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made topological reformulation explicit
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YCor
YCor
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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 $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ 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 $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ 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.

Edit (YCor Nov 2019): topological reformulation: let $X,Y$ be Hausdorff compact, totally disconnected spaces (the OP is mostly interested in the case when $X,Y$ are metrizable). Suppose that for every $n$, there are pairwise disjoint clopen subsets $Y_1,\dots,Y_n$ of $Y$, each homeomorphic to $X$. Does it follow that there exists a sequence $(Y_i)$ of pairwise disjoint clopen subsets of $Y$, each homeomorphic to $X$?

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 $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ 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 $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ 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.

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 $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ 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 $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ 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.

Edit (YCor Nov 2019): topological reformulation: let $X,Y$ be Hausdorff compact, totally disconnected spaces (the OP is mostly interested in the case when $X,Y$ are metrizable). Suppose that for every $n$, there are pairwise disjoint clopen subsets $Y_1,\dots,Y_n$ of $Y$, each homeomorphic to $X$. Does it follow that there exists a sequence $(Y_i)$ of pairwise disjoint clopen subsets of $Y$, each homeomorphic to $X$?
fixed notation, added gn tag because of Stone duality interpretation
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YCor
YCor
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Counting Copiescopies of a BA within a BA: Arbitrarily Many versus Infinitely Manyarbitrarily many vs infinitely many

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$$\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ for $1 \leq i \leq n$ and $b_i \wedge b_j = 0$ if $i \neq j$. Is there necessarily a sequence $\{ b_i \}_{i \in \omega} \in B$ with $\mathcal{A} \cong b_i$$\mathcal{A} \cong \mathcal{B}\upharpoonright 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.

Counting Copies of a BA within a BA: Arbitrarily Many versus Infinitely Many

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.

Counting copies of a BA within a BA: arbitrarily many vs infinitely many

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 $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ 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 $\mathcal{A} \cong \mathcal{B}\upharpoonright b_i$ 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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Asher M. Kach
Asher M. Kach
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Counting Copies of a BA within a BA: Arbitrarily Many versus Infinitely Many

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.