1
$\begingroup$

Let $\{X_n\}$ be a bounded sequence.

$a)$ Prove that there exists an $s$ such that for any $r>s$ there exists an $M \in \mathbb{N}$ such that for all $n \geq M$ we have $x_n < r$.

$b)$ If $s$ is a number as in $a)$ then prove $limsup \ X_n \leq s$

$c)$ . Show that if $S$ is the set of all $s$ as in $a)$ then $limsup \ X_n = inf \ S$.

My thoughts: Since $\{X_n\}$ is a bounded sequence there exists a convergent subsequences $\{X_{n_i}\}$ converging to $limsup\ X_n=L$. Therefore for all $\epsilon >0$ there exists an $M \in \mathbb{N}$ such that for if $n \in \mathbb{N}$ and $n \geq M$ then $|X_n - L| < \epsilon$. Now we can define $s=sup\{X_k : k \geq M\}$ then by definition $s$ is the least upper bound for $\{X_n\}$ for all $n \geq M$ and so $|X_n| \leq s < r$ and so $X_n < r$ as required.

Now from here I can't think of how to get $b)$ and $c)$ any help is appreciated, thanks!

$\endgroup$
  • $\begingroup$ Try to think it this way: $s$ satisfying property a) means that, for all $r>s$, $x_n<r$ definitely. What can you deduce about the limsup from this? Also, the $s$ you consider is perfectly correct, but it depends on the $M$ you take. What happens varying $M$? In particular, taking $M$ to infinity? $\endgroup$ – Lucio Oct 25 '17 at 19:00
  • $\begingroup$ If I take $M$ to infinity then I get $limsup\ X_n$ which would be less than any other upper bound namely $s$. Is that correct logic to get $b)$? $\endgroup$ – Justin Stevenson Oct 25 '17 at 19:03
  • $\begingroup$ Taking $M$ to infinity will be required for $c)$; my hint for $b)$ was the previous observation. Another hint: take $s$ as in $a)$, take $r>s$. Then $x_n<r$ definitely, which implies $\limsup x_n \leq r$. Now observe that this holds for any $r>s$. What can you conclude? $\endgroup$ – Lucio Oct 25 '17 at 19:32
  • $\begingroup$ Sorry I've thought about it for a bit, but I can't seem to conclude what I need to.. $\endgroup$ – Justin Stevenson Oct 25 '17 at 20:12
  • $\begingroup$ Okay, then I'll explain it: if $\limsup x_n\leq r$ for all $r>s$, then it must be $\limsup x_n\leq s$ as well (you can argue by contradiction, or take $r=s+\varepsilon$ and then take the limit for $\varepsilon\to 0$). In this way you get $a)\Rightarrow b)$. Now $b)$ tells you that $\limsup x_n$ is a minorant of $S$. In order to show that it's the infimum, if suffices to find a sequence of elements in $S$ converging to $\limsup x_n$. That's where the sequence $s_M$, where $s_M$ is how you defined $s$ for an $M$ fixed, comes into play: as you said, $s_M$ converges to $\limsup x_n$. $\endgroup$ – Lucio Oct 25 '17 at 20:23

Your Answer

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

Browse other questions tagged or ask your own question.