I need help showing that Dirichlet's function is discontinuous on the interval (0,1).
f(x)=
{1/q if x=p/q in lowest terms with p,q in N
0 if x is irrational}
This function is continuous at irrational $\displaystyle x$. To prove it's discontinuous at $\displaystyle \frac{p}{q}$, let $\displaystyle \epsilon=\frac{1}{2q}$. Since for all $\displaystyle \delta>0$, $\displaystyle (p/q-\delta,p/q+\delta)$ contains an irrational point, $\displaystyle \exists x$ such that $\displaystyle |f(p/q)-f(x)|=|1/q-0|=\frac{1}{q}>\frac{1}{2q}=\epsilon$.
There's a good picture of this function on MathWorld.
In fact f(x) is continuous at every irrational point, and discontinuous at rational ones. To prove this you'll prove that $\displaystyle \lim_{x\to x_0}f(x)=0$ at every point in $\displaystyle (0,1)$ (Hint: if a sequence of rational points $\displaystyle \left\{\frac{a_n}{b_n}\right\}\,,\,a_n\,,\,b_n\in \mathbb{Z}\,,\,\,b_n> 0\,\,\forall\,n\in\mathbb{N}$ , converges to an irrational number, then $\displaystyle \lim_{n\to\infty}b_n=\infty$
Tonio