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 $\delta.$