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Let ($\mathbb{X}$,d) and ($\mathbb{Y}$,d') metric spaces. $\mathbb{X}$ is compact. Prove that: if $\mathcal{F}$ is an equicontinuous family of function from $\mathbb{X}$ to $\mathbb{Y}$ then $\mathcal{F}$ is uniformly equicontinuous.

This is what i thought: For every secuence on $\mathbb{X}$, let say $(x_n)_n$, there is a subsequence $(x_{n_k})$ of $(x_n)$ that converges. And also i know that $\mathcal{F}$ is equicontinuous, so every function there is continuous, so i can use that $f(x_{n_k}) \rightarrow f(x)$...but, i that doesn't take to the right proof...Any suggestion?

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Pick $\varepsilon > 0$. For each $x \in X$ pick a neighbourhood $U_x$ such that $\mathcal{F}$ is equicontinuous at $x$ for $\frac{\varepsilon}{2}$. Let $\delta$ be the Lebesgue number for the cover $\{U_x: x \in X\}$. This $\delta$ works (check this!).

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