Showing a sequence of real numbers converges to a supremum

Question

I am trying to prove that any non-empty set $S \subset \mathbb{R}$ which is bounded above has a sequence of numbers $s_n \in S, n = 1, 2, 3, ...$ such that $s_n \rightarrow sup(S)$

My proposed method is to create the set of upper bounds and list them in reverse order i.e. largest upper bound is denoted $u_1$ where $u_1 > u_2>...>sup(S)$. And we list the elements of S in ascending order, i.e. $s_1 <s_2 < ...<s_n$. Since $|u_n|>|s_n|$ for some $s \in S$ we can say that as $n \rightarrow \infty$ $ u_n \rightarrow supS$ and due to the previous inequality we get that $s_n \rightarrow supS$.

My only problem with the proof is that I have not used the exact formal definition of when a sequence of numbers converges to a value. Any suggestions?

Answer 1

You cannot list the upper bounds: there are uncountably many of them. Let $a=\sup S$. You should now distinguish two cases: $a\in S$, and $a\notin S$. If $a\in S$, there is a trivial sequence in $S$ converging to $a$; if you don’t see it immediately, consider Tito Eliatron’s example in the comments.

If $a\notin S$, you will have to use the definition. Use the fact that $a=\sup S$ to show that for each $n\in\Bbb Z^+$ there is an $s\in S$ such that $a-s<\frac1n$, and then use this fact to get a sequence in $S$ converging to $a$.

Answer 2

If $s=\sup S$, then for every $n\in\mathbb N$, then number $s-1/n$ is not an upper bound of $S$, since it is smaller than the least upper bound $s$. Thus there exists an $s_n\in S$, such that $$ s-\frac{1}{n}<s_n\le s. $$ Clearly the sequence $\{s_n\}\subset S$ converges to $s=\sup S$.

Note. If $s=\sup S\not\in S$, then the sequence $\{s_n\}\subset S$ can be constructed to be strictly increasing.