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## Homework Statement

Prove that if f

_{n}-> f uniformly on a set S, and if g

_{n}-> g uniformly on S, then f

_{n}+ g

_{n}-> f + g uniformly on S.

## Homework Equations

## The Attempt at a Solution

f

_{n}-> f uniformly means that |f

_{n}(x) - f(x)| < [tex]\epsilon[/tex]/2 for n > N_1.

g

_{n}-> g uniformly means that |g

_{n}(x) - g(x)| < [tex]\epsilon[/tex]/2 for n > N_2.

By the triangle inequality, we have |f

_{n}(x) - f(x) + g

_{n}(x) - g(x)| <= |f

_{n}(x) - f(x)| + |g

_{n}(x) - g(x)| < [tex]\epsilon[/tex]/2 + [tex]\epsilon[/tex]/2 = [tex]\epsilon[/tex].

This implies |[f

_{n}(x) + g

_{n}(x)] - [f(x) + g(x)]| < [tex]\epsilon[/tex] for n > N_1, N_2.

Therefore f

_{n}+ g

_{n}-> f + g uniformly on S.

Is this correct? I'm pretty confident it's right, but I just want to make sure. Thanks!