Every nonempty set bounded below has an infimum

Given: A nonempty $S\subseteq \mathbb{R}$ that is bounded below, and its reflection $-S=\{-s:s\in S\}$.

The least-upper-bound property of $\mathbb{R}$: every nonempty subset of $\mathbb{R}$ that is bounded above has a least upper bound in $\mathbb{R}$, namely an upper bound that is $\le$ every upper bound (Complete ordered field (least-upper-bound property)).

Reflection: $-(-S)=S$; $S$ is nonempty exactly when $-S$ is; $u$ is an upper bound of a set $X$ exactly when $-u$ is a lower bound of $-X$; and $\mathrm{\ell }$ is a lower bound of $X$ exactly when $-\mathrm{\ell }$ is an upper bound of $-X$ (Reflection through zero exchanges upper and lower bounds).

Greatest lower bound (infimum): $\mathrm{\ell }$ is one for $S$ when $\mathrm{\ell }$ is a lower bound of $S$ and ${\mathrm{\ell }}^{\mathrm{\prime }}\le \mathrm{\ell }$ for every lower bound ${\mathrm{\ell }}^{\mathrm{\prime }}$ of $S$ (Greatest lower bound (infimum)).

A least upper bound and a greatest lower bound are unique when they exist, so the notations $\sup$ and $\inf$ are unambiguous (Suprema and infima are unique).

Negation reverses the order, elementwise: $-(-a)=a$, because $(-a)+a=0$ and additive inverses are unique (Field, Identities and inverses in a field are unique); and $a\le b$ if and only if $-b\le -a$, because translation invariance applied with the constant $-a-b$ turns $a<b$ into $-b<-a$ and, applied with the constant $a+b$, turns $-b<-a$ back into $a<b$, while $a=b$ holds exactly when $-a=-b$ (Order is preserved by adding a constant and by adding inequalities).