Set of Integers Bounded Below by Integer has Smallest Element
Jump to navigation
Jump to search
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:
\(\ds \) | \(\) | \(\ds s \in S: b_S - m \le s - m\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \forall s \in S: b_S \le s\) | Cancellability of elements of $\Z$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds b_S \in S \land \paren {\forall s \in S: b_S \le s}\) | Definition of Smallest Element |
So $b_S$ is the smallest element of $S$.
$\blacksquare$
Also see
- Set of Integers Bounded Below has Smallest Element
- Set of Integers Bounded Above by Integer has Greatest Element
- Well-Ordering Principle
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 5$: The system of integers: Exercise $1$
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 3$: Natural Numbers: Exercise $\S 3.6 \ (3)$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 10.1$: The well-ordering principle