0
$\begingroup$

If $(a_n)_{n\ge 1} \subset \mathbb{R}$ is a bounded sequence, what is the relation between $\limsup\limits_{n\to \infty} a_n$ and $\sup\limits_{n\in \mathbb{N}}a_n$? My intuition tells me that they should be equal, but when I looked this up on Wikipedia I only found that $\limsup\limits_{n\to \infty}a_n\le \sup\limits_{n\in \mathbb{N}}a_n$. I can't produce an example where the inequality is strict and I believe that since $\sup\limits_{n\in \mathbb{N}}a_n$ is the limit of some subsequence of $(a_n)_{n\ge 1}$ and $\limsup\limits_{n\to \infty}a_n$ is the maximum of the limit points we should always have equality. So, please show me an example where this isn't true or prove that it is always true.

$\endgroup$
1
$\begingroup$

Take any strictly decreasing and convergent sequence $(a_n)_{n\in\mathbb{N}}\subseteq\mathbb{R}$ and let $a$ be the limit of $(a_n)_{n\in\mathbb{N}}$. Then $a$ is the only accumulation point of $(a_n)_{n\in\mathbb{N}}$, hence $\limsup_{n\to \infty} a_n = \lim_{n\to\infty} a_n= a$. As $(a_n)_{n\in\mathbb{N}}$ is strictly decreasing by assumption, we have $a_1>a_2>...>a$, so that $\sup_{n\in \mathbb{N}} a_n =a_1>a=\limsup_{n\to \infty} a_n$.

For instance, the sequence $a_n:=\frac{1}{n}$, $n\in \mathbb{N}$ (with $a:=0$) has such properties.

$\endgroup$
7
  • $\begingroup$ thank you, that is a great example! However, I don't get it: shouldn't we have a subsequence of $a_n$ tend to $a$? $\endgroup$
    – TheZone
    Nov 10 '20 at 15:53
  • $\begingroup$ Yes, but trivially, $a_n$ is a subsequence of itself. $\endgroup$
    – ym94
    Nov 10 '20 at 15:58
  • $\begingroup$ Sorry, I misread your notations. I meant that we should have a subsequence of $a_n$ that tends to $\sup\limits_{n\in \mathbb{N}}a_n$ $\endgroup$
    – TheZone
    Nov 10 '20 at 16:04
  • $\begingroup$ In general, if $a_n$ is a sequence, you don't have a subsequence that tends to $\sup a_n$. The example above is a counter example: Consider any strictly decreasing sequence. Then $\sup a_n =a_1$ and there is no subsequence of $a_n$ converging to $a_1$. $\endgroup$
    – ym94
    Nov 10 '20 at 16:08
  • $\begingroup$ I think that the result I had in mind was that we can form a sequence of elements from $\{x_n | n \in \mathbb{N}\}$ that tends to the supremum, isn't it? $\endgroup$
    – TheZone
    Nov 10 '20 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.