Cantor space and cantor set I am trying to show that that the the cantor set is homeomorphic to cantor space and here is the function defined from $C$ cantor set to $\mathcal{C}$ Cantor space, let $a_{n}$ be $0$ or $2$ And for all $x\in C$, $x=\sum_{i=1}^{\infty}\frac{a_{n}}{3^{n}}$, the function $f(x)=(\frac{a_{1}}{2},\frac{a_{2}}{2},...)$
it is easy to check that $f$ is bijective, but How can I show that $f$ is continuous ?
If I can show that’s $f$ is continuos, $f^{-1}$ is easily continuos because the function is Fromm compact space to Hausdorff space.
Any help about the continuity will be greatly appreciated
 A: The Cantor space $\mathcal{C}$ has a basis of open sets $O_{IJ}$ parametrised by pairs $I$ and $J$ of finite sets of indexes. $O_{IJ}$ comprises all sequences $c$ of $0$s and $1$s satisfying the constraints $c_i = 0$ and $c_j = 1$ for $i \in I$ and $j \in J$. To show $f$ is continuous, it is sufficient to show that each $f^{-1}[O_{IJ}]$ is open in the Cantor set $C$. To see this, let $N = \max(I \cup J)$ and assume  $x=\sum_{n=1}^{\infty}\frac{a_{n}}{3^{n}} \in f^{-1}[O_{IJ}]$,  There are now two cases depending on whether $z = \sum_{n=N+1}^{\infty}\frac{a_{n}}{3^{n}} = 0$. If $z = 0$, let $X = C \cap (x - \frac{1}{3^{N+1}}, x + \frac{1}{3^{N+1}})$ while, if $z \neq 0$, let $X = C \cap (\sum_{n=1}^{N}\frac{a_{n}}{3^{n}}, \sum_{n=1}^{N}\frac{a_{n}}{3^{n}} + \frac{2}{3^{N+1}})$. In either case, if $y = \sum_{n=1}^{\infty}\frac{b_{n}}{3^{n}} \in X$, then $b_n = a_n$ for each $n \le N$, implying that $y \in f^{-1}[O_{IJ}]$. So $X$ is an open subset of $C$ contained in $f^{-1}[O_{IJ}]$. This implies that $f^{-1}[O_{IJ}]$ is open, which is what we wanted.
Note: in the case when $z = 0$, $X = C \cap (x - \frac{1}{3^{N+1}}, x + \frac{1}{3^{N+1}}) = C \cap [x, x + \frac{1}{3^{N+1}})$ because $C \cap (x - \frac{1}{3^{N+1}}, x) = \emptyset$. This is the case when $x$ is either $0$ or the right-hand upper bound of one of the open intervals removed from $[0, 1]$ to form $C$.
