Let $ \emptyset\ne A\subset \mathbb{R}^n$ and let $f:A\to \mathbb{R}$ be a bounded function. There is the following definition:
For every non-empty subset $B$ of $A$, the oscillation of $f$ on $B$ is the number osc$(f)_B:=\sup(f)_B - \inf(f)_B$.
I want to prove that osc$(f)=\sup\{|(f(x))-(f(y))|:x,y \in B\}.$
I'm not quite sure how to go about a proof, but here's my attempt.
$\sup_B(f)-\inf_B(f)=\sup_B(f)+\sup_B(-f)=\sup\{f(x):x\in B\}+\sup\{-f(y):y\in B\}$ $=\sup\{f(x)+(-f(y)):x,y\in B\}$ $=\sup\{|f(x)-f(y)|:x,y\in B\}$ (the absolute value can be applied, since $\sup_B(f)\ge \inf_B(f)$).
Do you think this is correct / rigorous enough?
The proof looks ok to me, except for the very last motivation "the absolute value can be applied since...". Note that in general $$ \sup_{x,y}g(x,y)\ne\sup_{x,y}|g(x,y)| $$ even if the LHS is positive. For example, for $-2\le g(x,y)\le 1$, the LHS is $1$, but the RHS is $|-2|=2$.
Here we get equality because of the sign symmetry of the expression $$ \sup_{x,y}(f(x)-f(y))=\sup_{x,y}(f(y)-f(x))=\sup_{x,y}(-(f(x)-f(y))), $$ so we can finish the proof as $$ ...=\sup_{x,y}(f(x)-f(y))=\max\sup_{x,y}(\pm(f(x)-f(y)))=\sup_{x,y}\max(\pm(f(x)-f(y)))=\sup_{x,y}|f(x)-f(y)|. $$