An independent squence of functions that are uniform on $[0,1]$ Suppose that $X$ is uniform in $[0,1]$. Find an infinite sequence of functions $f_{i}$ so that all $f_{i}(X)$ are independent and uniform $[0,1]$.
um I'm not really sure how to do this. I'm thinking taking translates mod 1 might work.
 A: 
Suppose that $X$ is uniform in $[0,1]$. Find an infinite sequence of functions $f_{i}$ so that all $f_{i}(X)$ are independent.

Consider $f_n(x)=(-1)^{\lfloor2^nx\rfloor}$ for every $n$ in $\mathbb N$.
One could furthermore impose every $f_n(X)$ to be uniform on $[0,1]$ but this would require writing a few more lines of explanation.
Added later on to the question:

Find an infinite sequence of functions $f_{i}$ so that all $f_{i}(X)$ are independent and uniform $[0,1]$.

Here is a classical construction showing this, based on the binary expansion of every real number.
One starts from some sequence $(\epsilon_i)_{i\in\mathbb N}$ i.i.d. Bernoulli uniform on $\{0,1\}$, then the random variable $$\xi=\sum\limits_{i\in\mathbb N}2^{-i}\epsilon_i$$ is uniform on $[0,1]$. Likewise, consider the collection $\mathcal K$ of infinite subsets of $\mathbb N$ and, for every $K$ in $\mathcal K$, write $K=\{k(i)\mid i\in\mathbb N\}$ with $k(i)\lt k(i+1)$ for every $i$. Then, for every $K$ in $\mathcal K$, the random variable $$\xi_K=\sum\limits_{i\in\mathbb N}2^{-i}\epsilon_{k(i)}$$ is uniform on $[0,1]$ since  $(\epsilon_{k(i)})_{i\in\mathbb N}$ is also i.i.d. Bernoulli uniform on $\{0,1\}$. Furthermore, for every such collection of disjoint $K(n)$ in $\mathcal K$, $(\xi_{K(n)})$ is independent since each collection $(\epsilon_k)_{k\in K(n)}$ is independent of all the others. 
For every $n$ in $\mathbb N$, call $f_n$ some measurable function such that $\xi_{K(n)}=f_n(\xi)$. This exists since, for every $K$ in $\mathcal K$, the maps $$\xi\mapsto(\epsilon_i)_{i\in\mathbb N}\mapsto(\epsilon_k(i))_{i\in\mathbb N}\mapsto\xi_K$$ are measurable (recall for example that each $\epsilon_i$ is the indicator function of the event that the integer part of $2^i\xi$ is odd). Then the sequence $(f_n(X))_{n\in\mathbb N}$ is i.i.d. uniform on $[0,1]$.
For a "constructive" example, note that the sets $$K(n)=\{(2n-1)2^{i-1}\mid i\in\mathbb N\},$$ are disjoint subsets of $\mathbb N$, for $n$ in $\mathbb N$, and consider the functions $f_n$ defined by$$f_n(x)=\sum_{i\in\mathbb N}2^{-i}\,\mathbf 1(\lfloor2^{(2n-1)2^{i-1}}x\rfloor\ \text{odd}).$$
