# subalgebra of a simple forcing

Let $\alpha > 0$ be any ordinal. Consider the forcing $Add(\omega,\alpha) * Add(\omega_1,1)$. Let $B$ be its boolean completion. Let $\dot{X}$ be the canonical $B$-name for the generic subset of $\omega_1$ added by the second iterand. By forcing folklore, there is a complete subalgebra $A \subseteq B$ that adds $X$, completely generated by the boolean values $|| \check{\beta} \in \dot{X} ||$ (see Jech p. 247). One can check that for each $\beta < \omega_1$, $|| \check{\beta} \in \dot{X} || = (1,\check{ \lbrace (\beta,1) \rbrace } )$.

Questions:

(1) What is the nature of $A$ in relation to $B$? Although $A$ adds all the reals of $B$, I think I can show that it is always a proper subset of $B$ (even for $\alpha = 1$). This leads to the second question.

(2) If $G$ is generic for $A$, what is the nature of the quotient algebra $B/G$? One can show that it is $(\omega,\infty)$-distributive. In the case where $\alpha$ is countable, it is atomic. Is it nonatomic for uncountable $\alpha$? Is it strategically closed?

(3) If $G$ is generic for $A$, does $V[G]$ always contain some $Add(\omega,\alpha)$ generic, like in the case $\alpha$ is countable?

Edit: I thank J.D. Hamkins for his answer, but I already knew about the case $\alpha < \omega_1$, as stated in the original post. I would really like some insight on the case $\alpha \geq \omega_1$.

NEW RELATED QUESTION

Consider instead $Col(\mu,<\kappa)*Add(\kappa,1)$, where $\mu< \kappa$ are regular. Let $B$ be the completion, and let $A$ be the complete subalgebra generated by the canonical name for the Cohen subset of $\kappa$ added by the second part. Let $H \subseteq A$ be generic.

Does $B/H$ have a $\mu$-closed dense subset in $V[H]$? Any proof or refutation would be great, or a discussion of other related structural aspects of $A,B$, and $B/H$. The properties of this forcing in relation to large cardinals are important to my research, so all nontrivial information is welcome.

• I strongly suspect you mean that $\Vert \check{\beta} \in \dot{X} \Vert = \{ (p,\dot{q}) \in Add(\omega,\alpha)\ast Add(\omega_1,1): p \Vdash (\beta, 1) \in \dot{q} \}$ – Not Mike Apr 2 '13 at 1:53
• (Where the completion of $Add(\omega,\alpha)\ast Add(\omega_1,1)$ is taken to be corresponding regular open algebra) – Not Mike Apr 2 '13 at 1:55
• Michael, let $e: Add(\omega,\alpha)∗Add(\omega_1,1) \to B$ be the canonical dense embedding. If $(p,\dot{q}) \Vdash \beta \in X$, then it must be that $p \Vdash \dot{q}(\beta)=1$. Thus $(p,\dot{q}) \leq (1,\lbrace (\beta,1) \rbrace )$. And clearly, $(1,\lbrace (\beta,1) \rbrace )$ forces $\beta \in X$. Therefore, $|| \beta \in X|| =e((1,\lbrace (\beta,1) \rbrace ))$. – Monroe Eskew Apr 3 '13 at 0:29
• While the new question is interesting, I find the phrasing to be bad. "Prove or disprove and/or discuss related structural aspects" sounds as if you are assigning us homework or a seminar lecture topic. You should instead be asking a question. – Asaf Karagila Nov 20 '13 at 20:55
• Sorry to offend, let me rephrase.......... – Monroe Eskew Nov 20 '13 at 21:00

Update. If $\alpha$ is countable, then I claim that $A$ and $B$ are forcing equivalent, the quotient forcing is atomic, isomorphic to $P(\omega_1)$, and every extension by $A$ adds a generic for $\text{Add}(\omega,\alpha)$.

Suppose that $\alpha$ is countable and we have $V[g][X]$, the extension by $B=\text{Add}(\omega,\alpha)\ast\text{Add}(\omega_1,1)$. Thus, $g$ is an $\alpha$ sequence of Cohen reals, and we may string them together end-to-end and make a binary sequence of length $\omega\cdot\alpha$. In the model $V[g]$, it is dense that this sequence appears explicitly as a block in the generic set $X$, since any condition in $V[g]$ can be extended to include it. Thus, it follows that $g\in V[X]$ and at least in this situation, the set $X$ explicitly gives us a $V$-generic $\alpha$-sequence of Cohen reals, namely $g$.

But more to the point, it follows that $V[g][X]=V[X]$. Thus, forcing with $A$ or $B$ gives rise to the same extensions, and so $A$ and $B$ are forcing equivalent. As Andreas pointed out in a comment to my earlier answer (and as the OP seems to be aware), this does not necessarily mean that $A=B$, although it does mean that the quotient forcing is atomic.

Note that every ordinal $\beta\lt\omega_1$ has a nonzero Boolean possibility in $B$ of being the first ordinal where $g$ appears as a block in $X$. Thus, we have the Boolean value $b_\beta$ which is the Boolean value in $B$ that this is the case. It follows that $\{b_\beta\mid\beta\lt\omega_1\}$ forms a maximal antichain in $B$. Furthermore, I claim that $A$ together with the $b_\beta$'s generate all of $B$, because if we know the values in $A$ and we also know which $b_\beta$ holds then we can compute any Boolean value in $B$, which is determined by the information about $g$ and the information about $X$. Finally, notice that the OP's automorphism argument in the comments shows that no $b_\beta$ is in $A$, since we can know $X$ fully and still not know $g$ exactly, since we might have had $\pi(g)$ instead. So no information about $X$ (which is all $A$ knows about) can tell us anything definite about where the block of $g$ starts.

Thus, the quotient forcing is the atomic forcing using the atoms $b_\alpha$, which is isomorphic to the power set $P(\omega_1)$.

Lastly, let me just mention---although you probably know this already---that when $\alpha$ is countable, then $\text{Add}(\omega,\alpha)$ is isomorphic to $\text{Add}(\omega,1)$.

• I don't see the inference from "the quotient forcing is trivial" to "$A=B$". A trivial forcing can be a big atomic Boolean algebra. Note that mbsq already mentioned in the question that, in the countable case, the quotient algebra is atomic, so the problem is whether this atomic algebra is just 2. – Andreas Blass Feb 16 '13 at 21:25
• Joel, doesn't your argument work even when $\alpha$ is uncountable? That is, every real in $V[g]$ will appear as a block in $X$, regardless of the size of $\mathbb{R}$. – jonasreitz Feb 16 '13 at 21:38
• ...I guess this just shows that $V[g][X]$ and $V[X]$ have the same reals - I'm not sure it follows that $g \in V[X]$. – jonasreitz Feb 16 '13 at 21:42
• I agree with Andreas. I think this gives an example of an atomic but not =2 quotient. For suppose $B = A$. Let $C \subseteq B$ be the complete subalgebra equivalent to $Add(\omega,\alpha)$. Now let $G \subseteq B$ be generic. Then we trivially compute $G \cap C$ as $(G \cap A) \cap C$. However, let $\pi$ any nontrivial automorphism of $C$. Let $H$ be $C$-generic, and let $X$ be $Add(\omega_1,1)$-generic over $V[H]$, and let $G \subseteq B$ be the corresponding generic.... – Monroe Eskew Feb 16 '13 at 21:55
• Then $V[G] = V[H][X] = V[\pi H][X]$. Furthermore, $G \cap A$ is fully determined by $X$, so our computation would yield $G \cap A \cap C = H$, but also $G \cap A \cap C = \pi H$, contradiction. – Monroe Eskew Feb 16 '13 at 21:56

I have answers the original questions, by way of counterexample, which I will sketch. The "new related question" remains unsolved. I am grateful to Mohammad Golshani for his answer to this question, which I realized helps to answer the present question.

Assume CH holds in the ground model $V$. Let $B_0$ be the completion of $Col(\omega,<\omega_2) * Add(\omega_2^V,1)$. Let $\mathbb{P}$ be the countable support product of $Col(\omega,\alpha)$ for $\alpha < \omega_2$, and let $B_1$ be the completion of $\mathbb{P} * Add(\omega_2^V,1)$.

Let $A_0 \subseteq B_0$ be the complete subalgebra generated by the name for the Cohen subset of $\omega_2^V$ from the second iterand, and let $A_1 \subseteq B_1$ be the same with respect to $B_1$. The following is a special case of a more general lemma that is part of my work in progress, to be submitted soon. (CH is used)

Claim: $A_0 \cong A_1$, and $X \subseteq \omega_2^V$ is generic for $A_0$ iff it is generic for $A_1$.

Now, Mohammad Golshani gives examples of two $\Sigma_1$ properties, $\varphi_0, \varphi_1$, in parameters from $V$, such that whenever $G_0 \subseteq Col(\omega,<\omega_2)$ is generic, and $G_1 \subseteq \mathbb{P}$ is generic, $V[G_0] \models \varphi_0 \wedge \neg \varphi_1$, and $V[G_1] \models \varphi_1 \wedge \neg \varphi_0$. By the Claim, if $X \subseteq \omega_2^V$ is generic for $A_0$ (and $A_1$), then we necessarily have $V[X] \models \neg \varphi_0 \wedge \neg \varphi_1$.

Now we note that $Col(\omega,<\omega_2)$ is equivalent to $Col(\omega,\omega_1) \times Add(\omega,\omega_2^V)$. Let $g: \omega \to \omega_1^V$ be generic for the left side. Then $B_0 / g$ is equivalent to $Add(\omega,\omega_1^{V[g]}) * Add(\omega_1^{V[g]},1)$, like in the original question. For any $G_0 * X$ generic for $B_0$ extending $g$, we have $g \in V[X]$, and $V[X]$ contains no filter over $Add(\omega,\omega_2^V)$ which is generic over $V[g]$, using Mohammad's lemma and the above factoring. This answers question (3).

It also answers question (2). The quotient in this case is indeed atomless, and it is not $<\omega_1$-strategically closed. If it were, then we could construct an $Add(\omega,\omega_1^{V[g]})$-generic (over $V[g]$) filter, by recursively deciding initial segments. This also sheds some light on question (1), although that is a broad and vague question.