2
$\begingroup$

let $A$ be an infinite subset of $\mathbb R$ that is bounded above and let $u=\sup A$. Show that there exists an increasing sequence $ (x_n) $ with $x_n \in A $ for all $n\in \mathbb N$ such that $u = \lim_{n\rightarrow\infty} x_n$.

If $u$ is in $A$ then the proof is trivial. If $u$ does not belong to $A$ then for any $ \epsilon > 0$ there exists an $ x_1$ in $A$ such that $ u-\epsilon < x_1<u$. By density theorem there exists an $r_1$ lies between $x_1$ and $u$, since $r_1$ is not an upper bound we will find an $x_2$ in A such that $ u-\epsilon < x_1 < r_1< x_2<u$. Continuing this way we will get a monotone increasing sequence and then applying monotone convergence theorem we will get desired result.

  I want to know whether I am right or I am wrong.
$\endgroup$
2
  • 3
    $\begingroup$ Yes, you are right. $\endgroup$ Aug 20, 2018 at 18:05
  • $\begingroup$ Your proof is flawed, take a look at @MohammadRiazi-Kermani's answer below. Also, consider accepting an answer by clicking on the tick mark below the vote buttons if an answer helped resolve your problem. $\endgroup$
    – user279515
    Aug 20, 2018 at 19:40

2 Answers 2

2
$\begingroup$

I am not convinced that your sequence converges to $u$

You have picked an epsilon and formed a sequence between $u-\epsilon $ and $u$.

How do you know that the sequence converges to u.

We know the sequence will converge to a number $l$ such that $u-\epsilon <l\le u$, but how do we know that $l=u$ ?

Why don't you pick the terms of your sequence between $u-1/n$ and $u$ ?

$\endgroup$
5
  • $\begingroup$ This is right, OP's proposed proof is not accurate. $\endgroup$
    – user279515
    Aug 20, 2018 at 19:38
  • $\begingroup$ @ Mohammad Riazi-Kermani Sir, since u is the supremum of the set A but not the sequence $x_n $ so that's why the limit will be some $ l\leq u$ not necessarily u. Is that you wanted to say Sir? $\endgroup$ Aug 21, 2018 at 14:49
  • $\begingroup$ True, that is what I was worried about. $\endgroup$ Aug 21, 2018 at 14:52
  • $\begingroup$ thank you very much Sir for pointing that. out ..but if I find a sequence such that u-1/$n_k $<$ x_k $ <u for all k in $\mathbb N $ such that $x_1 < x_2 <... $ and $ n_1 < n_2 <... $ then using squeeze theorem I can conclude that $(x_k) $ is the desired sequence with limit u.Am I right now sir ? $\endgroup$ Aug 21, 2018 at 15:36
  • $\begingroup$ @suchandaadhikari Yes, you are right with your new approach. $\endgroup$ Aug 21, 2018 at 15:39
0
$\begingroup$

Why is the proof when $u \in A$ trivial? How can we find a strictly increasing sequence that converges to $\sup A$ or to prove that such a sequence exists?

$\endgroup$
1
  • $\begingroup$ I am not asked to find strictly increasing sequence only increasing sequence if u is in A then I will take $ x_n $=u for all n and by the definition of increasing sequence it's not wrong. $\endgroup$ Aug 21, 2018 at 14:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .