Is it true that for $f,g: [a,b] \rightarrow \Bbb R$, and if $f,g$ are bounded then
$$ |f(x) - g(x)| < \epsilon, \forall x \in [a,b] \implies |\sup(f(x)) - \sup(g(x))| < \epsilon$$
This was a remark made inside a proof done in class by my professor for a different theorem. I'm trying to follow along but I can't find a theorem in the book that shows this and trying to prove by definition was a mess for me. Can anyone help me understand this?
Since $|f(x)-g(x)|<\epsilon$ you have $f(x)<g(x)+\epsilon$ and so $\sup f\le \sup g+\epsilon$. A similar inequality with $f$ and $g$ interchanged holds and so $$ \lvert\sup f-\sup g|\le\epsilon. $$