Prove a map of binary expansion is continuous

Question

Prove that the map: f: $\{0,1\}^\mathbb{N} \times \{0,1\}^\mathbb{N}$ $\to$ $[0,1] \times [0,1]$ is continuous.

I know that the map can be written as ($m_1,m_2,m_3,m_4,...$) $\times$ ($n_1,n_2,n_3,n_4,...$) $\mapsto$ ($0.m_1m_2m_3m_4...,0.n_1n_2n_3n_4...$) as binary expansions, and f is surjective, but how do I show it is continuous?

Answer 1

Hint: It suffices to check that $$F : \{ 0,1\}^{\mathbb N} \to [0,1],\ \ \ F(m_1, m_2, \cdots, ) = 0.m_1m_2\cdots $$ is continuous. Let $V\subset [0,1]$ be an open set. First of all, show that $F^{-1}(V)$ is open when

$$V = \left(\frac{m_1}{2} + \frac{m_2}{2^2} + \cdots + \frac{m_k}{2^k}, \frac{m_1}{2} + \frac{m_2}{2^2} + \cdots + \frac{m_k}{2^k} + \frac{1}{2^k} \right),$$

where $k\in \mathbb N$ and $m_1, \cdots, m_k \in \{0,1\}$. Then the general $U$ can be written as countable union of these $V$'s.

Answer 2

Here is how I would do it. First, note that as John Ma noted, it suffices to show continuity of $$ F : \{0,1\}^\Bbb{N}\to [0,1], (x_n)_n \mapsto \sum_{n=1}^\infty x_n/2^n. $$

To see this, show that each of the maps $$ F_N : \{0,1\}^\Bbb{N}\to [0,1], (x_n)_n \mapsto \sum_{n=1}^N x_n/2^n $$ is continuous (use that each projection on the individual components is continuous) and that $F_N \to F$ uniformly, e.g. by the Weierstrass M-Test.

You can even use a similar approach to show directly that your map is continuous.