Sequence converging to the supremum

Question

Please help me, I can't find the theorem anywhere that states something like:

For a bounded set $U\subset\mathbb R$ there exists a non-descreasing sequence $(a_n)_{n\in \mathbb N}$ with $a_n \in U$ with $\lim_{n\to\infty}a_n=\sup U$.

Thank you!

Answer 1

You can prove the theorem (under the assumption that $\sup U$ exists) directly from the definition of supremum. For each $n$ there is some point in $U$ within $1/n$ of the supremum. Use the axiom of choice to choose a sequence $(a_n)$ so that for each $n$, $a_n$ is within $1/n$ of the supremum. Then prove that this sequence converges to the supremum.

Answer 2

Because it follows directly from the definition of supremum. You can create the sequence $\{a_n\}$ easily: Let $M = \sup U $. Assume that $M \notin U$ (otherwise just let $a_i = M \, \forall i \in \mathbb{N}$). Fix $a_0 \in U$. Choose $a_i \in (\frac{M + a_{i-1}}{2}, M) \cap U$ for $i = 1, 2, \dots$ Such $a_i$ must exist since $M$ is the supremum. Otherwise $\frac{M + a_{i-1}}{2} < M$ would be a lower upper bound for $U$.