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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 $1<b\in \mathbb{R}$ and $x\in \mathbb{R}$.

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What have you tried? –  Alex Becker Jul 9 '12 at 3:27
    
is the variable $x$ fixed in the definition of your set ? –  saposcat Jul 9 '12 at 3:29
    
@Alex it's trivial that the supremum of the first one is that of the second one. When $x$ is irrational both sups are the same. When $x$ is rational, i have no idea how to show that the supremum of the second one is an upperbound of the first set.. –  Katlus Jul 9 '12 at 3:32
    
@jonas sorry that was a typo –  Katlus Jul 9 '12 at 3:33
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Would you please quantify $f$ and $x$? –  Neal Jul 9 '12 at 3:35
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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 $b^x=\sup\{b^r:r\leq x\}\geq \sup\{b^r:r<x\}$. To finish is to show that $b^x$ is the least upper bound of $\{b^r:r<x\}$, which means that no smaller number is an upper bound. Suppose that $0<y<b^x$. 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 $\{b^r:r<x\}$.

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Thanks. Is it right using bernoulli's inequality to show that for any rational 1<b, $b^{1/n}$ is convergent to 1? –  Katlus Jul 9 '12 at 4:19
    
@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. –  Jonas Meyer Jul 9 '12 at 4:24
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