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 $X$?

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    $\begingroup$ 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 $\leq b_i$? $\endgroup$ – Stefan Geschke Sep 28 '11 at 21:13
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    $\begingroup$ Stefan, I think you should refer instead to ${\cal B}\upharpoonright b_i$. $\endgroup$ – Joel David Hamkins Sep 29 '11 at 1:50
  • $\begingroup$ Yes, Joel, I meant $\mathcal B\restriction b_i$. $\endgroup$ – Stefan Geschke Sep 29 '11 at 5:26
  • $\begingroup$ Yes, I am using BA for Boolean algebra and $\mathcal{A} \cong b_i$ for $\mathcal{A} \cong \mathcal{B} \upharpoonright b_i$. $\endgroup$ – Asher M. Kach Sep 29 '11 at 10:51

This is a nice question from a while ago, and this is only a very partial answer, hoping it will motivate more complete ones.

I'll address the topological equivalent reformulation. It has a positive answer in a few cases described below. First, say that $X$ is good if the question has a positive answer for this given $X$, for every $Y$.

1) If $X$ has a clopen subset homeomorphic to $X\sqcup X$ then $X$ is good. (In particular the answer to the question is positive when $Y=X$). Indeed, in this case, define by induction pairwise disjoint clopen subsets $Y_1,\dots,Y_{n-1},Y'_{n}$ of $Y$ each homeomorphic to $X$, and then choose two disjoint clopen subsets $Y_n,Y'_{n+1}$ of $Y'_n$, each homeomorphic to $X$.

Similarly, the answer to the question is positive if $Y$ has a clopen subset homeomorphic to $Y\sqcup Y$ (or to $Y\sqcup X$).

2) If $X$ embeds as a clopen subset in a finite disjoint union $nX'$ and vice versa, then $X$ is good iff $X'$ is good.

3) If $X$ is countable then $X$ is good. Indeed, the case $X$ empty being trivial, suppose $X$ nonempty and let $\alpha=\alpha_X$ be the largest Cantor-Bendixson rank of a point in $X$; it is achieved by $n$ points, where $n=n_X$ is a positive integer, the pair $(\alpha,n)$ characterizing $X=X(\alpha,n)$ up to homeomorphism. Note that $X$ is then homeomorphic to the disjoint union of $n$ copies of $X(\alpha,1)$, so, in view of (2), let us focus on the case $n_X=1$.

So the condition that $Y$ has $k$ disjoint clopen copies of $X$ means that $Y$ has $\ge k$ points with a countable neighborhood and of Cantor-Bendixson rank $\alpha$. Hence the condition that it has this for arbitrary large $k$ means that $Y$ has infinitely many points with a countable neighborhood and of Cantor-Bendixson rank $\alpha$. Let $(x_n)$ be an injective sequence of such points. By induction let $Y_n$ be a countable clopen neighborhood of $x_n$ disjoint from $\bigcup_{i<n}V_i$, and with the additional condition that every point in $Y_n\smallsetminus\{x_n\}$ has Cantor-Bendixson rank $<\alpha$ (in particular, $x_i\notin Y_n$ for every $i>n$). Then $Y_n$ is homeomorphic to $X$ for each $n$, and these are pairwise disjoint clopen subsets.

4) Call $X$-point a point in a topological space a point with a basis of neighborhoods homeomorphic to $X$. If $X$ possesses an $X$-point that is isolated among $X$-points, then then $X$ is good (call this isolated $X$-point). This is actually a generalization of (3) since countable spaces $X$ with $n_X=1$ have this property (the point of maximal Cantor-Bendixson rank being an isolated $X$-point). The argument follows that of (3). An example not covered by the previous items is the case of the space obtained as union of a Cantor space and a discrete countable set accumulating at a single point of this Cantor space.

5) I don't know if $X,X'$ good imply $X\sqcup X'$ good; however this is clear if $X,X'$ have no nonempty homeomorphic subsets. This applies to $X$ Cantor and $X'$ countable.

Actually the few metrizable Stone spaces I can think of, so far, seem to satisfy close variants of the latter arguments. For the classification of metrizable Stone spaces, see notably

Reichbach, M. The power of topological types of some classes of 0-dimensional sets. Proc. Amer. Math. Soc. 13 1962 17-23 (Open link).


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