Showing that $U_x=\left\{a^p:p\in\mathbb{Q},p<x\right\}$ and $V_x=\left\{a^q:q\in\mathbb{Q},q>x\right\}$ are contiguous classes for $a>1$

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

Proof. Let $$U_x=\left\{a^p:\text{p\in\mathbb{Q} and p and $$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 $$p, for every real number $$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?

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 $$\rho = q-p$$.
• 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. – user457568 Dec 30 '18 at 18:21