corollary to the completeness axiom

Question

The corollary states "Every nonempty subset $S$ of $\mathbb{R}$ that is bounded below has a greatest lower bound inf S.

The part I don't get in the proof is from where they came up with the set $-S$ where $-S=(-s : s\in S)$? Did they create it or used it from somewhere?

Proof: Let $-S$ be the set $(-s : S\in S)$; $-S$ consists of the negatives of the numbers in $S$. Since $S$ is bounded below there is an $m$ in $\mathbb{R}$ such that $m < s$ for all $s\in S$. This implies that $-m \geq -s$ for all $s\in S $, so $m > u$ for all u in the set $-S$ Thus $-S$ is bounded above by $m$.. The Completeness Axiom 4.4 applies to $-S$, so $sup(S)$ exists... There is some more but those parts are left as an exercise.

Who is "they"? Is your question why would one think of considering this set, or something else? — Andrés E. Caicedo, May 20 '13 at 03:59

score 3 · Answer 1 · answered May 20 '13 at 03:39

3

Basically they created it. You are given a set $S$ that is bounded below. The completeness axiom refers to sets that are bounded above, so you need to find a way to use that. Since negation changes the sense of an inequality, it is a way to bring it into the picture.

answered May 20 '13 at 03:39

Ross Millikan

374,822

oh ok got it thanks – user60887 May 20 '13 at 03:40

score 0 · Answer 2 · answered May 20 '13 at 03:41

If you know that there exists a (unique) least upper bound of any set of reals that is bounded above, and you want to prove that there exists a (unique) greatest lower bound of a particular set of reals that is bounded below, then it's only natural to reflect the particular set about $0$ and apply what you already know to the reflected set.

corollary to the completeness axiom

2 Answers2