I've searched for the discussion of proving the continuity of exponential function, in most cases the function is defined by power series or inverse of log function where the log is defined by integration of $1/x$.
Does anyone know the prove (or sketch of prove) of $a^x$ being continuous when considered as real values function, where $a^x$ is defined the ordinary way, that is,
When $x$ is positive integer, $a^x$ is $a$ multiply x times, $a^{-x}=\frac{1}{a^x}$, $a^{1/x}$ is the unique number $b$ that satisfies $b^x=a$. For general real number r, define $a^r=sup\{a^q, q\in Q, q\leq r\}$
Here is a very rough outline: Take $a>1$. The case $a<1$ is dealt with similarly. Then the map $x\mapsto a^x$ is increasing on the rationals, and satisfies the group law $a^{x+y}=a^x a^y$ for rational $x$ and $y$. The only sort of discontinuity available to an increasing function is a jump discontinuity, with the jump being from a lower value on the left to a higher value on the right. If there is a jump discontinuity at any (possibly irrational) $x$, you can use the group law to infer a jump discontinuity at $qx$ for any rational $q$. You end up with more jumps than an increasing function (which must necessarily be locally bounded) can possibly accomodate. Details left to the reader as an exercise.