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Is there an uncountable collection of compact disjoint subsets of the real line such that each element of the collection is uncountable?

Thanks.

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    $\begingroup$ It wouldn't surprise me if you could fit an uncountable disjoint collection of Cantor sets into $\Bbb R$, though it might require the axiom of choice. $\endgroup$ – Arthur Jan 23 '14 at 21:55
  • $\begingroup$ @Arthur: In fact, given any Cantor set $C$ ($C$ is a nonempty perfect and nowhere dense subset of $\mathbb R$), you can fit a disjoint collection of cardinality $c = 2^{{\aleph}_{0}}$ of Cantor sets into $C.$ (Use the Hahn-Mazurkiewicz theorem to obtain a continuous map from $C$ onto $[0,1] \times [0,1],$ then look at the inverse images of $\{r\} \times [0,1]$ as $r$ varies over $[0,1].$) (moments later ... opps, that's what Asaf Karagila posted) $\endgroup$ – Dave L. Renfro Jan 23 '14 at 22:46
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There exists a continuous surjection from $[0,1]$ onto $[0,1]\times[0,1]$. For example, the Peano curve. Pick such function and denote it by $f$.

For every $r\in[0,1]$ consider $A_r=f^{-1}(\{r\}\times[0,1])$. This is the preimage of a closed set under a continuous function, therefore $A_r$ is closed, and it is a subset of $[0,1]$ so it is bounded. Therefore it compact.

Finally, $[0,1]$ is the union of all $A_r$'s and there are $2^{\aleph_0}$ of them, each can be mapped onto a set of size continuum, and therefore is uncountable.

So $\{A_r\mid r\in[0,1]\}$ is an uncountable collection of pairwise disjoint, uncountable compact sets.

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Yes. For $t=(t_1,t_2,\dots)\in\{1,2\}^\mathbb N$, let $A_t$ be the set of all real numbers whose continued fraction is $$ \frac1{a_1+\frac1{t_1+\frac1{a_2+\frac1{t_2+\frac1{\dots}}}}} $$ where each $a_i$ is $3$ or $4$. This is a Cantor set, and the $A_t$ are pairwise disjoint.

Of course, that I used only the numbers $1$-$4$ in the continued fractions is irrelevant. To illustrate the flexibility of the method, see here for a related result.

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