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 $$\inf S = 0$$.