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Let $B(x,y)$ be the set of all the bounded functions $f: X \to Y$ ($X,Y$ metric spaces). Prove that if $\mathcal F \subset B(x,y)$ is an equicontinuous family, then $\overline {\mathcal F}$ is equicontinuous.

Here's what I've tried:

Take $x_0 \in X$, $\epsilon>0$, I want to prove that there is $\delta>0$: $d_X(x_0,x)<\delta \implies \forall f \in \overline {\mathcal F} d_Y(f(x_0),f(x))<\epsilon$

So let $x_0 \in X$, $\epsilon>0$ and take an arbitrary $f \in \overline {\mathcal F}$. There is $\{f_n\}_{n \in \mathbb N} \subset B(X,Y)$: $f_n \to f$.

$d_Y(f(x_0),f(x)) \leq d(f(x_0),f_n(x_0))+d_Y(f_n(x_0),f_n(x))+d_Y(f_n(x),f(x))$. I know that there is $N \in \mathbb N$: $\forall n \geq N$ $d(f(x_0),f_n(x_0))+d_Y(f_n(x_0),f_n(x))+d_Y(f_n(x),f(x))<\dfrac{\epsilon}{3}+d_Y(f_n(x_0),f_n(x))+\dfrac{\epsilon}{3}$.

By the equicontinuous hypothesis, there is $\delta>0$: if $d_X(x,x_0)<\delta \implies \forall n \space d_Y(f_n(x),f_n(x_0))<\frac{\epsilon}{3}$. It follows that if $d_X(x,x_0)<\delta$, then $d(f(x_0),f(x))<\dfrac{\epsilon}{3}.$

I think I am doing something wrong since I am not using that each $f_n \in B(X,Y)$, but I don't know where my mistake is.

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Here, the fact that we require the functions to be bounded is not problematic (we can replace the metric on $Y$ by the equivalent one $\min\{d_Y,1\}$). We require boundedness when for example $Y$ is a normed space and we define the uniform norm as $\lVert f\rVert_\infty:=\sup_{x\in X}\lVert f(x)\rVert_Y$.

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