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I'm trying to show that $ (U_x,V_x) $ is a pair of contiguous classes.

Proof. Let $ U_x=\left\{a^p:\text{$p\in\mathbb{Q}$ and $p<x$}\right\} $ and $ V_x=\left\{a^q:\text{$p\in\mathbb{Q}$ and $q>x$}\right\} $; obviously is $ a^p\leqq a^q $ (because $ a\mapsto a^\rho $ is increasing for rational $ \rho $). Let $ \xi<\eta $ be two separators for $ U_x $ and $ V_x $: thus we have for all $ a^p $ and $ a^q $ the chain of inequalities $ a^q\leqq\xi $ and $ \eta\leqq a^q $. My textbook says that from there we can derive $ a^q/a^p\geqq\eta/\xi $ and therefore $ a^{q-p}>\eta/\xi>1 $. [...]

I'm okay with the $ >1 $ part, but I don't get from where the author derived the strict inequality between the first two members.

Secondly, assuming what claimed (that $ a^{1-p}>\xi/\eta>1 $) is true, the author states something like

Every positive rational number $ \rho $ can be expressed as $ \rho=q-p $, where $ p<x<q $, for every real number $ x $ (because of $ \mathbb{Q} $ is dense in $ \mathbb{R} $). We can now note that $ \inf\left\{a^\rho:\text{$ \rho\in\mathbb{Q} $ and $ \rho>0 $}\right\} $ equals $ 1 $, and derive a contradiction.

Could someone explain me this apparently tedious passage? How are the density of $ \mathbb{Q} $ and the $ \inf{E}=\inf\left\{a^\rho\dots\right\}=1 $ statements used?

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1) Given that $\xi<\eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $\hat{\xi}, \hat{\eta}$ such that $\xi < \hat{\xi} < \hat{\eta} < \eta$. Now, $ a^p \leqq \xi < \hat{\xi} < \hat{\eta} < \eta \leqq a^q$. Then, $a^{q-p}>\hat{\eta}/ \hat{\xi}$ follows.

2) Given a positive rational number $\rho$ and $x \in \mathbb{R}$, consider the open interval $] x, x+ \rho [$. Then, you can find a rational number $q \in ] x, x+ \rho [$ because $\mathbb{Q}$ is dense in $\mathbb{R}$. Set $p:= q -\rho$, then $\rho = q-p$.

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  • $\begingroup$ Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^p\leqq\xi<\eta\leqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}\geqq\eta/\xi $. I'm looking for how to derive that this inequality is in fact strict. $\endgroup$ – user457568 Dec 30 '18 at 18:21
  • $\begingroup$ Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps. $\endgroup$ – JAskgaard Dec 30 '18 at 18:33

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