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Are $C$ and $C\times C$ homeomorphic? (Here, $C$ denotes the "first middle third" Cantor set.)

Seems that they are but I can't come up with an idea how to show it.

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    $\begingroup$ Yes. They are. There is a characterization of the Cantor set as the unique-up-to-homeomorphism totally disconnected compact Polish space without isolated points. $\endgroup$
    – Asaf Karagila
    Commented Jan 23, 2017 at 8:43
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    $\begingroup$ Also, if you view $C$ as $2^ω$, it is almost obvious. $\endgroup$
    – user87690
    Commented Jan 23, 2017 at 10:06
  • $\begingroup$ See also math.stackexchange.com/q/3034989. $\endgroup$
    – Paul Frost
    Commented Dec 11, 2018 at 15:03

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Yes the Cantor set $C$ is homeomorphic to $C \times C$. The Wikipedia article on Cantor spaces seems to contain enough information to answer your question. Quoting parts of the article below:

[A] topological space is a Cantor space if it is homeomorphic to the Cantor set.

. . .

[T]he canonical example of a Cantor space is the countably infinite topological product of the discrete $2$-point space $\{0, 1\}$. This is usually written as $2^\mathbb{N}$ or $2^\omega$ (where $2$ denotes the $2$-element set $\{0,1\}$ with the discrete topology).

. . .

[M]any properties of Cantor spaces can be established using $2^\omega$, because its construction as a product makes it amenable to analysis.

Cantor spaces have the following properties:

  • . . .
  • The product of two (or even any finite or countable number of) Cantor spaces is a Cantor space.
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