Let $A \subset \mathbb{R}$ and bounded above. Prove that if $s = \sup(A)$ iff $s$ is an upper bound of $A$ and there exists a sequence $(s_n)$ in $A$ which converges to $s$.
($\implies$) Assume $s = \sup(A)$, then by definition $s$ is an upper bound of $A$. Now there are two cases for $A$, 1) it contains a finite number of elements, say $k$ of them. 2) there are an infinite number of elements in $A$. If there is only a finite number of elements, then let $A = \{a_1, a_2, a_3, \cdots, a_k\}$ where $a_i \in \mathbb{R}$ for $i = 1, 2, \cdots, k$. WLOG assume $a_1 < a_2 < \cdots < a_k$, then $\sup(A) = a_k = \max(A)$. So define the sequence $(s_n) = (a_1, a_2, \cdots, a_k, a_k, a_k, a_k \cdots)$, then $(s_n) \rightarrow a_k = \sup(A)$.
I am not exactly sure how to handle the infinite case nor how to approach the ($\Leftarrow$) proof either. If anyone can provide a solution (or a different approach that doesn't split it into cases for the ($\implies$) proof) that would be good.
Best Answer
($\Rightarrow$) We need not distinguish the cases $A$ finite or not finite:
By definition (characterization) of the least upper bound $s=\sup(A)$: $$\left\{\begin{array}\\a\le s,\quad\forall a\in A\;(*)\\\forall \epsilon>0,\quad \exists\; a\in A,\;\; s-\epsilon\le a\le s\\\end{array}\right.$$ so $(*)$ means that $s$ is an upper bound and let $\epsilon=\frac 1n$ then by $(**)$ there's $a_n\in A$ such that $$s-\frac1n\le a\le s$$ and by the squeeze theorem we see that $$\lim_{n\to\infty}a_n=s$$
($\Leftarrow$) It's pretty similar to the first proof and I leave it for you.