Binary expansion in $[0, 1)$ has no point-wise convergent subsequence For $x \in [0,1)$, let $\sum_1^{\infty} a_n(x)2^{-n}$, $a_n(x) = 0 $ or $1$, be the base-2 decimal expansion of $x$. (If $x$ is dyadic rational, choose the expansion such that $a_n(x) = 0$ for $n$ large). Then the sequence $<a_n>$ in $\lbrace 0, 1 \rbrace^{[0,1)}$ has no point-wise convergent subsequence. $ \lbrace 0, 1 \rbrace^{[0,1)}$ is equipped with the product topology arising from the discrete topology on $\lbrace 0, 1 \rbrace $.
 A: Let $\{n_k\}_{k\in\mathbb N}$ be any strictly increasing sequence of natural numbers. Let $x_0\in[0,1)$ be such that its base-2 expansion looks like this:
$$\underset{1}{0}\underset{2}{0}\ldots\underset{n_1-1}{0}\underset{n_1}{1}\underset{n_1+1}{0}\ldots\underset{n_2-1}{0}\underset{n_2}{0}\underset{n_2+1}{0}\ldots\underset{n_3-1}{0}\underset{n_3}{1}\underset{n_3+1}{0}\ldots$$
That is, the $n$th digit is $1$ if $n=n_k$ for $k$ odd, and $0$ for all other $n$'s. Then, $a_{n_k}(x_0)=1$ if $k$ is odd and $a_{n_k}(x_0)=0$ if $k$ is even. Therefore, $\{a_{n_k}(x_0)\}_{k\in\mathbb N}$ does not converge in $\{0,1\}$ in the discrete topology, so that the subsequence $\{\langle a_{n_k}(x)\rangle_{x\in[0,1)}\}_{k\in\mathbb N}$ fails to converge in the product topology (i.e., it does not converge pointwise). Since this subsequence was arbitrary, the claim follows.
A remark: it is clear that $x_0$ is not a dyadic rational, because $0$'s and $1$'s alternate ad infinitum. Hence, no ambiguity concerning its base-2 representation arises.

Folland (1999, pp. 130–131) Exercise 4.43. Best. Textbook. Ever.
