For a function $f:A \to \mathbb{R}$, I need to prove the following three statements are equivalent:
- f is not uniformly continuous on $A$.
- There exists an $\epsilon_0 \geq 0$ such that for every $\delta(\epsilon_0) \geq 0$ there are points $x_\delta, y_\delta$ in $A$ such $\left|x_\delta - y_\delta \right| < \delta $ and $\left|f(x_\delta) - f(y_\delta)\right| \geq \epsilon_0$
- There exists an $\epsilon_0$ and two sequences $(x_n), (y_n)$ in $A$ such that $\lim\left(x_n-y_n\right) = 0$ and $\left|f(x_n) - f(y_n)\right| \geq \epsilon_0$ for all $n \in \mathbb{N}$
Proof
$(1 \iff 2)$
Statement two is simply the negation of the definition of uniform continuity.
$(2 \implies 3)$
We are given that for any $\delta > 0$ there exists and $x$ and $y$ that satisfy the stated condition. Then consider a monotonic decreasing sequence of such deltas $\left|x_n - y_n\right| < \delta_n$. Where
$$\delta_1 > \delta_2 > \cdots > \delta_n > \delta_{n+1}> \cdots$$
and
$$\left|f(x_n) - f(y_n)\right| \geq \epsilon_0, \, n \in \mathbb{N}$$
Clearly, for one of these fixed $\delta_{n =f}$ the following inequality is satsified
$$\left|x_n - y_n\right| < \delta_f \,, \, \text{ where } \, n\geq f.$$
This truth for arbitrary $\delta_f$ implies the existence of a Cauchy sequence such that
$$\left|f(x_n) - f(y_n)\right| \geq \epsilon_0, \, n \in \mathbb{N}$$
$(3 \implies 2)$
I think this simply involves reversing the $(2\to3)$ steps.
Besides not fully hashing out the $(3 \implies 2)$ is this proof valid?