Proving Supremums

Proving with Archimedean Property

Question

Prove that $sup(A)=2$ where $A=\{1+\frac{1}{n}|n\in N\}$

Proving with Density

Process

If we want to prove supremum $sup(S)=\sqrt{2}$

1. Show that $2$ is an Upper Bound of $S$
2. Show that for all $b$‘s that are Upper Bound, then $\sqrt{2}\le b$

Proof

1. Prove upper bound For all $q\in S$, we have $q^2<2$ implying $q<\sqrt{2}$ (since $\sqrt{\ast }$) is increasing on $[0,\infty )$ Thus $\sqrt{2}$ is an upper bound of $S$
2. Prove least upper bound Assume $b\in R$ is an upper bound of $S$ To derive a contradiction, suppose $b<\sqrt{2}$ Since $Q$ is dense in $R$, there exists $q\in Q$ such that $b<q<\sqrt{2}$ Then $q^2<2$, thus $q\in S$. (Hence $b<q$ for some $q\in S$) This contradicts $b$ being an upper bound of $S$ Therefore, it must be that $\sqrt{2}\le b$ Therefore, $sup(S)=\sqrt{2}$