Prove that $s=\sup A$ iff $a_{n}\rightarrow s$

Question

Let $A$ be a bounded subset of $R$ and $s$ an uperbound of $A$. Show that $s=\sup A$ if and only if there is a sequence $a_{n}$ of points of $A$ such that $a_n\rightarrow s$

Any ideas?

Answer 1

Suppose that $s\neq\sup A$. Let $\varepsilon=s-\sup A$. Then, for each $a\in A$,\begin{align}|s-a|&=s-a\\&=s-\sup A+\sup A-a\\&\geqslant s-\sup A\\&=\varepsilon.\end{align}Therefore, $s$ cannot be the limit of a sequence of elements of $A$.

On the other hand, if $s=\sup A$, then let $n\in\mathbb N$. Take $a_n\in A$ such that $s-a_n<\frac1n$. Then$$(\forall n\in\mathbb{N}):|s-a_n|<\frac1n$$and therefore $s=\lim_{n\to\infty}a_n$.

Answer 2

Suppose $s = \text{sup}(A)$, then for $\epsilon = \dfrac{1}{n}$, there exists $a_n$ such that $a_n > s -\dfrac{1}{n}$, and also $a_n \le s < s+\dfrac{1}{n}$. Thus $|a_n - s| < \dfrac{1}{n}$. This shows $a_n \to s$. Conversely, if $a_n \to s$. Let $\epsilon > 0$ be given. Choose $N_1$ such that $N_1\epsilon > 1$. And choose $N_2$ such that if $n > N_2$ then $a_n - s > -1/n$. Thus if we take: $m = \text{max}(N_1,N_2)$, then: if $n > m$ we have:$a_n \in A$ and $a_n > s - \dfrac{1}{n} > s - \epsilon$. Thus if we let $x = a_{m+1}$, then $x > s -\epsilon$. This proves that $s = \text{sup}(A)$.