Let $A$ be a set in $\mathbb{R}$, non-empty and bounded above. Prove that $s = \sup A$ if and only if $s$ is an upper bound of $A$ and there exists a sequence $(s_n)$ in $A$ which converges to $s$.
I am a little uncertain, given the fact there is no statement suggesting sequences in $A$ are monotone in anyway, so how can we say they are bounded?
Next, I realised it says "there exists a sequence". So this would mean we need only show that there is at least one sequence in the set $A$ that converges?
I started as:
$\Longrightarrow$
Assume $ s = \sup A$
Then let $\epsilon> 0 $ be arbitrary, meaning $|s – \epsilon| < s$ is no longer an upper-bound for the set $A$ since, there exists some $s_N \in A$ such that $|s – \epsilon| < s_N \leq s$
Now here is where I run into trouble, since I can't make the statement
"For some $n \geq N, s_n \geq s_N$ such that
$s – \epsilon< s_N < s_n \leq s < s+\epsilon$"
I can't make a statement about convergence, since there is no condition that we have an increasing sequence in this set.
But since the question says "there exists a sequence which converges", am I allowed to simply say "take some increasing sequence in $A$ that…"? Or am I missing some sort of idea here, and my method/approach is entirely wrong?
Best Answer
Suppose $s=\sup A$. Then, the supremum, being the least upper bound, is certainly an upper bound. To construct the sequence, note that $s-\frac{1}{n}$ is no longer an upper bound and hence we can select
$$s_n \in (s-\frac{1}{n},s) \cap A .$$
The sequence $(s_n) \subset A$ convergences to $s$ because $|s-s_m| \leq \frac{1}{n}$ for all $m \geq n$.
Conversely, suppose $s$ is an upper bound and that $(s_n) \subset A$ converges to $s$. We just need to prove that no number smaller than $s$ remains an upper bound. So let $\epsilon>0$. We must show that
$$A \cap (s,s-\epsilon) \neq \emptyset .$$
But there exists an $N \in \mathbb{N}$ such that $n \geq N \Rightarrow |s_n - s| < \epsilon.$ In particular
$$s_N \in A \cap (s,s-\epsilon) .$$