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.

  • 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

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 .


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