Set of Numbers of form n - 1 over n is Bounded Above

Theorem
Let $S$ be the subset of the set of real numbers $\R$ defined as:


 * $S = \set {\dfrac {n - 1} n: n \in \Z_{>0} }$

$S$ is bounded above with supremum $1$.

$S$ has no greatest element.

Proof
We have that:


 * $\dfrac {n - 1} n = 1 - \dfrac 1 n$

As $n > 0$ it follows from Reciprocal of Strictly Positive Real Number is Strictly Positive that $\dfrac 1 n > 0$.

Thus $1 - \dfrac 1 n < 1$ and so $S$ is bounded above by $1$.

Next it is to be shown that $1$ is the supremum of $S$.

Suppose $x$ is the supremum of $S$ such that $x < 1$.

Then:
 * $1 - x = \epsilon$

where $\epsilon \in \R_{>0}$.

By the Axiom of Archimedes we have that:
 * $\exists m \in \Z_{>0}: n > \dfrac 1 \epsilon$

and so from Reciprocal Function is Strictly Decreasing:
 * $\exists m \in \Z_{>0}: \dfrac 1 n < \epsilon$

Thus:
 * $1 - \dfrac 1 m > 1 - \epsilon = x$

and so $x$ is not the supremum of $S$ after all.

Thus $\sup S$ cannot be less than $1$.

It follows that:


 * $\sup S = 1$

Next it is noted that:
 * $\forall n \in \Z_{>0}: \dfrac 1 n > 0$

and so:
 * $\forall x \in S: x < 1$

Thus as $\sup S \notin S$ it follows from Greatest Element is Supremum that $S$ has no greatest element.