# Proof

• By hypothesis, $D\subset\mathbb R$ and let $f_n:D\to\mathbb R$ are continuous functions on $D.$
• Assume, the sequence $(f_n)_{n\to\mathbb N}$ is uniformly convergent to a limit function $f:D\to\mathbb R.$
• By definition of uniformly convergent, there is an index $N\in\mathbb N$ such that $|f_N(x)-f(x)| < \frac {\epsilon}{3}$ for all $x\in D.$
• Since all $f_n:D\to\mathbb R$ are continuous on $D,$ so is $f_N.$
• By definition of continuous, there is a $\delta > 0$ such that $|f_N(x)-f_N(x^\prime)|<\frac {\epsilon}{3}$ for all $x^\prime\in D$ with $|x-x^\prime| < \delta.$
• By the triangle inequality, for all $x^\prime\in D$ with $|x-x^\prime| < \delta$: $$|f(x)-f(x^\prime)|\le |f(x)-f_N(x)|+|f_N(x)-f_N(x^\prime)|+|f_N(x^\prime)-f(x^\prime)| < \frac {\epsilon}{3}+\frac {\epsilon}{3}+\frac {\epsilon}{3}=\epsilon.$$
• It follows by definition of continous, that $f:D\to\mathbb R$ is continous.
• Now, assume the sequence $(f_n)_{n\to\mathbb N}$ is only pointwise convergent to a limit function $f:D\to\mathbb R.$
• As an example, consider $f_n(x):=x^n$ on the closed real interval $[0,1].$
• Clearly, the functions $x^n$ are continuous on $[0,1].$
• Moreover, for all $x\in[0,1)$, $\lim_{n\to\infty} x^n=0,$ while for $x=1$, $\lim_{n\to\infty} x^n=1.$
• The function $f$ is therefore not continous at $x=1\in[0,1].$
• Dispite the fact that for every $x\in[0,1]$, $\lim_{n\to\infty} f_n=f,$ i.e. the continous functions $f_n$ converge only pointwise to $f.$

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
2. Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition