I read "A Pictorial Proof of Uniform Continuity" by D. M. Bloom and I had a question.
Here is the proof:
Suppose that $\varepsilon$ is any positive real number, $I = [a, b], f(x)$ is continuous on $I$ , $S = \{(x,y) \in I \times I \vert \vert f(x) - f(y) \vert \ge \varepsilon \}$ and $F(x,y) = \vert x - y \vert$.
Then, $\{(x,y) \in I \times I \vert x = y \} \cap S = \emptyset$. and $S$ is a bounded closed set. So $F(x,y)$ has a positive minimum value on $S$.
Let the positive minimum value be $\delta$.Then $\vert f(x) - f(y) \vert \ge \varepsilon \Rightarrow \vert x - y \vert \ge \delta$, i.e. $\vert x - y \vert < \delta \Rightarrow \vert f(x) - f(y) \vert < \varepsilon$
This shows $f(x)$ is uniformly continuous on $I$.
My question is the following:
If S is an empty set, this proof doesn't work.
Am I correct?