Apostol's proof on uniform convergence and continuity

Question

Assume that $f_n\to f$ on $S$ uniformly. If each $f_n$ is continuous at $c\in S$. Then the limit function $f$ is also continuous at $c$.

In the proof it has been stated that the theorem is obvious for an isolated point. That is, if $c$ is an isolated point then $f$ is automatically continuous at $c$. So we only consider accumulation points in the proof. Why is it so?

Answer 1

Beacause every function is continuous at an isolated point. If $c$ is such a point and $f$ is your function, saying that $f$ is continuous at $c$ means that$$(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in D_f):d(x,c)<\delta\implies d\bigl(f(x),f(c)\bigr)<\varepsilon.$$Let $\varepsilon>0$ and take $\delta>0$ such that $d(x,c)<\delta\implies x=c$; such a $\delta$ must exist, since $c$ is an isolated point. But then$$d(x,c)<\delta\implies x=c\implies d\bigl(f(x),f(c)\bigr)=0<\varepsilon.$$

Answer 2

Because $c$ is an isolated point, there exist $\delta>0$ such that $B(c,\delta)=\{c\}$. Therefore; \begin{equation} |f(x)-f(c)| = 0 < \epsilon \quad if \quad x\in B(c,\delta) \end{equation} where $B(c,\delta)$ is the ball centered at $c$.