Define a complete embedding of Boolean algebra as an homomorphism of Boolean algebras which preserves also the sup and inf operations. Notice that if $\mathbb{B}$ and $\mathbb{D}$ are complete boolean algebras, $i:\mathbb{B}\to\mathbb{D}$ is a complete embedding and $G$ is $V$-generic for $\mathbb{D}$, then $H=i^{-1}[G]$ is $V$-generic for $\mathbb{B}$.

I'm curious to know if the following can be the case:

Assume $\mathbb{B}$ and $\mathbb{D}$ are complete stationary set preserving boolean algebras. Can there be two distinct complete embeddings $i_0:\mathbb{B}\to\mathbb{D}$, $i_1:\mathbb{B}\to\mathbb{D}$ such that if $G$ is $V$-generic for $\mathbb{D}$ and $H_j=i_j^{-1}[G]$ are the corresponding $V$-generic filters for $\mathbb{B}$ induced by the respective $i_j$, we can have that there is a name $\tau$ in the forcing language for $\mathbb{B}$ such that:

$\|\tau$ is a stationary subset of $\omega_1\|=1_{\mathbb{B}}$

$V[G]\models\sigma_{H_0}(\tau)$ is a stationary subset of $\omega_1$

$V[G]\models\sigma_{H_1}(\tau)$ is non-stationary

  • $\begingroup$ I should say that my interest in this question lies in the fact that a positive answer would provide an example of two substatntially different ways to embed $\mathbb{B}$ into $\mathbb{D}$ as a complete boolean sub-algebra. So it relates to the problem of under which conditions it is possible to have inside $\mathbb{D}$ two distinct complete subalgebras which are isomorphic. $\endgroup$ – matteo viale Jul 22 '11 at 4:06
  • $\begingroup$ Hi Matteo! Nice to see you in MO. $\endgroup$ – Andrés E. Caicedo Jul 22 '11 at 7:02
  • $\begingroup$ What is $\sigma$? Or $\sigma_{H_i}(\tau)$? $\endgroup$ – Stefan Geschke Jul 22 '11 at 7:33
  • $\begingroup$ Matteo, when you say "stationary-set preserving", are you talking just about subsets of $\omega_1$, or do you mean stationary subsets of any cardinal, or generalized stationarity? Stefan, I think that is Matteo's notation for the value of a name by a filter. i.e. what is elsewhere denoted $\text{val}(\tau,H_i)$ or $(\tau)_{H_i}$ or $(\tau)^{H_i}$. $\endgroup$ – Joel David Hamkins Jul 22 '11 at 9:41
  • $\begingroup$ I apologize for my unclear notation, it is exactly meaning what Joel was suggesting..... $\endgroup$ – matteo viale Jul 22 '11 at 16:17

Your situation can happen.

Let $\mathbb{B}=\text{Add}(\omega_1,1)$ be the forcing to add a Cohen subset $S\subset \omega_1$, and let $\mathbb{D}$ be the forcing that first adds such a set $S$, and then shoots a club through it $C\subset S$. Note that $\mathbb{B}$ is countably closed in $V$ and therefore stationary-set preserving, and the generic Cohen set $S$ that is added is both stationary and co-stationary. Further, the forcing $\mathbb{D}$ is stationary-set preserving over $V$, because by a bootstrap argument we may find a dense set of conditions $(s,c)$, where $s\subset \omega_1$ is bounded and $c\subset s\cup\text{sup}(s)$ is closed, and the set of such conditions in $\mathbb{D}$ is countably closed. Thus, $V[S][C]$ is stationary-set-preserving over $V$, even though it is not stationary-set-preserving over $V[S]$.

Notice that we may completely embed $\mathbb{B}$ into $\mathbb{D}$ in the natural way, since $\mathbb{D}$ was described as first adding $S$, and then shooting a club through it.

But we may also embed $\mathbb{B}$ into $\mathbb{D}$ in a different way: by first applying the automorphism of $\mathbb{B}$ that flips all bits. This automorphism in effect replaces $S$ with its complement, so that under this embedding, the club gets added to the complement of $S$.

Thus, if $\tau$ is the name of the generic set $S$ added by $\mathbb{B}$, then $1_{\mathbb{B}}$ forces that $\tau$ is stationary, and with the first embedding we have that $\text{val}(\tau,H_0)=S$, which remains stationary and in fact containing a club in $V[S][C]$, but with the second embedding we have $\text{val}(\tau,H_1)=\omega_1\setminus S$, which is non-stationary in $V[S][C]$.

The two embeddings correspond as you said to the two fundamentally different ways we can think about the Cohen set being treated by the club-shooting forcing, since either we shoot the club through the set, or through its complement, and this difference radically affects the stationarity of this set. But meanwhile, all the ground model stationary sets are preserved, since the composition forcing has a countably closed dense set.

  • $\begingroup$ thanks, this was exactly the kind of answer I was expecting, neat and clear to the point.... $\endgroup$ – matteo viale Jul 22 '11 at 16:16

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