Rudin proof verification Excercise 1.6c

Question

$\bf Exercise\, 1.6$

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

1. Show that a number is the upper-bound.
2. Show that number is the least upper-bound.

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

1. $b^r = b^tb^{r-t}\geq b^t$
2. 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.)

Answer 1

Since $b>1$, the function $t\mapsto b^t$ is increasing. So$$b^r=\sup_{t\leqslant r}b^t=B(r).$$

Answer 2

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 $b^x; x \not \in \mathbb Q$.