], Q)$". If additionally $\lambda$ is a cardinal in $V[G]$ then $N[G] \preccurlyeq (H(\lambda))^{V[G]}$. Therefore the $p^{*}$ is as desired.


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