Please help me with the proof that $\sup\{b^r\in \mathbb{R}\mid x\geq r\in \mathbb{Q}\} = \sup\{b^r\in \mathbb{R}\mid x\gt r\in \mathbb{Q}\}$ where $1 and $x\in \mathbb{R}$.
Proving the suprema of $\{b^r\mid x\geq r\in\mathbb{Q}\}$ and $\{b^r\mid x\gt r\in\mathbb{Q}\}$ are equal if $b\gt 1$
-1
$\begingroup$
real-analysis
-
0@Alex then would you please tell me how to prove that for $f(r)=b^r$ since i don't know continuity of $\mathbb{Q}$ – 2012-07-09
1 Answers
0
I'm guessing you've already observed that $r\mapsto b^r$ is increasing (otherwise you can show this), and as you mentioned in a comment, there is nothing to show if $x$ is not in $\mathbb Q$. Assume that $x$ is rational, and note that $b^x=\sup\{b^r:r\leq x\}\geq \sup\{b^r:r
-
0@Katlus: There is definitely at least one right way to use Bernoulli's inequality to show that. You don't need $b$ to be rational. – 2012-07-09