Fix $b>1$.
If $m$, $n$, $p$, $q$ are integers, $n > 0$, $q > 0$, $r = m/n = p/q$, then I can prove that $(b^m)^{1/n} = (b^p)^{1/q}$. Hence it makes sense to define $b^r = (b^m)^{1/n}$.
I can also show that $b^{r+s} = b^r b^s$ if $r$ and $s$ are rational.
If $x$ is real, let us define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t \leq x$.
Then we can also show that $b^r = \sup B(r)$, where $r$ is rational.
Hence it makes sense to define $b^x = \sup B(x)$ for every real $x$.
Now here is my question:
How to prove, using the above scheme, that $b^{x+y} = b^x b^y$ for all real $x$ and $y$?
My effort:
It can be shown that if $A$ and $B$ are two non-empty sets of positive real numbers, if $A$ and $B$ are both bounded above in $\mathbb{R}$, and if the set $C$ is defined as $$C \colon= \{ab \colon a \in A, b \in B \}, $$ then we have $$\sup C = \sup A \cdot \sup B.$$
So using this result, we obtain $$ b^x b^y = \sup B(x) \cdot \sup B(y) = \sup \{ b^r \colon r \in \mathbb{Q}, r \leq x \} \cdot \sup \{ b^s \colon s \in \mathbb{Q}, s \leq y \} = \sup \{ b^{r+s} \colon r \in \mathbb{Q}, s \in \mathbb{Q}, r \leq x, s \leq y \} \leq \sup \{ b^t \colon t\in \mathbb{Q}, t \leq x+y \} = \sup B(x+y) = b^{x+y}. $$
Now how to prove the reverse inequality?