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Consider a bounded sequence $\{A_n\}_n$ and a subsequence $\{A_{n_k}\}_k \subseteq \{A_n\}_n$. Is it true that $$ \liminf_{n\rightarrow \infty}A_n \geq \liminf_{k\rightarrow \infty}A_{n_k} $$ and $$ \limsup_{n\rightarrow \infty}A_n \geq \limsup_{k\rightarrow \infty}A_{n_k} $$

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  • $\begingroup$ The second one is ok, the first one should be with "$\le$" instead. You can derive it from the second one by using the identity $\liminf_n a_n=-\limsup_n(-a_n)$, for instance. $\endgroup$ – user228113 Dec 15 '15 at 12:45
  • $\begingroup$ In general $\lim \sup A_n ≥ \lim \sup A_{n_k}$, but with $\lim \inf$ you have $≤$ instead of $≥$. $\endgroup$ – s.harp Dec 15 '15 at 12:45
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    $\begingroup$ Check your inequalities, since $A_{n_k}$ is a subset of $A_n$, then $A_n$, with more elements, should have a smaller lower bound. $\endgroup$ – Michael Burr Dec 15 '15 at 12:46
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If the sub-sequence has an accumulation point (infinitely many elements are arbitrarily close to it) then it is an accumulation of the original sequence. So, the lowest accumulation point of the sequence is as low as the lowest accumulation of the sub-sequence and can be even lower. Similarly, the highest accumulation point of the sequence is as high as the highest accumulation point of the sub-sequence and can be even higher.

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