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I am trying to show the following:

For each $n \in \mathbb N$, let $f_n:X \to Y$, where $(Y,d)$ is a complete metric. Suppose that for every $\epsilon>0$, there exists $n_0 \in \mathbb N$ such that for all $x \in X$, if $n,m \geq n_0 \implies$ $d(f_n(x),f_m(x))< \epsilon$.

Show that there exists $f:X \to Y$ such that $f_n \to f$ uniformly.

My attempt at a solution

For a given $x \in X$, it is easy to see that $(f_n(x))_{n \in \mathbb N}$ is a Cauchy sequence in $Y$. Since $Y$ is complete, there exists $y=\lim f_n(x)$. Define $$f:X \to Y$$$$x \to \lim f_n(x)$$

By definition, it is immediate that $f_n \to f$ pointwise. Now, I don't know how to prove the uniformly convergence, how can I control $d(f(x),f_n(x))$ for all $x$ at the same time given $n$ sufficiently large?

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  • $\begingroup$ say that $d(f_n(x) , y) \le \sup_m d(f_n(x) , f_m(x))$ so it is immediate that $f_n \to f$ uniformly $\endgroup$ – reuns Jul 25 '16 at 21:05
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The distance function $d: (Y \times Y, d_{max}) \to \Bbb R$ is continuous. Hence, for all $n \ge n_0$, we have for all $x \in X$,

$$d(f(x), f_n(x)) = d(\lim_{m \to \infty} f_m(x), f_n(x)) = d( \lim_{m \to \infty} (f_m(x), f_n(x))) = \lim_{m \to \infty} d(f_m(x), f_n(x)) \le \epsilon$$

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