Set of Integers Bounded Below by Integer has Smallest Element

Theorem
Let $\Z$ be the set of integers.

Let $\le$ be the ordering on the integers.

Let $\O \subset S \subseteq \Z$ such that $S$ is bounded below in $\struct {\Z, \le}$.

Then $S$ has a smallest element.

Proof
We have that $S$ is bounded below in $\Z$.

So:
 * $\exists m \in \Z: \forall s \in S: m \le s$

Hence:
 * $\forall s \in S: 0 \le s - m$

Thus:
 * $T = \set {s - m: s \in S} \subseteq \N$

The Well-Ordering Principle gives that $T$ has a smallest element, which we can call $b_T \in T$.

Hence:
 * $\paren {\forall s \in S: b_T \le s - m} \land \paren {\exists b_S \in S: b_T = b_S - m}$

So:

So $b_S$ is the smallest element of $S$.

Also see

 * Set of Integers Bounded Below has Smallest Element
 * Set of Integers Bounded Above by Integer has Greatest Element
 * Well-Ordering Principle