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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 $x\in \mathbb{R}$.

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    @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

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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. To finish is to show that $b^x$ is the least upper bound of $\{b^r:r, which means that no smaller number is an upper bound. Suppose that $0. Let $n$ be a positive integer such that $b^{1/n}<\dfrac{b^x}{y}$ (showing that such $n$ exists is a good exercise, the point being that $\dfrac{b^x}{y}>1$). Then $b^{x-1/n}>y$, so $y$ is not an upper bound for $\{b^r:r.

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    @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