Boundedness of Nth Powers

Theorem

Let $x \in \R$ be a real number.

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.

• If $x = 1$, then of course $x^n = 1$ for all $n \in \N$.

Proof

$x$ greater than $1$

First, let $x > 1$.

Suppose $S$ were bounded above.

Then $S$ has a supremum $B$.

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

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

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.

$\blacksquare$

$x$ less than $1$

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

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

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

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

Hence $\inf S = 0$.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 3.6 \ (2)$