Given $x\in[1,+\infty)$ and $\alpha\in\mathbb{R}$, we define:
$$x^\alpha=\sup\{x^p|p\in\mathbb{Q},p\leq\alpha\}.$$
I want to prove that for all $x\in[1,+\infty)$ and $\alpha,\beta\in\mathbb{R}$, we have that $x^\alpha\cdot x^\beta=x^{\alpha+\beta}$. The proof goes as follows:
Using that the supremum of product of sets of positive numbers is the product of supremums, we've got the following:
$$x^\alpha\cdot x^\beta=\sup\{x^p\cdot x^q|p,q\in\mathbb{Q},p\leq\alpha,q\leq\beta\}=\sup\{x^{p+q}|p,q\in\mathbb{Q},p\leq\alpha,q\leq\beta\}\leq\sup\{x^r|r\in\mathbb{Q},r\leq\alpha+\beta\}.$$
We know the last inequality is true because $\{p+q|p,q\in\mathbb{Q},p\leq\alpha,q\leq\beta\}\subseteq\{r\in\mathbb{Q}|r\leq\alpha+\beta\}$. We need to prove that given $r\in\mathbb{Q}$ such that $r\leq\alpha+\beta$, then there exists $p,q\in\mathbb{Q}$ with $p\leq\alpha$ and $q\leq\beta$ verifying that $r=p+q$ (or at least $r\leq p+q$). This is the part I got stuck.
Thanks in advance.
Let $r<\alpha +\beta$ for a rational $r.$ Equivalently $$ {r\over 2}-\alpha <\beta-{r\over 2}$$ There exists a rational number $u$ such that $$ {r\over 2}-\alpha <u<\beta-{r\over 2}$$ Let $p={r\over 2}-u$ and $q={r\over 2}+u.$ Then $p+q=r,$ $p<\alpha$ and $q<\beta.$