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

Asked Jan 10 '17 at 09:13

Active Jan 10 '17 at 09:13

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

edited Apr 13 '17 at 12:21

Community

asked Jan 10 '17 at 09:13

Little Rookie

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 $L$) – Laray Jan 10 '17 at 09:23
@Laray They don't always mean the same thing. It depends on the sequence and subsequence. – Little Rookie Jan 10 '17 at 09:39
When you take the limit as $n_k\to\infty$, It's sufficient to find one and not necessary be smallest. – Nosrati Jan 10 '17 at 09:54
@MyGlasses yes, it does not have to be the smallest, but that's not the point here. – Little Rookie Jan 10 '17 at 10:00

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