Rudin proof verification Excercise 1.6c 
$\bf Exercise\, 1.6$
Fix $b>1.$
If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t\leq x$. Prove that
$$b^r =\sup B(r)$$ where $r$ is rational. Hence it makes sense to define
$$b^x =\sup B(x)$$ where $x$ is real.

Problem
The method to such proofs is generally

*

*Show that a number is the upper-bound.

*Show that number is the least upper-bound.

Much of the proofs I have seen tackle the following in this manner.

*

*$b^r = b^tb^{r-t}\geq b^t$

*Recall $b^r \in B(r) $ hence $b^r = \max B(r) = \sup B(r) $
But this proof (the 2. part) cannot be extrapolated to $x$ which is not necessarily rational since for any irrational $x$ $b^x \notin B(x)$ hence $b^x \neq \max B(x)$
Are my concerns correct? If yes, can somebody provide an alternative route for proving for $r$?
(It would be very much helpful if somebody can prove it using just the established axioms till the first chapter in Baby Rudin.)
 A: You are missing the point of the exercise.
The exercise is to extend the definition of $b^n; n \in \mathbb N$ to $b^x; x \in \mathbb R$.
At this point $b^n$ is defined to mean "$b$ multiplied by itself $n$ times".  This fine but it's not a very useful definition.  What if $x \not \in \mathbb N$.  What does $b^x$ mean then?
So we extend the definition.  Well we defined $b^{\frac 1n}$ to mean the positive number $c$ so that $c^n = b$.  This had NOTHING whatsoever to do with $b^n$ equaling $b*b*b... *b$ and the fact that they both looked like $b^{something}$ was entirely coincidental.
So we have $b^n = b*b*...*b$ by definition. And $b^x$ is utterly undefined.  We'll define $b^{\frac nm}$ as $c^n$ where $c^m = b$.  Now, wait, you ought to be saying.  That's an ENTIRELY different definition and has nothing to do with the old definition $b^n = b*b*b....*b$.
But that's okay, because we are extending a definition.  We just have to prove that with the new definition of $b^{\frac nm} = c^n; c^m =b$ that IF $\frac nm = k \in \mathbb N$ then $b^{\frac nm} = c^n$ will also be so that $b^{\frac nm} = b*b*b*...*b$ $k$ times.
If so, the new definition does 1) agrees with the old one and 2) allows for the term $b^x$ to be defined for more cases of $x$.
It does.
Okay.  So $b^r; r\in \mathbb Q$ is defined.  But $b^x; x \not \in \mathbb Q$ is NOT defined.
We need to extend the definition.
So we do that by DEFINING $b^x := \sup \{b^r| r\in \mathbb Q; r \le x\}$.
That will be our new definition and it will be a good one if it does the two things:
1) agrees with the old one and 2) allows for the term $b^x$ to be defined for more cases of $x$.
Well, it certainly does number 2) but does it do number 1)?  That's what we have to prove.
If $r \in \mathbb Q$ does $b^r = \sup B(r)$.  If so then the new definition agrees with the old one.
So you prove it for $r\in \mathbb Q$.
!!!!YOU DO NOT HAVE TO PROVE $b^x = \sup B(x)$ if $x \not \in \mathbb Q$.  You CAN'T prove it even if you wanted to.... because that is the DEFINITION of $b^x; x \not \in \mathbb Q$.
A: Since $b>1$, the function $t\mapsto b^t$ is increasing. So$$b^r=\sup_{t\leqslant r}b^t=B(r).$$
