Cauchy sequences in $\mathbb{C}$ does not imply that the function converges.

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I am currently working on a proof and I have the current properties:

A Cauchy sequence $\{ f_n\}_{n\in \mathbb{N}} $ of functions in some space $X$ with the property that $|(f_n - f_m)(x)|<\epsilon$ for any given $\epsilon > 0$ which I have used to deduce that the sequence $ \{ |f_n(x)| \} \subset \mathbb{C}$ is Cauchy.

Now since we know that $\mathbb{C}$ is complete, I may deduce that $|f_n(x)|\to |f(x)|$. However a similar proof that I saw online then deduced that $f_n(x)\to f(x)$ for some unique function $f$. This doesn't quite sit right with me since can we not consider the counter example $f_n(x) = e^{i x g(n)}$ for some $g$? We will always have $f_n$ having modulus $1$ but without $f_n$ necessarily converging...

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Your condition exactly says that $\{f_n(x)\}$ is a Cauchy sequence for all $x \in X$, so by completeness of $\mathbb{C}$ you can define

$$f: X \to \mathbb{C}: x \mapsto \lim_n f_n(x)$$