Let $x \in \mathbb{R}$ and let $S = \{q \in \mathbb{Q} : q < x\} \subset \mathbb{Q}$.
Prove that $x = \sup{S}$.
Proof: Assume for sake of contradiction that $x \neq \sup{S}$. Then by definition of supremum, there exists $x' \in \mathbb{R}$ such that $x' \geq q$ for every $q \in S$, and if $y$ is an upper bound of $S$ then $x' \leq y$. Since $x$ is an upper bound of $S$ but $x \neq \sup{S}$ we must have $x' \neq x \implies x' < x$. Then by the density of $\mathbb{Q}$ in $\mathbb{R}$ there exists $q' \in \mathbb{Q}$ such that $x' < q' < x$. Clearly then $q' \in S$ but since $x' < q'$ therefore $x'$ is not an upper bound of $S$, contradicting our assumption. Conclude that $x = \sup{S}$. $\Box$
Is this proof correct?
I have shown that there cannot be an upper bound which is less than $x$, so therefore $x$ must be the supremum.
Best Answer
Yes, it is correct. Another way of proving this is noting that $x$ is an upper bound of $S$. Assume there is another upper bound $x'$ such that $x' < x$. Do the same contradiction by density of $\mathbb Q$ in $\mathbb R$. Then $x$ is the least upper bound, so $x = \sup S$.