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Prove: If a sequence converges, then every subsequence converges to the same limit.

Instead of saying that $n_{k}\geq k> N\implies |a_{n_{k}}-L|<\epsilon$

Can i say that

let $i$ be the smallest positive integer such that $n_{i}\geq N$,

so $n_{k}> n_{i}\geq N\implies |a_{n_{k}}-L|<\epsilon$

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    Didn't you just change the $k$ with an $n_i$? They basically mean the same thing (index of element/element, that is sufficiently close to $L$) – 2017-01-10
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    @Laray They don't always mean the same thing. It depends on the sequence and subsequence. – 2017-01-10
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    When you take the limit as $n_k\to\infty$, It's sufficient to find _one_ and not necessary be smallest. – 2017-01-10
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    @MyGlasses yes, it does not have to be the smallest, but that's not the point here. – 2017-01-10

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