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Let $(f_n)$ be a sequence of functions defined on $[0,1]$. Show that if $(f_n)$ converges to zero uniformly on $[0,1]$, then for any sequence of points $(x_n)$ with $x_n \in [0,1]$ for every $n$, the sequence $f_n(x_n)$ has limit zero.

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    You can prove, as a previous exercise, that $f_n\to f$ uniformly in $A$ if and only if the sequence of numbers $\sup_{x\in A} \vert f_n(x)-f(x)\vert$ converges to $0$. Then, look at the Nate's comment.2012-01-25

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Since $\{f_n\}$ converges uniformly to $0$, given $\epsilon>0$, there is an $N$ so that $|f_m(x)|<\epsilon$ for all $m>N$ and all $x\in[0,1]$.

To show that the sequence $\{f_n(x_n)\}$ has limit $0$, you need to show that for every $\epsilon>0$, there is an $N$ so that $|f_m(x_m)|<\epsilon$ for all $m>N$.

Can you see how to put these together to get what you want?

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    @phillips0023 I'm sorry; what function? The limit function is the zero function: $|f_m(x_m)-0|=|f_m(x_m)|$. From the first paragraph, you can make $|f_n(x)|$ as small as you like $for\ all$ numbers $x\in[0,1]$ as long as $n$ is sufficiently large, and that includes the $x$'s in the given sequence.2012-01-25