Let $g:[-M,M]\to\mathbb{R}$ be a uniformly continuous function. By definition, for every $\varepsilon>0$ there exists $\delta>0$ such that for every $x,y\in[-M,M]$ that satisfy $|x-y|<\delta:|g(x)-g(y)|<\varepsilon$. My lecturer said that because of this, the following relation holds: For every $\varepsilon>0$, there exists $\delta>0$ such that:
$$\{x,y\in[-M,M]\mid |g(x)-g(y)|\geq\varepsilon\}\subseteq\{x,y\in[-M,M]\mid|x-y|\geq\delta\}$$
I have been trying to understand why is this true, but to no avail. I'm pretty new to sets at general, so it might be the reason why this is a little more difficult to me to understand than I expected it to be.
Thanks!
Assume that $\{x,y\in[-M,M]:|g(x)-g(y)|\geq\epsilon\}\nsubseteq\{x,y\in[-M,M]:|x-y|\geq\delta\}$, so there are $x,y\in\{x,y\in[-M,M]:|g(x)-g(y)|\geq\epsilon\}$ but $x,y\notin\{x,y\in[-M,M]:|x-y|\geq\delta\}$.
So $x,y$ fail to satisfy the property described in the set $\{x,y\in[-M,M]:|x-y|\geq\delta\}$, which is $|x-y|\geq\delta$, the negation of $|x-y|\geq\delta$ is simply $|x-y|<\delta$, so $x,y$ are such that $|x-y|<\delta$.
So you have $|x-y|<\delta$ together with $|g(x)-g(y)|\geq\epsilon$, which contradicts the statement that being uniformly continuous.