Cantor Space as Countably Infinite Product
Theorem
Let $A_n = \struct {\set {0, 2}, \tau_n}$ be the discrete space of the two points $0$ and $2$.
Let $\ds A = \prod_{n \mathop = 1}^\infty A_n$.
Let $\struct {A, \tau}$ be the product space where $\tau$ is the product topology on $A$.
Then $A$ is homeomorphic to the Cantor space.
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Proof 1
In $\CC$, basis elements are sets of the form $\set {y: \size {x - y} < \epsilon}$ for $x \in \CC$ and some $\epsilon \in \R_{>0}$.
In $\ds \prod_{n \mathop = 1}^\infty A_n$, sets of the form $\set {\sequence {a_i} \in \prod A_n: a_i \text { is fixed for } 1 \le i \le n}$ forms a basis for the product topology.
Consider the function $f$ taking each point from $\tuple {a_1, a_2, \ldots, a_n, \ldots}$ in $\prod A_n$ to the point $0 \cdotp a_1 a_2 \ldots a_n \ldots_3$ in $\CC$.
As $f$ takes basis elements to basis elements, both $f$ and $f^{-1}$ are seen to be continuous.
$\blacksquare$
Proof 2
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Since $\CC$ is a metric space, $\CC$ is Hausdorff. By Tychonoff's Theorem, $\ds A = \prod_{n \mathop = 1}^\infty A_n$ is compact.
Consider the function $f$ taking each point from $\tuple {a_1, a_2, \ldots, a_n, \ldots}$ in $\prod A_n$ to the point $0 \cdotp a_1 a_2 \ldots a_n \ldots_3$ in $\CC$.
$f$ is a bijection.
In $\CC$, the basis elements are sets of the form $\ds B_{N,x} = \set {y \in \CC: \size {x - y} < 3^{-N}}$, where $N \in \N_{>0}$ and $x \in \CC$.
Let $\ds x=\sum_{j \mathop = 1}^\infty a_j 3^{-j}$
where $\forall j: a_j \in \set{0,2}$
$\ds f^{-1} \sqbrk {B_{N,x}} = \set{\tuple {b_1, b_2, \ldots, b_n, \ldots} \in A: \size {\sum_{j \mathop = 1}^\infty (a_j-b_j) 3^{-j}} < 3^{-N} }$
The equation $\ds \size {\sum_{j \mathop = 1}^\infty (a_j-b_j) 3^{-j}} < 3^{-N}$, where $\forall j: a_j, b_j \in \set{0,2}$ holds true if and only if $a_j = b_j$ for all $j \in \set{1,2, \ldots, N}$ and it is not the case $a_j-b_j = 2$ for all $ j >N$ and it is not the case $b_j-a_j =2$ for all $ j >N$.
If $\ds x=\sum_{j \mathop = 1}^N a_j 3^{-j}$,
$\ds f^{-1} \sqbrk {B_{N,x}} = \set{\tuple {b_1, b_2, \ldots,b_N, b_{N+1}, \ldots} \in A: a_j = b_j, \forall j \in \set{1,2,...,N} } \setminus \set{\tuple {a_1, a_2, \ldots, a_N, 2,2,2, \ldots} }$
If $\ds x=3^{-N} + \sum_{j \mathop = 1}^N a_j 3^{-j}$,
$\ds f^{-1} \sqbrk {B_{N,x}} = \set{\tuple {b_1, b_2, \ldots,b_N, b_{N+1}, \ldots} \in A: a_j = b_j, \forall j \in \set{1,2,...,N} } \setminus \set{\tuple {a_1, a_2, \ldots, a_N, 0,0,0, \ldots} }$
Else,
$\ds f^{-1} \sqbrk {B_{N,x}} = \set{\tuple {b_1, b_2, \ldots,b_N, b_{N+1}, \ldots} \in A: a_j = b_j, \forall j \in \set{1,2,...,N} } $
In all cases, $f^{-1} \sqbrk {B_{N,x}}$ is open since Product of Hausdorff Factor Spaces is Hausdorff, singleton sets are closed.
Thus $f$ is continuous.
Since $f$ is continuous, it follows from Continuous Bijection from Compact to Hausdorff is Homeomorphism that $f$ is a homeomorphism.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $8$