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

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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.

$\blacksquare$


Sources

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