# Math Help - showing dirichlet's function is discontinuous on (0,1)

1. ## showing dirichlet's function is discontinuous on (0,1)

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}

2. Originally Posted by dannyboycurtis
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 $x$. To prove it's discontinuous at $\frac{p}{q}$, let $\epsilon=\frac{1}{2q}$. Since for all $\delta>0$, $(p/q-\delta,p/q+\delta)$ contains an irrational point, $\exists x$ such that $|f(p/q)-f(x)|=|1/q-0|=\frac{1}{q}>\frac{1}{2q}=\epsilon$.

There's a good picture of this function on MathWorld.

3. Originally Posted by dannyboycurtis
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}

In fact f(x) is continuous at every irrational point, and discontinuous at rational ones. To prove this you'll prove that $\lim_{x\to x_0}f(x)=0$ at every point in $(0,1)$ (Hint: if a sequence of rational points $\left\{\frac{a_n}{b_n}\right\}\,,\,a_n\,,\,b_n\in \mathbb{Z}\,,\,\,b_n> 0\,\,\forall\,n\in\mathbb{N}$ , converges to an irrational number, then $\lim_{n\to\infty}b_n=\infty$

Tonio