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Informally, I am wondering if a Boolean algebra $\mathcal{B}$ contains infinitely many disjoint copies of a Boolean algebra $\mathcal{A}$ whenever it contains arbitrarily many disjoint copies of $\mathcal{A}$.

More formally, fix Boolean algebras $\mathcal{A}$ and $\mathcal{B}$. Assume that, for each $n \in \omega$, there are $b_1,\dots,b_n \in B$ with $\mathcal{A} \cong b_i$ for $1 \leq i \leq n$ and $b_i \wedge b_j = 0$ if $i \neq j$. Is there necessarily a sequence $\{ b_i \}_{i \in \omega} \in B$ with $\mathcal{A} \cong b_i$ for all $i \in \omega$ and $b_i \wedge b_j = 0$ if $i \neq j$?

Though I am curious about the question in the general setting, my primary interest is when $\mathcal{A}$ and $\mathcal{B}$ are both countable.

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Do I assume correctly that BA stand for Boolean algebra? And when you write $\mathcal A\cong b_i$, do you mean $\mathcal A\cong\mathcal A\restriction b_i$ where $\mathcal A\restriction b_i$ is the BA of all elements of $\mathcal A$ that are $\leq b_i$? – Stefan Geschke Sep 28 '11 at 21:13
Stefan, I think you should refer instead to ${\cal B}\upharpoonright b_i$. – Joel David Hamkins Sep 29 '11 at 1:50
Yes, Joel, I meant $\mathcal B\restriction b_i$. – Stefan Geschke Sep 29 '11 at 5:26
Yes, I am using BA for Boolean algebra and $\mathcal{A} \cong b_i$ for $\mathcal{A} \cong \mathcal{B} \upharpoonright b_i$. – Asher M. Kach Sep 29 '11 at 10:51

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