(Theorem) Elements in a subset of the reals can be arbitrarily close to the least upper bound

From Honors Mathematics B, Homework 1.

Theorem: Elements in a subset of \mathbb R can be arbitrarily close to the supremum and infimum

Suppose $S\subset \mathbb{R}$ is a set of real numbers, and that $x\in \mathbb{R}$ is an upper bound for this set (i.e., for all $s\in S$, we have $s\le x$). Then

$$x=\mathsf{lub}(S)⟺{\forall }_{\begin{array}{c}ϵ\in \mathbb{R}\\ ϵ>0\end{array}}{\exists }_{s\in S}x\le s+ϵ.$$

That is, the given upper bound $x$ is the least upper bound of $S$ if and only if, for all real numbers $ϵ>0$, there exists some $s\in S$ for which $x\le s+ϵ.$

Proof. We can prove this statement by proving the contrapositive

$$x\ne \mathsf{lub}(S)⟺{\exists }_{\begin{array}{c}ϵ\in \mathbb{R}\\ ϵ>0\end{array}}{\forall }_{s\in S}[x>s+ϵ].$$

For the forward implication, let $\ell =\mathsf{lub}(S)$ with $x>\ell$, so $x$ is not the least upper bound for $S$ and $x-\ell >0$ is strictly positive. Observing that the average of $0$ and $x-\ell$ has $0<\frac{x-\ell }{2}<x-\ell$, we can choose $ϵ=\frac{x-\ell }{2}$. Since the upper bound $\ell$ is defined to have $s\le \ell$ for all $s\in S$, we have

$$s+\frac{x-\ell }{2}\le \ell +\frac{x-\ell }{2}<\ell +(x-\ell )=x.$$

This shows that there exists $ϵ$ for which $s+ϵ<x$ is a strict inequality, which completes the forward proof.

For the reverse implication, we can show the given upper bound $x$ is greater than some other real upper bound for $S$. The hypothesis $x>s+ϵ$ implies that the average of $s+ϵ$ and $x$ has

$$s+ϵ<\frac{s+ϵ+x}{2}<x.$$

We see the average is a real number strictly greater than all of $s$, so it precisely an real upper bound for $S$ that is strictly smaller than $x$. Since $x$ is greater than some other upper bound for $S$, it cannot be the least upper bound, which completes the reverse proof.