# Showing the set with a supremum has an increasing sequence converging to that supremum.

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.
• Yes, you are right. Aug 20, 2018 at 18:05
• 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.
– user279515
Aug 20, 2018 at 19:40

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 $u$ ?

• This is right, OP's proposed proof is not accurate.
– user279515
Aug 20, 2018 at 19:38
• @ 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? Aug 21, 2018 at 14:49
• True, that is what I was worried about. Aug 21, 2018 at 14:52
• 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 ? Aug 21, 2018 at 15:36
• @suchandaadhikari Yes, you are right with your new approach. Aug 21, 2018 at 15:39

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?

• 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. Aug 21, 2018 at 14:33