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A few hours ago a user posted a link to this pdf:

There was a discussion about Proposition 3.2.8. I read it, and near the end, there is a map given $$ \bigcap_{i_1,\dots,i_n,\dots\in\{0,1\}}X_{i_1,\dots,i_n,\dots}\mapsto (i_1,\dots,i_n,\dots). $$ And it says this is a homeomorphism. Is there an more explicit explanation why it's a homeomorphism?

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just draw the pictures! Note that the sequence tells you which direction to go in the construction in the cantor set to find your point. – Alexander Thumm Sep 29 '11 at 6:04
up vote 1 down vote accepted

If you examine the construction of $C$, you’ll see that each set $Y_{i_1,\dots,i_n}$ is the closure of a certain open ball; to simplify the notation, let $B_{i_1,\dots,i_n}$ be that open ball. The map in question is a bijection that takes $B_{i_1,\dots,i_n}\cap C$ to $$\{(j_1,j_2,\dots)\in\{0,1\}^{\mathbb{Z}^+}: j_1=i_1, j_2=i_2,\dots,j_n=i_n\},$$ which is a basic open set in the product $\{0,1\}^{\mathbb{Z}^+}$.

Every open subset of $C$ is a union of sets of the form $B_{i_1,\dots,i_n}\cap C$, so the map is open. Every open set in the product $\{0,1\}^{\mathbb{Z}^+}$ is a union of sets of the form $$\{(j_1,j_2,\dots)\in\{0,1\}^{\mathbb{Z}^+}: j_1=i_1, j_2=i_2,\dots,j_n=i_n\},$$ so the map is continuous. Finally, a continuous, open bijection is a homeomorphism.

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This makes it more clear. Thanks. – Wickham Sep 29 '11 at 16:14

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