Lower bound, bounded below, bounded set

Throughout, $\mathbb{R}$ denotes the complete ordered field (Complete ordered field (least-upper-bound property)) and $S \subseteq \mathbb{R}$ is a subset of it.

The notions upper bound and bounded above are already fixed by Complete ordered field (least-upper-bound property) and are only recalled here, never redefined: $u \in \mathbb{R}$ is an upper bound of $S$ if $s \le u$ for all $s \in S$, and $S$ is bounded above if it has at least one upper bound. The dual notions are:

• $\ell \in \mathbb{R}$ is a lower bound of $S$ if $\ell \le s$ for all $s \in S$.
• $S$ is bounded below if it has at least one lower bound.
• $S$ is bounded if it is both bounded above and bounded below, that is, if there are $\ell, u \in \mathbb{R}$ with $\ell \le s \le u$ for every $s \in S$.

• A bound is an element of $\mathbb{R}$ and is not required to lie in $S$. A bound that does lie in $S$ is a maximum or a minimum (Maximum and minimum of a set), and that is a strictly stronger condition (FALSE: the supremum of a set belongs to the set).
• Bounds come in half-lines: if $\ell$ is a lower bound of $S$ then so is every $\ell'' \le \ell$, and if $u$ is an upper bound then so is every $u'' \ge u$. Consequently a set that has one bound of a given kind has infinitely many, and the interesting question is whether the collection of them has a best element, which is what a supremum (Complete ordered field (least-upper-bound property)) or an infimum (Greatest lower bound (infimum)) is.
• Bounded above and bounded below are independent conditions. The set $A = \{\, n \cdot 1_{\mathbb{R}} : n \ge 1 \,\}$ of canonical naturals of $\mathbb{R}$ is bounded below by $0$ (Canonical naturals are positive and strictly increasing) and is not bounded above (Every complete ordered field is Archimedean); its reflection $-A$ is bounded above and not bounded below (Reflection through zero exchanges upper and lower bounds).
• The empty set is bounded, and vacuously so: every real number is both an upper bound and a lower bound of $\emptyset$, since the defining condition quantifies over no elements. Having bounds is therefore much weaker than having a least upper bound or a greatest lower bound (FALSE: every subset of $\mathbb{R}$ has a supremum).