Let $\sqsubset=\bigcup_n \sqsubset_n$ be a relation on $\omega^\omega$ where each $\sqsubset_n$ is arithmetic and $\{f: f\sqsubset_n g\}$ is closed for each $g\in \omega^\omega, n\in \omega, i.e. \Pi_1^0(g)$, a typical example is domination past $n$. In the following fix $\chi $ large enough regular cardinal.

Definition [Goldstern]. $\mathbb{Q}$ almost preserves $\sqsubset$ (or is almost preserving) if for all countable $N\prec H(\chi)$ containing $\mathbb{Q}, p\in \mathbb{Q}\cap N, g\in \omega^\omega$ and $N$ is $\sqsubset$-covered by $g$ (any $f\in N, f\sqsubset g$). There exists $q\leq p$ $(N,\mathbb{Q})$-generic such that $q\Vdash \forall f\in N[G] f\sqsubset g$.

We can check that this notion is preserved under iterations of finite length, more precisely, if $P$ is almost preserving and $1\Vdash_P \dot{Q}$ is almost preserving, then $P*\dot{Q}$ is almost preserving. My question is: is there any known example (references) that this notion is not preserved at limit stages (countable support iteration of proper forcings)?

  • $\begingroup$ Do we require N to be countable? $\endgroup$
    – Zoorado
    Sep 28, 2016 at 10:02
  • 1
    $\begingroup$ Yes N is countable! And I recognize the username ;-) Hi there! $\endgroup$
    – Jing Zhang
    Sep 28, 2016 at 12:50

1 Answer 1


We can do something like setting $f\sqsubset_n g$ if and only if $f(m)=g(n^m)$ for all $m$ (so $f$ looks like $g$ restricted to the powers of $n$).

For each $g$, the set $\{f:f\sqsubset_n g\}$ is a singleton, hence closed. Also, given countably many $f$ we can find a single $g$ covering them all using powers of primes.

Now we use the fact that there are countable support iterations of length $\omega$ where no new reals are added at each finite stage, but new reals are added in the limit.

If $g$ $\sqsubset$-covers $N$ in the ground model and forcing with $P$ adds a new real, then $g$ no longer covers $N$ in the extension.


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