Let $B(x) = \left\{b^t \ | \ t \in \mathbb{Q} ,\ x \in \mathbb{R} \land t \leq x \ \right\}$. Prove that $b^r = \sup B(r)$ if $r$ is rational.
This was my proof:
Proof:
It is clear that $b^r = \max\{B(r)\}$, and is this an upper bound of $B(r)$
The set of all upper bounds of $B(r)$ is
$$U(r) = \left\{x \ | \ x \geq b^r\ \land x \in \mathbb{R}\right\}$$
It thus follows that any $x < b^r$ is not an upper bound of $B(r)$. Therefore $b^r$ must be the least upper bound of $B(r)$, and $b^r = \sup B(r)$
$$q.e.d$$
Are there any logical errors in my proof? Is it fairly rigorous?