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.
Best Answer
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 $u$ ?