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Does there exists an uncountable family $\mathcal{A}$ consisting of uncountable compact subsets of $\mathbb{R}$ and pairwise disjoint?

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Let $K$ be the set of reals in $[0,1]$ with only $0$s and $1$s in their decimal expansions. For each $x \in K \setminus\{1\}$ the set $2x + K$ is compact, and this family is pairwise disjoint (since $x$ is recoverable from any element of $2x + K$).

Edit to make the last claim more explicit: suppose $y \in (2x_0 + K) \cap (2x_1 + K)$. The $n$th digit of the decimal expansion of $y$ is at least $2$ iff the $n$th digit of the decimal expansion of $x_0$ is $1$ iff the $n$th digit of the decimal expansion of $x_1$ is $1$. Since $x_0$ and $x_1$ have only $0$s and $1$s in their expansions, this forces them to be equal.

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Let $C$ be any Cantor set in the line. $C \times C$ is again a Cantor set, so let $h:C \times C \cong C$ be a homeomorphism. For each $x \in C$ let $C_x = \{x\} \times C$; clearly each $C_x$ is yet another Cantor set, and $\{h[C_x]:x \in C\}$ is a partition of $C$ into $2^\omega$ Cantor sets.

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