# Thread: Proving a function is discontinuous

1. ## Proving a function is discontinuous

My task is to show that the following function is discontinuous. $\displaystyle f(x) = 1 \, \text{for } x>0 \wedge f(x) = 0 \, \text{for } x \leq 0, x_0 =0.$
-The Doctor

2. We say a function is continuous at a point $\displaystyle x_0$ if

$\displaystyle \underset{\epsilon>0}{\forall} \; \underset{\delta>0}{\exists} \; \underset{x\in \mathbb R}{\forall} \; |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon$

Thus to prove your function (called the Heaviside function and usually noted $\displaystyle H$) is discontinuous at $\displaystyle x_0=0$, we must prove that

$\displaystyle \underset{\epsilon >0}{\exists} \; \underset{\delta >0}{\forall} \; \underset{x\in \mathbb R}{\exists} \; |x|<\delta \wedge |f(x)|>\epsilon$ (since $\displaystyle x_0=f(x_0)=0$)

Choose $\displaystyle \epsilon = \frac{1}{2}$, and let $\displaystyle \delta >0$. We then have

$\displaystyle |\frac{\delta}{2}|=\frac{\delta}{2}<\delta$

but also

$\displaystyle |f(\frac{\delta}{2})|=|1-0|=1>\frac{1}{2}=\epsilon$,

so $\displaystyle H$ is discontinuous at $\displaystyle x_0=0$.

3. Thanks so much for your help. I really appreciated it.