Infinitely many discontinuous points Give an example of a function $f: [0,1] \to \mathbb R$ with infinitely many discontinuous points. You can use the fact that:
$\displaystyle\lim\limits_{k\to\infty}$ $\displaystyle\sum_{i=1}^k \displaystyle\frac {1}{2^i} = 1$
I don't even know how to start. 

This might be what the problem is trying to hint you toward. Since the rationals are countable, we can construct a bijection $h : \mathbb{N} \to \mathbb{Q} \cap [0, 1]$, that is an enumeration of the rationals in $[0, 1]$. Now define $f: [0, 1] \to [0, 1]$ by $$f(x) = \sum_{i \in \{k \in \mathbb{N} : h(k) \leq x \}} 2^{- i} .$$
Then $f$ is discontinuous on the rationals and continuous on the irrationals (and is indeed monotonic).
