Cantor Space as Countably Infinite Product

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Theorem

Let $A_n = \left({\left\{{0, 2}\right\}, \tau_n}\right)$ be the discrete space of the two points $0$ and $2$.

Let $\displaystyle A = \prod_{n \mathop = 1}^\infty A_n$.

Let $\left({A, \tau}\right)$ be the product space where $\tau$ is the Tychonoff topology on $A$.


Then $A$ is homeomorphic to the Cantor space.


Proof

In $\mathcal C$, basis elements are sets of the form $\left\{{y: \left|{x - y}\right| < \epsilon}\right\}$ for $x \in \mathcal C$ and some $\epsilon \in \R_{>0}$.

In $\displaystyle \prod_{n \mathop = 1}^\infty A_n$, sets of the form $\left\{{\left\langle{a_i}\right\rangle \in \prod A_n: a_i \text { is fixed for } 1 \le i \le n}\right\}$ forms a basis for the Tychonoff topology.

Consider the function $f$ taking each point from $\left({a_1, a_2, \ldots, a_n, \ldots}\right)$ in $\prod A_n$ to the point $0.a_1 a_2 \ldots a_n \ldots_3$ in $\mathcal C$.

As $f$ takes basis elements to basis elements, both $f$ and $f^{-1}$ are seen to be continuous.

$\blacksquare$


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