0
$\begingroup$

Where $H:\mathbb{R} \to \mathbb{R}$ is defined by $H(x)=\begin{cases}1&\quad \text{if }x\text{$\geq$0}\\0&\quad\text{if } x \text { otherwise}\end{cases}$. 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.

$\endgroup$
  • 2
    $\begingroup$ You need to show that the left and right limit do not agree, in this case $0\ne 1$. $\endgroup$ – lightxbulb Feb 28 '19 at 20:18
  • $\begingroup$ Maybe the OP explictly wants an $\epsilon$-$\delta$-argument? $\endgroup$ – Mars Plastic Feb 28 '19 at 21:23
  • $\begingroup$ yes, I need an explicit delta/epsilon proof $\endgroup$ – A.A. Feb 28 '19 at 22:42
1
$\begingroup$

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 .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.