I keep being told how trivial uniform continuity is for $f(x) = x$ on $\Bbb R$, but sometimes those simplest things are hardest for me to see. I haven't been able to find a proof for this anywhere, most likely BECAUSE it is just so simple... in lecture it was given to us as fact. I'd like to see how to get it.
In my head it follows easily from the definition I was given:
"A function $f: D \to \Bbb R$ is said to be uniformly continuous on $D \iff \forall \epsilon >0 , \exists \ \delta$ such that $|x-y| < \delta \implies |f(x) - f(y)| < \epsilon$"
But I know I can't use that in place of a proof so it's making me wonder if I truly do know what the definition is telling me. I thought I could simply USE $|x-y| < \delta \implies |f(x) - f(y)| < \epsilon$ since $f(x) = x$ but I'm second guessing? That doesn't seem right...
Let $\epsilon>0$. What we wish to show that there exists a $\delta >0$ such that when we assume $|x-y| < \delta$, this implies $|f(x)-f(y)|< \epsilon$. But, note that $$|f(x) - f(y)|=|x-y|.$$ So the "trivial" part is that we just choose $\delta = \epsilon$, because then we have $$|x-y| < \delta=\epsilon \implies |x-y|<\epsilon \implies |f(x) - f(y)|<\epsilon$$ as needed!