Recall the definition of the Cantor set. We start with $Y_0 := [0, 1]$ (with the Euclidean topology). Let $Y_1$ be obtained from $Y_0$ by removing the open interval $(1/3,2/3)$. Let $Y_2$ be obtained from $Y_1$ by removing the open intervals $(1/9,2/9)$ and $(7/9,8/9)$. Recursively, let $Y_n$ be obtained from $Y_{n−1}$ by removing the open middle third of each of the $2^{n−1}$ components of $Y_{n−1}$. The Cantor set is the intersection $Y := \cap_{n=0}^∞ Y_n$.
$(a)$ Use compactness to argue that $Y$ is non-empty.
$(b)$ Show that $Y$ is homeomorphic to the space $X :=$ {$0, 1$}$^{\mathbb N}$ with the product topology.
Hint. Show that $ϕ : X → Y$ defined by $ϕ(x) := \sum_{n=1}^∞(2/3)^nx_n$ is a homeomorphism. You can understand $ϕ$ by thinking of $x$ as an “address” for $ϕ(x)$.
$(c)$ The above representation resembles the binary representation of the points in $[0, 1]$, where a point $z ∈ [0, 1]$ is represented by its binary expansion $0.z1z2z3 · · · $, so that $z_k ∈$ {$0, 1$} and $z = \sum_{n=1}^∞(1/2)^nz_n$. Explain why a similar argument as in the previous part would not show that $[0, 1]$ is homeomorphic to $X$.
I already proved parts $(a)$ and $(b)$. But how to explain part $(c)$ as similar arguments to part $(b)$?
In part $(b)$, I showed that $ϕ$ is one-to-one, i.e. if $x=y \in X$, then $ϕ(x)=ϕ(y)$. I also showed that $ϕ$ is onto, i.e., if $y \in Y$, and $y=\sum_{n=1}^∞(2/3)^ny_n$, then $ϕ(\sum_{n=1}^∞(2/3)^ny_n = y$. Which means that we have a unique pre-image for every element of $Y$ in $X$. Moreover, $ϕ$ is continuous, cause from part $(a)$ we see that each point of Cantor set is a limit point. And also, we know thag Cantor set is uncountable. Finally, $ϕ^{-1}$ is continuous since Cantor set is closed and bounded, and so inverse of $ϕ$ is closed and bounded and hence contains all of its limit points.
So how to show by a similar argument that if $z = \sum_{n=1}^∞(1/2)^nz_n$, then $[0, 1]$ is not homeomorphic to $X$. (part $(c)$).
The map $(z_1, z_2, \dots) \mapsto \sum_{n=1}^{\infty} 2^{-n} z_n$ is continuous but it's not injective, so it can't be a homeomorphism. For example the sequences $(1,0,0,0\dots)$ and $(0,1,1,1,\dots)$ both map to $1/2$. Of course this only proves that this particular map isn't a homeomorphism, not that $[0,1]$ and the Cantor set aren't homeomorphic in general. But to prove they can't be homeomorphic at all, one only needs to observe (for example) that $[0,1]$ is connected and the Cantor set is not.