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If we know that a sequence $f_n$ of real-valued functions converges uniformly on the interval $[0,\epsilon]$ for every $\epsilon$ $\in (0,1)$, can we say that the sequence converges uniformly on the half-open interval $[0,1)$? If so, how to prove this fact rigorously? If not, what counter-exmple(s) can we give?

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Hint: Let $f_n(x) = x^n{}{}{}{}$.

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How to rigorously prove that this sequence does not converge uniformly on $[0,1)$? It is evident that it does converge uniformaly on $[0,\epsilon]$ for every $\epsilon \in (0,1)$. – Saaqib Mahmuud Feb 7 '13 at 20:48
@SaaqibMahmuud $\sup_{0\le x <1} x^n = 1$ for every $n$. – mrf Feb 7 '13 at 20:49

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