Given $x>0$ and $\alpha\in\mathbb{R}$, we define:
\begin{equation} x^{\alpha}=\left\{\begin{array}{ll} \inf\{x^p\mid p\in\mathbb{Q}_{<\alpha}\} & \text{if $0<x<1$}\\ \sup\{x^p\mid p\in\mathbb{Q}_{<\alpha}\} & \text{if $x\geq 1$}\\ \end{array}\right. \end{equation}
I'll prove here that we can get the following result (what I need is someone to check that it's correct the proof):
\begin{equation} x^{\alpha}=\left\{\begin{array}{ll} \sup\{x^p\mid p\in\mathbb{Q}_{>\alpha}\} & \text{if $0<x<1$}\\ \inf\{x^p\mid p\in\mathbb{Q}_{>\alpha}\} & \text{if $x\geq 1$}\\ \end{array}\right. \end{equation}
The proof goes as follows:
First, suppose $x\geq 1$. Let $p\in\mathbb{Q}_{<\alpha}$ and $q\in\mathbb{Q}_{>\alpha}$. Since $p<q$ and $x\geq 1$, we know that $x^p<x^q$. Fixing $q$ and taking supremums in $p$ we have $x^\alpha\leq x^q$. Taking infimums in $q$, we get $x^\alpha\leq\inf\{x^q\mid q\in\mathbb{Q}_{>\alpha}\}$.
By reduction to the absurd, supose that there's $\varepsilon>0$ such that $x^\alpha\leq x^q-\varepsilon$ for all $q\in\mathbb{Q}_{>\alpha}$. Then for all $p\in\mathbb{Q}_{<\alpha}$ and $q\in\mathbb{Q}_{>\alpha}$, $x^p<x^q-\varepsilon$. Let's take any $p\in\mathbb{Q}_{<\alpha}$ and denote by $p:=\dfrac{m}{n}$ for certains $m\in\mathbb{Z}$ and $n\in\mathbb{N}$. Then:
\begin{equation} \dfrac{x^{-n\cdot p}}{x^{-n\cdot q}}\geq\left(1-\dfrac{\varepsilon}{x^q}\right)^{-n}=\left(\dfrac{x^q}{x^q-\varepsilon}\right)^n=\left(1+\dfrac{\varepsilon}{x^q-\varepsilon}\right)^n\geq 1+n\cdot\dfrac{\varepsilon}{x^q-\varepsilon}, \end{equation}
where the last inequality is obtained using Bernouilli's inequality. By Arquimedian Property, we can choose $n:=n(p,q)$ big enough such that $1+n\cdot\dfrac{\varepsilon}{x^q-\varepsilon}>x$. Let $q>\max\left\{\alpha,\dfrac{n\cdot p+1}{n}\right\}$. Then:
$$x<\dfrac{x^{-n\cdot p}}{x^{-n\cdot q}}<\dfrac{x^{-n\cdot p}}{x^{-n\cdot p-1}}.$$
Isolating, we would obtain $x^{-n\cdot p}<x^{-n\cdot p}$ which is a contradiction. Therefore, $x^{\alpha}=\inf\{x^p\mid p\in\mathbb{Q}_{>\alpha}\}$.
Now, suppose $x\in (0,1)$. Using the fact that $\left(y^{-1}\right)^{\alpha}=y^{-\alpha}$ for $y\in(1,\infty)$ (this is proved), we have:
\begin{equation} \begin{split} x^{\alpha} & =\left(\left(x^{-1}\right)^{-1}\right)^{\alpha}=\left(x^{-1}\right)^{-\alpha}=\sup\left\{\left(x^{-1}\right)^p\mid p\in\mathbb{Q}_{<-\alpha}\right\}\\ & =\sup\left\{x^{-p}\mid p\in\mathbb{Q}_{<-\alpha}\right\}=\sup\left\{x^p\mid p\in\mathbb{Q}_{>\alpha}\right\}. \end{split} \end{equation}