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Let $\{f_n\}$ be a sequence of continuous functions $f_n\colon X \to R$ where $X \subset R$. Prove that if every point $x_0 \in X$ has an interval $ I_{x_0}=B\left({x_0 ,\varepsilon _{x_0}}\right)\cap X$ for some $\varepsilon _{x_0 } >0$, such that $\{f_n\}$ converges uniformly in $I_{x_0 } \cap X$, then $\{f_n\}$ it´s also equicontinuous.

Is this result true for a non-countable set of continuous functions? Is the reciprocal true? Is this result true for other kind of topological, or metric spaces? Please help me with this, to start, it's the only that i could not do.

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    Definition of Equicontinuity: We say that the set of continuous functions $ E = \left\{ {f\,:X \to R\,\,\,\,\,f\,continuous} \right\} $ is equicontinuous, if for every $ x_0 \in X $ and $ \forall \varepsilon > 0\,\,\,\exists \,\,\delta > 0\,\,:\,\forall \,x \in \,B\left( {x_0 ,\delta } \right) $ such that $ \left| {f\left( x \right) - f\left( {x_0 } \right)} \right| < \varepsilon $2011-11-12
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    @DavideGiraudo: Why don't you write your comment as a solution? (Also, it seems to me that your argument could be generalized to more general spaces.)2011-11-12
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    @JesseMadnick Yes, it's seems it works in any metric space (at least, we didn't use neither separability nor local compactness of $\mathbb R$).2011-11-12

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