For $n\in\mathbb{N}$, let $f_n\colon X\rightarrow \mathbb{R}$ be a sequence of functions that converges uniformly to $f\colon X\rightarrow \mathbb{R}$. Is it then true that $|f_n(x)−f(x)|$ for all $x\in X$ is bounded above by some function only depending on $n$?
The reason for the question is that if $f_n$ converges uniformly to $f$, then given any $\epsilon>0$, one should be able to find an $N(\epsilon)\in \mathbb{N}$ so that $n>N(\epsilon)$ implies $|f_n(x)−f(x)|<\epsilon$ for all $x\in X$ . By analogy with showing some function is continuous, I would try to find $N(\epsilon)$ by bounding $|f_n(x)−f(x)|$ from above. To satisfy the definition, it seems like this bound can only depend on $n$, so one can in turn bound it from above by $\epsilon$ and solve for $n$.
Yes the function may be equal to $$\varepsilon (n) =\sup_{x\in X} |f_n (x) - f(x) |$$