Set of Integers Bounded Above by Integer has Greatest Element

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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 above in $\struct {\Z, \le}$.


Then $S$ has a greatest element.


Proof 1

$S$ is bounded above, so $\exists M \in \Z: \forall s \in S: s \le M$.

Hence $\forall s \in S: 0 \le M - s$.

Thus the set $T = \set {M - s: s \in S} \subseteq \N$.

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

Hence:

$\paren {\forall s \in S: b_T \le M - s} \land \paren {\exists g_S \in S: b_T = M - g_S}$

So:

\(\ds \) \(\) \(\ds \forall s \in S: M - g_S \le M - s\)
\(\ds \) \(\leadsto\) \(\ds \forall s \in S: -g_S \le -s\) Cancellability of elements of $\Z$
\(\ds \) \(\leadsto\) \(\ds \forall s \in S: g_S \ge s\)
\(\ds \) \(\leadsto\) \(\ds g_S \in S \land \paren {\forall s \in S: g_S \ge s}\) which is how the greatest element is defined.


So $g_S$ is the greatest element of $S$.

$\blacksquare$


Proof 2

Since $S$ is bounded above, $\exists M \in \Z: \forall s \in S: s \le M$.

Hence we can define the set $S' = \set {-s: s \in S}$.

$S'$ is bounded below by $-M$.

So from Set of Integers Bounded Below by Integer has Smallest Element, $S'$ has a smallest element, $-g_S$, say, where $\forall s \in S: -g_S \le -s$.

Therefore $g_S \in S$ (by definition of $S'$) and $\forall s \in S: s \le g_S$.

So $g_S$ is the greatest element of $S$.

$\blacksquare$


Also see


Sources

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