Sequence of Powers of Number less than One/Necessary Condition/Proof 4

Proof
Define:
 * $\ds L = \inf_{n \mathop \in \N} \size x^n$

By the Continuum Property, such an $L$ exists in $\R$.

Clearly, $L \ge 0$.

$L > 0$.

Then, by the definition of the infimum, we can choose $n \in \N$ such that $\size x^n < L \size x^{-1}$.

But then $\size x^{n + 1} < L$, which contradicts the definition of $L$.

Therefore, $L = 0$.

Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

By the definition of the infimum, there exists an $N \in \N$ such that $\size x^N < \epsilon$.

It follows that:
 * $\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$

where Absolute Value Function is Completely Multiplicative is applied.

Hence the result, by the definition of a limit.