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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?

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2 Answers 2

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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$$d(x,c)<\delta\implies x=c\implies d\bigl(f(x),f(c)\bigr)=0<\varepsilon.$$

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  • $\begingroup$ Also sir is it correct to say that the points inside $S$ are either limit points or isolated (i.e, these 2 concepts are mutually exclusive for the points inside $S$)? And that is why Apostol moved over to limit points after stating the theorem to be obvious for isolated points? $\endgroup$
    – S.S
    Nov 19, 2017 at 12:33
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    $\begingroup$ @ShatabdiSinha I don't have access to Apostol's book now, but if his definition of limit point is this one, then you are right. $\endgroup$ Nov 19, 2017 at 12:36
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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 $c$.

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