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If $(f_n)$ is a sequence of uniformly continuous functions on an interval $(a,b)$ and if $f_n \rightarrow f$ uniformly on $(a,b)$, then $f$ is also uniformly continuous on $(a,b)$.

my solution: Say $\epsilon > 0$. By hypothesis, Take $N \in \mathbb{N}$ such that

$ |f_n(x) - f(x)| < \frac{\epsilon}{3}, \forall n > N \text{ and } \forall x \in (a,b) $

in particular, the statement above holds for $n = N + 1$ say. We know also by hypothesis that $f_{N+1}$ is uniformly continuous, therefore we can take $\delta > 0$ such that if $|x - x_0| < \delta$, then

$ |f_{N+1}(x) - f_{N+1}(x_0)| < \frac{\epsilon}{3} ,\forall x,x_0 \in (a,b) $. Now if $|x - x_0| < \delta$, then

$|f(x) - f(x_0)| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(x_0)| + |f_n(x_0) - f(x_0)| < \epsilon $

for every $x,x_0 \in (a,b) $. So $f$ is uniformly continous on $(a,b)$.

Can someone give me feedback please?? Thanks a lot!

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    Your proof will be correct if you remove $\forall x,x_0\in (a,b)$ since this is not true. Instead, it's \forall x,x_0\in (a,b):\left|x-x_0\right|<\delta.2012-12-09

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To make your proof correct, all you need to do is remove "$\forall x,x_0 \in (a,b)$." You're already quantifying over $x$ when you say $|x-x_0| < \delta$, so this is unnecessary and confuses the proof.