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:
 * $\forall \epsilon \in \R_{>0}: \size {a^{x y} - \paren {a^x}^y} < \epsilon$

, suppose that $x < y$.

Consider $I := \closedint x y$.

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

Let $M = \max \set {\size x, \size y}$

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

From Real Polynomial Function is Continuous:
 * $\exists \delta' \in \R_{>0}: \size {a^x - a^{x'} } < \delta' \leadsto \size {\paren {a^x}^{y'} - \paren {a^{x'} }^{y'} } < \dfrac \epsilon 4$

From Power Function on Strictly Positive Base is Continuous:

Further:

and:

Let $\delta = \max \set {\dfrac {\delta_1} {\size x}, \dfrac {\delta_2} M, \delta_3, \delta_4}$.

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

Thus:

Hence the result, by Real Plus Epsilon.