Power of Real Number between Zero and One is Bounded

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Theorem

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

Let $0 < x < 1$.

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


Then:

$\inf S = 0$

and:

$\sup S = 1$

where $\inf S$ and $\sup S$ are the infimum and supremum of $S$ respectively.


Proof

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

Let $x^k \in S$.

As $x < 1$, it follows from Real Number Ordering is Compatible with Multiplication that:

$x^{k + 1} < x^k$

So:

$\forall x \in S: x \le 1$

Hence it follows that $\sup S = 1$.


As $x > 0$, it follows 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:

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

and:

$\forall n \in \N: \paren {\dfrac 1 x}^n \le \dfrac 1 h$

But as $0 < x < 1$ it follows that:

$\dfrac 1 x > 1$

Thus $\dfrac 1 h$ is an upper bound for $\set {\paren {\dfrac 1 x}^n: n \in \N}$.

From Power of Real Number greater than One is Unbounded Above, this 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

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