Understanding a statement: Uniform convergence of functions I'm trying to understand the following statement:

Let $f$ be a real function defined on an interval $I$ and let $\{f_n\}$ a sequence of functions that converges uniformly to $f$ on $I$. For each $k \in \Bbb N$, there is a subsequence $\{f_{n_k}\}$ such that $|f_{n_k}(x)-f(x)|<2^{-k}$ for all $x \in I$.

I can't understand it because they say that for every fixed $k \in \Bbb N$, there is a subsequence $\{f_{n_k}\}$, and here the subsequence's index is $k$, so I think it doesn't make sense, shouldn't they use an index other than $k$? Maybe I'm misinterpreting the statement, but I'm not sure.
Could someone please explain this to me?
Thank you so much in advace. 

You're right, the way it is written is simply wrong. What they are saying is that we can find a subsequence $\langle f_{n_k}\rangle$ such that $|f_{n_k}-f|<2^{-k}$ over $I$.
Proof Since $f_n\to f$ uniformly on $I$, for each $\varepsilon >0$ there exists $n_\varepsilon$ such that $n\geqslant n_\varepsilon$ implies $$|f_n(t)-f(t)|<\varepsilon$$ over $I$. Now choose $\varepsilon =2^{-k}$. Then you get for each $k=1,2,3,\ldots$ an integer $n_k$ such that...? You have to take care in making $n_1<n_2<n_3<\cdots$, but that is not hard: if $n_1$ has been chosen and you get a $n_2'$, let $n_2=\max\{n_1,n_2' \}+1$. Continue in this fashion.
