MATH 3150 Class Notes 01-30-2025

notereal-analysis

Going back to the Archimedean Property

How to Prove $M$ is a Supremum

For more abstract proofs we can prove $M$ is a supremum of $S$ by showing that

1. $M$ is an upper bound of $S$.
2. If $\bar{M}$ is another upper bound of $S$, then $\bar{M}\ge M$.

For more concrete examples, we can instead show,

1. $M$ is an upper bound of $S$.
2. $\forall \epsilon >0,\exists x\in S$ such that $x\le M\le x+\epsilon$.

Proof Examples

Proof

"$⟹$" Prove that if $M=\sup S$, then $\forall \epsilon >0,\exists x\in S$ so that $x\le M\le x+\epsilon$.

If $\bar{M}<M$ then $\bar{M}$ is NOT an upper bound of $S$ $⟹\exists x\in S$ so that $\bar{M}<x$.

Apply this for $\bar{M}=M-\epsilon <M⟹\exists x\in S$ so that $\bar{M}<x$ i.e. $M-\epsilon <x⟹M<x+\epsilon$.

"$⟸$" We know $M$ is an upper bound and $\forall \epsilon >0,\exists x\in S$ such that $x\le M\le x+\epsilon$.

We want to show that if $\bar{M}$ is an upper bond of $S$ then $\bar{M}\ge M$.

So let $\bar{M}$ be an upper bound of $S$, so $\bar{M}\ge x\forall x\in S$. Since $M\le x+\epsilon$ and $x\le \bar{M}$, we have $M\le \bar{M}+\epsilon \forall \epsilon >0$. Thus, $M\le \bar{M}$.