Prove $\text{sup}\{x\}^{a}=\text{sup}\{x^{a}\}$ for $a\in\mathbb{Q}$ Assume $X=\{x:x\in\mathbb{Q}\text{ and }x\le t\}$. Here is the proof I've tried for $a\ge 0$:

If $a\gt 0$, then $x^{a}\le t^{a}$. Hence, $t^{a}$ is an upper bound of $\{x^{a} :x\in X\}$. Suppose there is another upper bound $c$ such that $x^{a}\le c\lt t^{a}$. It contradicts the assumption that $t$ is a supremum of $X$, because it follows $x\le c^{\frac{1}{a}}\lt t$.

However, in an attempt to prove for $a<0$, I've got stuck from the beginning because $x^{a}\ge t^{a}$ provides no information of $\text{sup}\{x^{a}:x\in X\}$ in the first place. Then, how can I proceed the proof?

EDIT : 


*

*I'm yet to learn the concept of continuity (I'm on the 3rd chapter of Rudin's Principles of Mathematical Analaysis) . Can you confine yourself to this?

*If the claim is false for a rational $\color{red}{a<0}$ (as PhoemueX suggests in his comment) , this answer does not hold (which requires the claim to be true) ?

 A: This cannot work as you have stated it. The difference between your problem and Rudin's definition is present, yet subtle. First to disprove your statement for $a<0$. Consider the set $\{1,2,3,4\}$. Let $a = -2$. Then your left hand side is going to be $\sup\{1,2,3,4\}^{-2} = 4^{-2} = \frac{1}{16}$. Your right hand side however is $\sup\{1^{-2}, 2^{-2}, 3^{-2}, 4^{-2}\} = \sup\{1,\frac{1}{4},\frac{1}{9},\frac{1}{16}\} =1$. They are clearly not equal. 
Let us now recall Rudin's definition, 
Let $b,x\in\mathbb{R}, b > 0,$. Let $B(b,x) = \{b^t\mid t\le x\}$. Then $b^x = \sup B(x)$. Note here the supremum is taken by varying the exponents themselves.
Your question is concerning the supremum over the bases which is a different concept. It is in fact true that if $x > 0$ and $b\in \mathbb{R}$, $$(x^{a})^b = \sup\{B(x,a)\}^b = \sup \{x^t\mid t\le a\}^b = \sup\{\sup\{x^t\mid t\le a\}^s\mid s\le b\} = \sup\{\sup\{x^{ts}\mid t\le a\}\mid s\le b\} = x^{ab}.$$ So then we need to prove that 
$\sup\{x^t\mid t\le a\}^s = \sup\{x^{ts}\mid t\le a\}$. But this is easy. We may assume that $s\in \mathbb{Q}$ and replace all of our previous sets with the added condition that $t,s\in\mathbb{Q}$. So we rewrite $s = \frac{m}{n}$. Then as per an exercise in rudin Ch. 1, $\sup\{x^t\mid t\le a\}^s = \sup\{x^t\mid t\le a\}^{m/n} = \left(\sup\{x^t\mid t\le a\}^m\right)^{1/n} $. We know also the product of supremums is the supremum of a product, provided those supremums exist. So,  $\left(\sup\{x^t\mid t\le a\}^m\right)^{1/n} = \left(\sup\{x^{t(m)}\mid t\le a\}\right)^{1/n} =(x^{am})^{1/n}$. but this again is $x^{a(m/n)}$ (this step may take a second to see). Thus we have gotten to where we need to be. Just keep in mind that the base is fixed and the exponent varies.
