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## Homework Statement

Let ##X \subset \mathbb{C}##, and let ##f_n : X \rightarrow \mathbb{C}## be a sequence of functions. Show if ##f_n## is uniformly Cauchy, then ##f_n## converges uniformly to some ##f: X \rightarrow \mathbb{C}##.

## Homework Equations

Uniform convergence: for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that for any ##x \epsilon X## if ##n > N(\varepsilon)## then ##\vert f_n(x) - f(x) \vert < \varepsilon##.

Uniformly Cauchy: for all ##\varepsilon > 0## there exists ##N(\varepsilon)## such that for any ##x \epsilon X##: if ##n, k > N(\varepsilon)## then ##\vert f_n(x) - f_k(x) \vert < \varepsilon##.

## The Attempt at a Solution

Proof: Suppose ##f_n## is uniformly Cauchy. First we show ##f_n## converges point wise to some ##f : X \rightarrow \mathbb{C}##.

Fix some ##x \epsilon X##. Then ##f_n(x)## is a Cauchy sequence of complex values. By completeness, ##f_n(x)## converges to some ##f(x): X \rightarrow \mathbb{C}##. Thus, ##f_n## converges point wise to some ##f## on ##X##.

Let ##\varepsilon/2 > 0##. By definition of uniformly Cauchy, there exists some ##N_1 = N_1(\varepsilon/2)## such that for any ##x \epsilon X##: if ##l, k > N_1## then ##\vert f_l(x) - f_k(x) \vert < \varepsilon / 2##. By point wise convergence, there exists ##N_2 = N_2(\varepsilon, x)## such that if ##n > N_2## then ##\vert f_n(x) - f(x) \vert < \varepsilon / 2##.

..

Not sure, I think I'm on the wrong track, I was going to try to pick max(N_1, N_2) and use triangle inequality but I don't see how to do this/how this would help to show uniform conv.

Any hints?, please

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