If $\sup S\not\in S$ then $\{x \in S: x > \sup{S} - \epsilon\}$ is infinite

Question

Let $S\subseteq\mathbb{R}$ be nonempty and bounded above, and let $\beta = \sup{S}$. Suppose that $\beta \not\in S$. Prove that for each $\epsilon > 0,$ the set $\{x \in S: x > \beta - \epsilon\}$ is infinite.

So far I had started with the fact that $\beta$ was the supremum of $S$ and that by definition $\beta - \epsilon$ must be an element of $S.$ Since $S\subseteq\mathbb{R},$ which is infinite, and this set is a subset of $S,$ it seems like this set should be infinite for each $\epsilon.$ But how do I show this?

Answer 1

By definition of $\sup$, each such set is non-empty. Assume that for some specific $\epsilon>0$, the set is finite. Then let $m$ be its maximum. By assumption, $m<\sup S$. What can you say about $\{\,x\in S:x>\sup S-(\sup S-m)\,\}$?

Answer 2

Proposition. Let $ S\subseteq \mathbb{R} $ be nonempty and bounded above. (Because $ \mathbb{R} $ has the least upper bound property) Let $ \beta $ be the least upper bound of $ S $. If $ \beta\notin S $, then $ \left\{x\in S:x>\beta-\epsilon\right\} $ is infinite for every $ \epsilon>0 $.

Proof. Assume $ \beta\notin S $. Let $ \epsilon>0 $ be arbitrary. By contradiction, suppose $ \left\{x\in S:x>\beta-\epsilon\right\} $ is finite. That set is nonempty because otherwise $\beta$ is not the least upper bound of $S$. Because every finite (nonempty) totally ordered set has a greatest element, let $ \beta_0 $ be the greatest element in $ \left\{x\in S:x>\beta-\epsilon\right\} $. Notice $ \beta_0 $ is an upper bound of $ S $. Because $ \beta $ is the least upper bound of $ S $, $ \beta\le \beta_0 $. Notice $ \beta_0\in S $. Because $ \beta $ is an upper bound of $ S $, $ \beta_0\le S $. So $ \beta_0=\beta $. Therefore $ \beta\in S $---a contradiction. As a result, $ \left\{x\in S:x>\beta-\epsilon\right\} $ is infinite.