Why must there exist a sequence $r_k ∈ R$ such that $r_k→r$?

Question

If $R⊂ℝ$ is bounded, then there exists some $r=\sup R$ . As a supremum, why must there exist a sequence $r_k ∈ R$ such that $r_k→r$?

Answer 1

Well I think you should recall that $\sup(R)$ is the least upper bound of $R$. So it is an upper bound of $R$ such that for any $a < \sup(R)$, there is an $x \in R$ with $a < x \leq \sup(R)$.

Then for each $n \in \mathbb{N}$, let $ \sup(R)- \frac{1}{n}< x_n \leq \sup(R)$. Where does $(x_n)$ converge to?

Answer 2

Pick $\epsilon > 0$. Suppose there is no $x\in R$ such that $|r-x| = r-x < \epsilon$. Then $x \leq r-\epsilon$ for all $x\in R$, so by definition, $r = \sup R \leq r-\epsilon \implies -\epsilon \geq 0$, a contradiction.

To show such a sequence exists, for all $k\in \Bbb{N}$ just pick $r_k\in R$ such that $r-r_k < \frac{1}{k}$. Then $r_k\to r$.