I am working on the following exercise: Use the $\epsilon-\delta$-criterion to show that the following function is discontinuous in $0\\$
$sgn(x) = \begin{cases} -1&,\text{ for }x<0\\ 0&,\text{ for }x=0 \\ 1&,\text{ for }x>0 \end{cases} $
Intuitively this makes sense but I am struggling with arguing for this. More specifically, do I have to argue for each of the cases? Also, how would you advise choosing delta in the beginning? What I have so far is:
$\epsilon-\delta$ criterion: Let $f : D \rightarrow \mathbb{R}$ and $x_{0} \in D.$ Then, $f$ is continuous at $x_{0}$ if and only if $\forall \epsilon > 0 \quad \exists \delta > 0 \quad \forall x \in D : |x - x_{0} | < \delta \implies |f(x) - f(x_{0}| < \epsilon$
Let $\epsilon > 0$ Set $\delta = 1 \\$
Case 1: $|x - 0| < 1 \implies |-1 - 0 | < \epsilon \\|x| < 1 \implies 1 < \epsilon$
Not true if we set $\epsilon = 2\delta = 2$
Case 2: $|x - 0| < 1 \implies |1 - 0 | < \epsilon \\|x| < 1 \implies 1 < \epsilon$
Not true with the same argument
Edit: Sorry, the original post had two typos. (In the definitions)
Take $\varepsilon=1$. Then, for any $\delta>0$, you have$$\left|\frac\delta2-0\right|<\delta\quad\text{and}\quad\left|\operatorname{sgn}\left(\frac\delta2\right)-\operatorname{sgn}(0)\right|=1\geqslant\varepsilon.$$So, this proves that$$(\exists\varepsilon>0)(\forall\delta>0)(\exists x\in\Bbb R):|x-0|<\delta\text{ and }\bigl|\operatorname{sgn}(x)-\operatorname{sgn}(0)\bigr|\geqslant\varepsilon,$$which is the same thing as asserting that $\operatorname{sgn}$ is discontinuous at $0$.