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 $b\gt 1$
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real-analysis
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0is the variable $x$ fixed in the definition of your set ? – 2012-07-09
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0@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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0@jonas sorry that was a typo – 2012-07-09
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1Would you please quantify $f$ and $x$? – 2012-07-09
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0@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 sorry – 2012-07-09
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0@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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0@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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0@Katlus Are you familiar with continuous functions? – 2012-07-09
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0@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 continuity – 2012-07-09
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0@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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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
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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 $b^x=\sup\{b^r:r\leq x\}\geq \sup\{b^r:r
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0Thanks. Is it right using bernoulli's inequality to show that for any rational 1$b^{1/n}$ is convergent to 1? – 2012-07-09
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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