Decide which of the following indicated functions $f$ are homeomorphisms and which are not. Show clearly why or why not.

Question

My partial solution is wrong, can someone please help me with this? I know we need to show bijection and continuous but not sure how.

Let $C$ denote the Cantor set, $\{0,1\}^{\mathbb{N}}$ with the product topology. For $\sigma \in C$, write $\sigma = (\sigma_1, \sigma_2, \dots)$ where $\sigma_i \in \{0, 1\}.$ Decide which of the following indicated functions $f$ are homeomorphisms and which are not. Show clearly why or why not.

(a) $f: C\to C$ given by $f((\sigma_1, \sigma_2,\sigma_3, \dots)) = (1-\sigma_1, 1-\sigma_2, 1-\sigma_3, \dots)$

(b) $f:C\to C\times C$ given by $f((\sigma_1, \sigma_2,\sigma_3, \dots)) = ((\sigma_1, \sigma_3, \sigma_5, \dots),(\sigma_2, \sigma_4, \sigma_6,\dots))$.

$\textbf{Solution:}$ To show homeomorphic, we must show that the function $f$ is injective and surjective. Our Cantor set C, can be described as the set of real numbers in $[0,1]$ having ternary expansions using only the digits $0$ and $2$; real numbers of the form $$\sum_{n=1}^\infty \frac{a_n}{3^n},$$ where each $a_n$ is either $0$ or $2$.

Answer 1

Hint: you have to show that the functions are injective, surjective and that the functions and their inverses are continuous with respect to the relevant topologies. I leave injectivity and surjectivity to you. As for continuity, the open sets in the product topology on the Cantor space $C = \{0, 1\}^{\Bbb{N}}$ are those of the form $A = X \times \{0, 1\}^{\Bbb{N}}$ where, for some $n \in \Bbb{N}$, $X \subseteq \{0, 1\}^n$, i.e., $A$ is the set of sequences $(a_i)$ determined by some finite set of constraints on the $a_i$. Intuitively, this means that a continuous function $f : C \to C$ is a function that only needs to "read" a finite part of its input to determine any given finite part of its output (and similarly for $f : C \to C \times C$ or $f : C \times C \to C$).

Answer 2

Forget the ternary expansion, that is a justification why we can call $C=\{0,1\}^{\Bbb N}$ the Cantor set (which is traditionally defined as a subspace of $[0,1]$, not an infinite product).

A fundamental fact about the product topology:

A function $F:X \to Y=\prod_i Y_i$ is continuous, where $Y$ has the product topology (and projections $\pi_i: Y \to Y_i$) iff $\pi_i \circ F: X \to Y_i$ is continuous for all $i$.

For your map in $a)$ note that $\pi_n \circ f$ is just the map $1-\pi_n$, which is indeed continuous for all $n$, and for $b)$ note that each component map is just itself an infinite "product" of projections, so again continuous (by a second application of the fact). Both maps are clearly bijections and continuity is now clear, so compactness does the rest.