Map from binary sequences on $\{0,1\}$ into the Cantor set $C$ respects order I am aware of the bijection between binary sequences and the Cantor set $C$ which is used to prove that it is an uncountable set. In fact, there is a website that the binary sequences can be given an order topology based on "lexicographic order." My question is how do you actually verify that the map from the sequences into $C$ respects order? Once this can be verified, it follows that $2^{\omega}$ and $C$ are in fact homeomorphic. Wouldn't this necessarily make our map increasing?
Math Reference
 A: The usual bijection between the set of binary sequences and the Cantor set is the following: given a binary sequence $(a_n)$ ($n\geq 1$), we map
$$a_n \longmapsto \sum_{n=1}^{\infty}\frac{2a_n}{3^n}.$$
Call this map $\Psi$.
Now, suppose that $(a_n)\lt (b_n)$, and let $k$ be the smallest positive integer such that $a_k\lt b_k$. Then $a_k=0$, $b_k=1$. Thus
$$\begin{align*}
\Psi(b_n) - \Psi(a_n) &= \sum_{n=1}^{\infty}\frac{2b_n-2a_n}{3^n}\\
&= \sum_{n=k}^{\infty}\frac{2b_n - 2a_n}{3^n}\\
&= \frac{2}{3^k} + \sum_{n=k+1}^{\infty}\frac{2b_n-2a_n}{3^n}\\
&\geq \frac{2}{3^k} - \sum_{n=k+1}^{\infty}\frac{2}{3^n}\\
&= \frac{2}{3^k} - 2\sum_{n=k+1}^{\infty}\frac{1}{3^n}.
\end{align*}$$
Since
$$\begin{align*}
\sum_{n=k+1}^{\infty}\frac{1}{3^n} &= \sum_{m=0}^{\infty}\frac{1}{3^{k+1}}\left(\frac{1}{3}\right)^n \\
&= \frac{1}{3^{k+1}}\left(\frac{3}{2}\right) = \frac{1}{2(3^k)},
\end{align*}$$
it follows that
$$\Psi(b_n) - \Psi(a_n) \geq \frac{2}{3^k} - \frac{2}{2(3^k)} = \frac{1}{3^k}\gt 0;$$
therefore, $\Psi$ respects the lexicographic order on the set of binary sequences.
