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Is it possible to embed uncountably many copies of the Cantor set in the unit interval so that any two are disjoint?

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    $\begingroup$ The answer is yes: if $C$ is the Cantor set, then $C\times C$ is made up of uncountably copies of $C$, and is homeomorphic to $C$. So $C\times C$ embeds into the unit interval. $\endgroup$
    – YCor
    Jun 22, 2015 at 13:25
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    $\begingroup$ I'm voting to close this question as off-topic because answered in the comments. $\endgroup$ Jun 22, 2015 at 14:14
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    $\begingroup$ @YCor Recent discussion on meta suggests that migration to MSE is not always optimal thing to do. Link: Migrating to Math.SE: too many close reasons $\endgroup$ Jun 22, 2015 at 15:56
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    $\begingroup$ @YCor to emphasis the above comment, this question is not suitable for Mathematics as is, as it lacks context. It might well be closed and in any case is not a good on-topic question there. $\endgroup$
    – user9072
    Jun 22, 2015 at 19:18
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    $\begingroup$ @MartinSleziak I'm aware that the choice of migrating to MathSE is nontrivial but I considered this post as a reasonable math question, correctly written, and not stupid (well, the product trick makes the answer trivial but it's understandable not to find it if one's not familiar to the abstract definitions of Cantor set), and definitely not of research level. Anyway if the opinion of the other who closed is different, I'm fine with this. $\endgroup$
    – YCor
    Jun 23, 2015 at 0:11

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Yes. The Cantor set is homeomorphic to the space $C=\{0,1\}^{\mathbb N}$, where $\{0,1\}$ has the discrete topology and the product gets the product topology. Now $\{0,1\}^{\mathbb N}$ is homeomorphic to the square $C^2$. Hence $C^2$ embeds into the unit interval. Clearly, $C^2$ is the disjoint union of uncountably many Cantor sets.

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