# Prove continuity with Heaviside function

Where $$H:\mathbb{R} \to \mathbb{R}$$ is defined by $$H(x)=\begin{cases}1&\quad \text{if }x\text{\geq0}\\0&\quad\text{if } x \text { otherwise}\end{cases}$$. Prove that for all $$a\neq0$$ the function $$H$$ is continuous at $$a$$. Also prove that $$H$$ is not continuous at 0.

I tried where $$\delta = \epsilon$$ but that doesn't work for $$a=1/3$$ $$\epsilon=2/3$$ $$x=-1/4$$. With a correct $$\delta$$ proving it is continuous at $$a$$ I should be able to do but not sure how to prove something is not continuous at a point, have never done that before.

• You need to show that the left and right limit do not agree, in this case $0\ne 1$. – lightxbulb Feb 28 '19 at 20:18
• Maybe the OP explictly wants an $\epsilon$-$\delta$-argument? – Mars Plastic Feb 28 '19 at 21:23
• yes, I need an explicit delta/epsilon proof – A.A. Feb 28 '19 at 22:42

For $$a>0$$ take $$\delta =a$$ and verify that $$|x-a| <\delta$$ implies $$x>0$$ so $$H(x)=H(a)=1$$ and $$|H(x)-H(a)| <\epsilon$$. For $$a<0$$ take $$\delta =-a$$ and verify that $$|x-a| <\delta$$ implies $$x<0$$ so $$H(x)=H(a)=0$$ and $$|H(x)-H(a)| <\epsilon$$.
Suppose $$H$$ is continuous at $$0$$. There exists $$\delta >0$$ such that $$|H(x)-H(0)| <1$$ if $$|x-0| <\delta$$. Take $$x =-\delta /2$$ to see that $$|0-1| <1$$ which is a contradiction .