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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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    is the variable $x$ fixed in the definition of your set ?2012-07-09
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    @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..2012-07-09
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    @jonas sorry that was a typo2012-07-09
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    Would you please quantify $f$ and $x$?2012-07-09
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    @jonas i just noticed that i didn't type sup there.. oh my.. Neal $f$:$\mathbb{Q} →\mathbb{R}$ and $x$ is any real number sorry2012-07-09
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    @Katlus: This would be false for many $f:\mathbb Q\to \mathbb R$ (edit: I see the question has been made more specific, but this is in response to the previous comment of 03:39:55.)2012-07-09
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    @Jonas actually i need to show the equality when $f(r)=b^r$, but i wanted some generalized one. I just rather changed it to my original question thanks!2012-07-09
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    @Katlus Are you familiar with continuous functions?2012-07-09
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    @Alex yeah but only with the ones with domain $\mathbb{R}$, not $\mathbb{Q}$ and etc. This is on the first chapter so i think it can be solved without any concept of continuity2012-07-09
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    @Katlus It can, but continuity is the generalization you're looking for. If $x\in\mathbb Q$, your equality holds whenever $f:\mathbb Q\to \mathbb R$ is continuous from below at $x$.2012-07-09
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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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    Thanks. Is it right using bernoulli's inequality to show that for any rational 1$b^{1/n}$ is convergent to 1? – 2012-07-09
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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