# $\liminf a_n\leq \lim a_{n_{k}} \leq \limsup a_n$

So we need to prove that the limit of any subsequence is sandwiched between the $\limsup$ and $\liminf$ of the original sequence.

The professor started off by showing that there is always a subsequence that converges to $\limsup a_n$, and also there is always a subsequence that converges to $\liminf a_n$. He also showed that $\limsup a_{n_{k}}$ is less than or equal to $\limsup a_n$. This part I also understand. But then how do you use these two facts to conclude that $\lim a_{n_{k}} \leq \limsup a_n$?

• Mode of attack: Assume $\lim a_{n_k} > \limsup a_n$ and try to obtain a contradiction. Oct 9, 2015 at 20:59

$\lim a_{n_k} \le \limsup a_{n_k}$!

• Can you elaborate?
– LKSR
Oct 9, 2015 at 20:35
• As a sequence in $k$, the limit of $a_{n_k}$ as $k$ goes to infinity is at most the limit supremum, with respect to $k$, of the sequence $a_{n_k}$. Maybe it will help you to write $b_k = a_{n_k}$. There is really nothing mysterious going on here!
– Unit
Oct 9, 2015 at 20:43
• Why is the limit at most the limit supremum for any sequence?
– LKSR
Oct 9, 2015 at 20:57
• Oh, because for every $n$, we have $a_n \le \sup \{a_m : m \ge n\}$ because $a_n$ is one of the elements of that set. Now take the limit as $n$ goes to infinity: the inequality is still preserved. By definition, $\lim_{n\to\infty} \sup\{a_m: m\ge n\} = \limsup a_n$.
– Unit
Oct 9, 2015 at 21:02
• assume for sake of contradiction that limsup $a_n$ $<$ lim $a_n$ = $\alpha$. Then there exists $N$ such that for all $n > N$ we have $|a_n - \alpha| < \alpha - \text{limsup} ~ a_n$. This is a contradiction by the definition of limsup since we have elements larger than it as n tends to infinity. Is this proof correct?
– LKSR
Oct 9, 2015 at 21:04