Ok so this is an absolutely beautiful proof.
Suppose {f_n} is a sequence of say real or complex valued continuous
functions. We could even assume f_n is a sequence of functions from
R^n->R where it's understood the || means the usual euclidean norm.
Suppose also that f_n -> f uniformly on some set say A.
Let us show f is continuous.
So take some x_0 in A and let e > 0.
Since
f_n->f uniformly on A, there is a positive integer N such that for
all n > N we have |f_n(x)-f(x)|< e/3 for all x in A.
Fix m
> N. Since f_m is continuous at x_0, there is a delta > 0 such
that for all x in A with |x-x_0| < delta we have |f_m(x) - f_m(x_0)|
< e/3.
Then for all x in A with |x-x_0| < delta, we have by the triangle inequality
|f(x)-f(x_0)| <= |f(x) - f_m(x)| + |f_m(x) - f_m(x_0)| + |f_m(x_0) - f(x_0)| < e/3 + e/3 + e/3 = e.
Done!
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