I am studying from Rudin's Principles of Mathematical Analysis 2nd Edition rather than the much more common 3rd Edition. There are some different problems between the two, and as far as I can find, no solutions available anywhere.
Let $x \in \mathbb{R}$, $x > 1$, $y \in \mathbb{R}$, $y > 0$. And Let $A$ be the set of all $x^q$, $q\in \mathbb{Q}$, $q < y$. We define $x^y = sup(A)$.
We have to prove that for $x,y,z \in \mathbb{R}$, $1 < x < z$, $y > 0$ then $x^y < z^y$.
I have tried a couple of strategies to prove this, but can't quite push it over the finish line. I think the closest I came without using a bunch of stuff not covered in the chapter was the following.
Proof Attempt: Let $x,y,z \in \mathbb{R}$, $1 < x < z$, $y > 0$. And let $A$ be the set of all $x^p$ for $p \in \mathbb{Q}$, $p < y$. Then $x^y = sup(A)$ by definition. And let $B$ be the set of all $z^q$ for $q \in \mathbb{Q}$, $q < y$. Then $z^y = sup(B)$ by definition. We have $x^p < z^p$ for all rational $p < y$. We also have $z^p \leq sup(B) = z^y$. So, $z^y$ is an upper bound of $A$.
Now, I run into an issue here, because I need to show that $z^y \neq x^y$. I think by doing so, we would know that $z^y > x^y$ since $x^y$ is the least upper bound of $A$. But I'm not sure how to go about doing this.
I am really just looking for a hint to point me in the right direction.
EDIT 1:
Following the suggestion of @Kenta. I attempted to prove $(a^b)^c = a^{bc}$ for $a,b,c\in \mathbb{R}, a > 1, b,c > 0$
Proof Attempt
Let $x \in \mathbb{R}, x > 1$. And let $x,y \in \mathbb{R}, x,y > 0$.
Let $A = \{x^p : p \in \mathbb{Q} \land p < y\}$. Then $x^y = sup(A)$.
Let $B = \{x^q : q \in \mathbb{Q} \land q < yz\}$. Then $x^{yz} = sup(B)$.
Let $C = \{(x^y)^r : r \in \mathbb{Q} \land r < z\}$. Then $(x^y)^z = sup(C)$.
We can write $q = pr$ for some $p,r \in \mathbb{Q}, p < y, r < z$ then for every element in $B$, we can write $x^q = x^{pr} = (x^p)^r$ since we have already proven that $x^{pr} = (x^p)^r$ for $p,r \in \mathbb{Q}$ And we have $x^p < x^y$ since $p < y$ and $(x^p)^r < (x^y)^r < (x^y)^z$ since $x^p < x^y$ and since $r < z$. So, we know that $(x^y)^z$ is an upper bound of $B$. But again, here I run into issues proving that $sup(B) = (x^y)^z$.
EDIT 2
I did see this post. But, I don't think I can prove that
$$sup\{(sup\{a^p : p \in \mathbb{Q} \land p < y\})^r : r \in \mathbb{Q} \land r < z\}$$
$$ = sup\{sup\{(x^p)^r : p \in \mathbb{Q} \land p < y\} : r \in \mathbb{Q} \land r < z\}$$
My trouble with proving this equality is that we use some approach like this:
Let $A = \{x^p : p \in \mathbb{Q} \land p < y\}$ then $x^y = sup(A)$. And let $B = \{(x^p)^r : p \in \mathbb{Q} \land p < y\}$ for some $r \in \mathbb{Q}, r < z$. Then $\forall (x^p)^r \in B$, we have $(x^p)^r < (x^y)^r$ since $p < y$. So, $(x^y)^r = (sup(A))^r$ is an upper bound of $B$. So, $(x^y)^r \geq sup(B)$. Then I want to do something like:
Suppose $\alpha < (x^y)^r$. Then we can write $\alpha = x^\beta$ for some $\beta \in \mathbb{R}$ and then somehow I want to get to $\exists (x^s)^t, s \in \mathbb{Q}_+ : \alpha < (x^s)^r < (x^y)^r$ which would mean $s < y$ and $(x^s)^r \in B$ so that any $\alpha < (x^y)^r$ is not an upper bound of $B$, so $(x^y)^r = sup(B)$. However, I don't think I can make that step to find $\alpha < (x^s)^t < (x^y)^t$.
Any help is much appreciated.
