1
$\begingroup$

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?

  • 0
    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.2011-09-29

1 Answers 1

2

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.

  • 0
    This makes it more clear. Thanks.2011-09-29