A clarification regarding the definition of uniform continuity of a function defined in a subset of $\mathbb R.$

Question

Let $f:A\to \Bbb R$ where $A\subseteq \Bbb R$. We say that, $f$ is uniformly continuous on $A$ if for any $\epsilon\gt 0$ there exists $\delta(\epsilon)=\delta\gt 0$ such that for any $x_1,x_2\in A$ and satisfying $|x_2-x_1|\lt \delta$ we have, $|f(x_2)-f(x_1)|\lt\epsilon.$

Now, my question is: Say, for a particular $\epsilon_0\gt 0$ there exists a $\delta\gt 0$ such that for any $x_1,x_2\in A$ and satisfying $|x_2-x_1|\lt \delta$ we have, $|f(x_2)-f(x_1)|\lt\epsilon_0.$ But, what if, no two distinct points in the domain of $f$ say, $A$ has a distance of $\delta$ or, in other words, what if every pair of distinct points in $A$ has a distance strictly greater than $\delta$ ? Will $f$ be still uniformly continuous?

My answer is "yes". This is because, the definition of uniform continuity says, that if any two points say, $x_1,x_2$ have the distance between them the required $\delta$ or even less than $\delta$ (, for some choice of $\epsilon$) then $|f(x_2)-f(x_1)|\lt \epsilon$ must hold, but NEVER in the definition of uniform continuity it assumes that there must exist two distinct points in the domain of the function ,$A$ such that the distance between them is at most the required $\delta.$

Is my reasoning correct?

Answer 1

This is true, but you can think of this even in the continuous case. This is because a domain may not have any accumulation point, or even the domain is discrete, for example, $$f:\mathbb{N}\to\mathbb{R}, f(x)=\max\{x,\pi\}$$ is continuous everywhere, because $|x-y|\ge1$ for any $x\ne y$, so you can pick $\delta=\dfrac{1}{2}$. And of course it is uniformly continuous also.

You can somehow consider it as the vacuously true in logic, so that since no such $x,y$ satisfies this condition, $f$ is also (uniformly) continuous.