Let $S$ be a bounded set. Prove there is an increasing sequence ($s_n$) of points in $S$ such that $\lim s_n = \sup S$. Note: If $\sup S$ is in $S$, it’s sufficient to define $s_n = \sup S$ for all $n$.
What I have tried so far:
Let $M=\sup S, L = \lim\, s_n,\,|s_n-L|<\epsilon/2$, then $M-\epsilon < L- \epsilon/2 < s_n < M < M+\epsilon$. But I get this based on $M <= \epsilon+L$, because $\epsilon+L$ is an upper bound for $s_n$.
However, as $s_n$ are just points in $S$, so I don't know if it's sufficient to say $M$, which is the $\sup$ of $S$, is smaller than $L+\epsilon$.
Best Answer
Since the problem is trivial when $\sup S\in S$ let us assume that this is not the case. Then we can see that we have a member $x_{1} \in S $ such that $\sup S - x_{1} < 1$. Now we construct the sequence $x_{n}$ inductively. For $n > 1$ let us choose $x_{n} \in S$ such that $\sup S - x_{n} < \min((\sup S - x_{n - 1}), 1/n)$. This will ensure that $x_{n}$ is increasing and at the same time it will ensure that $x_{n} \to \sup S$ as $n \to \infty$.