Let $\{a_n\}$ be a bounded sequence of real numbers and let $P$ be the set of limit points of $\{a_n\}$. Prove that $\limsup a_n = \sup P$ and $\liminf a_n = \inf P$

my work:

Since ${a_n}$ is bounded, then there is an $M$ such that ${a_n} \leq M$ for all $n$. By Bolzano-Weierstrass theorem, ${a_n}$ must have a subsequence that converges (in this case to $P$).
Then, $P \leq M$.
$\liminf {a_n} \leq \limsup {a_n} \leq M$.
Since $P \leq M$, then $\liminf {a_n} \leq \inf P \leq \sup P \leq\limsup {a_n} \leq M$.

I'm not sure how to finish though, or if what I've put is correct

Help is greatly appreciated!

  • $\begingroup$ $P$ is a set, one which may contain more than one point. What do you mean when you say that $a_n$ converges to $P$? $\endgroup$ Nov 18, 2013 at 3:05
  • $\begingroup$ subsequences which converges to a set of limit points P. $\endgroup$ Nov 18, 2013 at 3:08
  • $\begingroup$ Sequences don't "converge" to a set of points. If a sequence converges, it converges to a unique limit. $\endgroup$ Nov 18, 2013 at 3:13
  • $\begingroup$ that's why im saying that a subsequence of a_n converges to a certain element of the set of limit points P. $\endgroup$ Nov 18, 2013 at 3:15
  • $\begingroup$ I see, much clearer now, thank you. That's not how things are usually said, though. Honestly, I'm having a hard time of understanding your line of thought. $\endgroup$ Nov 18, 2013 at 3:22

1 Answer 1


This question can be answered by showing that the following two things are true:

  1. If $a_{n_k}\to a \in P$, then $\liminf a_n \leq a \leq \limsup a_n$.
  2. There is a subsequence that converges to $\liminf a_n$, and one that converges to $\limsup a_n$.

From there, we may conclude that since the liminf and limsup are lower and upper bounds of $P$ that are elements of $P$, they must also be the greatest lower bound and least upper bound.

  • $\begingroup$ I appreciate the answer but I don't think using the term "baffles" is necessary. I don't know how I'm suppose to present the properties of lim inf/lim sup in conjunction with the problem. Yes, I understand property 1 and property 2 of what you said. I don't understand why lim inf a_n has to be the inf of P? $\endgroup$ Nov 18, 2013 at 3:36
  • $\begingroup$ I'm sorry if I've offended you. Do you understand why if the lim inf is both a lower bound of $P$ and a member of $P$, it must also be the inf of $P$? $\endgroup$ Nov 18, 2013 at 3:39
  • $\begingroup$ Yes. definitely haha. why is lim inf of a_n a lower bound of P? this is because P is the set of limit points, therefore, ...yeah I get it.. It's just like.. feels somehow wrong, even though it's clearly correct. maybe because it literally employs the definition lim inf/sup and that's it, rather than trying to use some kind of algebra like I did. (if that makes sense). regardless, thank you! $\endgroup$ Nov 18, 2013 at 3:45
  • 1
    $\begingroup$ I do see what you mean. I'm wondering though, what is your definition of $\limsup$? Depending on the definition you start with, this question is either a trivial application of a definition (as you suggest) or something a bit more subtle. Is your definition $\limsup a_n = \lim_{n\to \infty} \sup_{k>n}\{a_n\}$? $\endgroup$ Nov 18, 2013 at 4:02

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