Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be the function defined by the rule that $f(x)=1$ if $x<0$ and $f(x)=2$ if $x \geq 0$. A friend asserts that $f$ is continuous at the point $x=0$, and you of course disagree. A symbolic formulation of your friend's assertions is:
$\forall \epsilon > 0$ $\exists\delta > 0$ $\forall x \left( |x-0| < \delta \Rightarrow | f(x) - f(0)| < \epsilon \right)$.
My job is to prove this friend wrong.
I figure the solution is to prove the negation.
i.e. prove:$\exists \epsilon > 0$ $\forall\delta > 0$ $\exists x \left( |x-0| < \delta \Rightarrow | f(x) - f(0)| \geq \epsilon \right)$.
If I am correct in my reasoning then I am not sure how to prove the negation as I have not proved an epsilon delta type proof before and we have not gone over it in lecture nor do we have a text to use.
Take $\epsilon_0 < 1$. Then $|f(x) - f(0)| = |2 - f(x)| = |2 - 1| = 1$ for any $x < 0$. And clearly $1 > \epsilon_0$.