This problem comes from baby Rudin, Chapter 1, Number 6(c).
For the problem, given the condition $b > 1$, one is asked to show that for rational $r$, $b^r = \sup B(r),$
where $B(x) = \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$.
I have demonstrated this property. However, the problem statement concludes,
Hence, it makes sense to define $b^x = \sup B(x)$ for every real $x$.
I am not sure I entirely follow this statement. Why does it make sense to define $b^x$ in this way?
What I have tried to do so far is the following:
For $x$ rational, the statement is given by the above analysis for $b^r$. For $x$ irrational, define $x$ by its Dedekind cut, and let $A = \left\{ s \mid s \in \Bbb Q, s < x \right\}$. Then, for any $a \in A$, $b^a = \sup B(a)$, and since for $a' < a$, $b^{a'} \in B(a)$, then $b^a > b^{a'}$ by the definition of supremum. If $b^x$ is not the supremum, then there exists some $y$ such that $b^a \le b^y for all $a$.
If $y$ is rational, then by the previous argument, $y = a$. But because $A$ has no greatest element, then there is some $y' > y$ that contradicts that $b^y$ is an upper bound. Thus, $y$ must be irrational. However, if $y
However, I have not yet shown that $b^{s'} > b^y$, because the current exercise is to define irrational exponents, so I cannot legitimately make that argument.
Am I on the right track? Have I over-complicated it, or am I missing something?
Edit: I should add, this is not homework.
Second edit: $b > 1$.