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](https://projecteuclid.org/ebooks/perspectives-in-logic/Proper-and-Improper-Forcing/toc/pl/1235419814)


Much thanks in advance!



Why does an $\varepsilon$-incomplete $Q$ remain so in the forcing extension of any $\varepsilon$-complete $P$?
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. 


