I'm self-studying from Abbott's Real Analysis. They begin with an attempt to prove that the pointwise limit of continuous functions is continuous, and show that it can't be done.
Assume $(f_n)$ is a sequence of continuous functions on $A\subseteq \mathbb{R}$ converging pointwise to limit $f$. Fix $c\in A$, and let $\epsilon >0$.
$|f(x) - f(c) \leq |f(x) - f_n(x)| + |f_n(x) - f_n(c)| + |f_n(c) - f(c)|$.
We can choose an $N$ such that $|f_N(c) - f(c)| < \epsilon /3$ because $(f_n) \to f$.
By continuity of $f_N$, $\exists \delta>0$ such that: $|f_N(x) - f_N(c)| < \epsilon /3$ for all $x$ satisfying $|x-c|<\delta$.
But the problem is that we also need $|f_N(x) - f(x)| <\epsilon/3$ for all $x$ satisfying $|x-c| < \delta$.
It seems (naively), that we could have approached this differently and started with a $\delta$-ball around $c$ and then found the $N$ such that $|f_N(x) - f(x)| < \epsilon /3$ for any $x$ in the $\delta$-ball. Ie, for each $x$, there is some $N_x$ that works, by pointwise convergence, and then we just take $N=sup\{N_x: x\in \delta \text{-ball}\}$. Is the issue with this approach that this supremum might not exist, and that for any $N$, there is another $x$ that requires a larger $N$ for $|f_N(x) - f(x)|<\epsilon/3$ ?
Thank you, I'm just learning about sequence of functions so apologies for the basic question.