Existence, Uniqueness and Counterexamples · Challenge problem

There is an $\varepsilon > 0$ such that for every $\delta > 0$ there is an $x$ with $|x - a| \ge \delta$ and $|f(x) - f(a)| \ge \varepsilon$.

For every $\varepsilon > 0$ and every $\delta > 0$ there is an $x$ with $|x - a| < \delta$ and $|f(x) - f(a)| \ge \varepsilon$.

There is an $\varepsilon > 0$ such that for every $\delta > 0$ there is an $x$ with $|x - a| < \delta$ and $|f(x) - f(a)| \ge \varepsilon$.

There is an $\varepsilon > 0$ and a $\delta > 0$ such that for every $x$ with $|x - a| < \delta$, $|f(x) - f(a)| \ge \varepsilon$.