analysis - Isn't that proof going the wrong way?

I'm currently working on the very well written book Understanding Analysis, by Stephen Abbott.

But I found a proof that looks wrong, I think that it going the wrong way (showing that A $\implies$ B while it should demonstrate that $B \implies A$).

Here is the theorem :

A function f : A $\to$ R, fails to be uniformly continuous if there exists a particular $\epsilon_0$ and two sequences $(x_n)$ and $(y_n)$ such that : $\lim(|x_n - y_n|) = 0$ and $|f(x_n) - f(y_n)| \geq \epsilon_0$.

Here is the proof :

Negating the definition of uniform continuity gives the following: A function $f : A → R$ fails to be uniformly continuous on A if there exists $ε_0 > 0$ such that for all $δ > 0$ we can find two points $x$ and $y$ satisfying $|x − y| < δ$ but with $|f(x) − f(y)| ≥ \epsilon_0$.

The fact that no $δ$ “works” means that if we were to try $δ = 1$, we would be able to find points $x_1$ and $y_1$ where $|x_1 − y_1| < 1$ but $|f(x_1) − f(y_1)| ≥ ε_0.$ In a similar way, if we try $δ = 1/n$ where $n ∈ N$, it follows that there exist points $x_n$ and $y_n$ with $|x_n − y_n| < 1/n$ but where $|f(x_n) − f(y_n)| ≥ \epsilon_0$. The sequences $(x_n)$ and $(y_n)$ are precisely the ones described in theorem.

Comment :

I think that the proof demonstrated that : f not uniformly continuous $\implies$ $(x_n)$ and $(y_n)$ exist. While it should have been the other way around.

Do you agree that the proof is wrong ?

Answer

Note that "$f\colon A\to \mathbb R$ is uniformly continuous" is in fact by definition equivalent to

$$\forall \epsilon>0\colon \exists \delta>0\colon\forall x,y\colon (|x-y|<\delta\to |f(x)-f(y)|<\epsilon )$$ Hence the negation $$\exists \epsilon>0\colon \forall \delta>0\colon\exists x,y\colon (|x-y|<\delta\land|f(x)-f(y)|\ge\epsilon )$$ is also equivalent to the negation "$f\colon A\to \mathbb R$ fails to be uniformly continuous". So on the one hand you may have tripped over the custom that "if" in a definition is conceptually an "iff" and that on the other hand the argument does indeed show something stronger than "if", namely "iff".