Boundedness of Nth Powers

Theorem
Let $x \in \R$ be a real number such that $x > 1$.

Let set $S = \left\{{x^n: n \in \N}\right\}$.

Then:
 * If $x > 1$ then $S$ is unbounded above.


 * If $0 < x < 1$ then $\inf S = 0$ and $\sup S = 1$, where $\inf S$ and $\sup S$ are the infimum and supremum of $S$ respectively.

Proof

 * First, let $x > 1$.

Suppose $S$ were bounded above.

Then $S$ has a supremum $B$.

As $x > 1$, it follows that $\frac B x < B$ and so therefore $\frac B x$ can not be an upper bound.

Therefore $\exists n \in \N: x^n > \frac B x \Longrightarrow x^{n+1} > B$.

So $B$ can not be an upper bound.

From that contradiction it can be concluded that $S$ can not have an upper bound and therefore $S$ is unbounded above.


 * Now suppose $0 < x < 1$.

When $n = 0$ it follows that $x^n = 1$ and so $\sup S \ge 1$.

Now let $x^k \in S$.

Then as $x < 1$, we have $x^{k+1} < x^k$ from Ordering is Compatible with Multiplication.

So $\forall x \in S: x \le 1$ hence it follows that $\sup S = 1$.

Also, note that as $x > 0$, it follows again by Ordering is Compatible with Multiplication that $\forall x \in S: x \ge 0$.

Therefore $x$ is a lower bound of $S$.

Now suppose $h > 0$ is also a lower bound of $S$.

Then $\forall n \in \N: x^n \ge h$.

Then $\forall n \in \N: \left({\frac 1 x}\right)^n \le \frac 1 h$

But as $0 < x < 1$ it follows that $\frac 1 x > 1$.

Thus $\frac 1 h$ is an upper bound for $\left\{{\left({\frac 1 x}\right)^n: n \in \N}\right\}$ which has been shown to be unbounded above.

Therefore there can be no such lower bound $h > 0$ of $S$.

Hence $\inf S = 0$.