# Are $C$ and $C\times C$ homeomorphic?

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.

• 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. Commented Jan 23, 2017 at 8:43
• Also, if you view $C$ as $2^ω$, it is almost obvious. Commented Jan 23, 2017 at 10:06

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.