Exponent Combination Laws/Power of Power/Proof 2

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

Proof
We will show that:
 * $\forall \epsilon \in \R_{>0}: \left\vert{a^{xy} - \left({a^x}\right)^y}\right\vert < \epsilon$

, suppose that $x < y$.

Consider $I := \left[{x \,.\,.\, y}\right]$.

Let $I_\Q = I \cap \Q$.

Let $M = \max \left\{ {\left\vert{x}\right\vert, \left\vert{y}\right\vert} \right\}$

Fix $\epsilon \in \R_{>0}$.

From Polynomial is Continuous:
 * $\exists \delta' \in \R_{>0} : \left\vert{a^x - a^{x'} }\right\vert < \delta' \implies \left\vert{\left({a^x}\right)^{y'} - \left({ a^{x'} }\right)^{y'} }\right\vert < \dfrac \epsilon 4$

From Power Function on Strictly Positive Base is Continuous:

Further:

And:

Let $\delta = \max \left\{ {\dfrac {\delta_1} {\left\vert{x}\right\vert}, \dfrac {\delta_2} M, \delta_3, \delta_4} \right\}$.

From Closure of Rational Interval is Closed Real Interval:
 * $\exists r, s \in I_\Q: \left\vert{x - r}\right\vert < \delta \land \left\vert{y - s}\right\vert < \delta$

Thus:

Hence the result, by Real Plus Epsilon.