# Why is $b^x \overset{\mathrm{def}}{=} \sup \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$ for $b > 1$ a sensible definition?

This problem comes from baby Rudin, Chapter 1, Number 6(c).

For the problem, given the condition $b > 1$, one is asked to show that for rational $r$, $$b^r = \sup B(r),$$

where $B(x) = \left\{ b^t \mid t \in \Bbb Q,\ t \le x \right\}$.

I have demonstrated this property. However, the problem statement concludes,

Hence, it makes sense to define $$b^x = \sup B(x)$$ for every real $x$.

I am not sure I entirely follow this statement. Why does it make sense to define $b^x$ in this way?

What I have tried to do so far is the following:

For $x$ rational, the statement is given by the above analysis for $b^r$. For $x$ irrational, define $x$ by its Dedekind cut, and let $A = \left\{ s \mid s \in \Bbb Q, s < x \right\}$. Then, for any $a \in A$, $b^a = \sup B(a)$, and since for $a' < a$, $b^{a'} \in B(a)$, then $b^a > b^{a'}$ by the definition of supremum. If $b^x$ is not the supremum, then there exists some $y$ such that $b^a \le b^y <b^x$ for all $a$.

If $y$ is rational, then by the previous argument, $y = a$. But because $A$ has no greatest element, then there is some $y' > y$ that contradicts that $b^y$ is an upper bound. Thus, $y$ must be irrational. However, if $y<x$, then by density of the rationals, there is some $s' \in \Bbb Q$ such that $y < s' < x$.

However, I have not yet shown that $b^{s'} > b^y$, because the current exercise is to define irrational exponents, so I cannot legitimately make that argument.

Am I on the right track? Have I over-complicated it, or am I missing something?

Edit: I should add, this is not homework.

Second edit: $b > 1$.

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I'm sorry, but what does $B(x)$ mean in general? – funktor Aug 14 '12 at 16:01
It is defined implicitly in the title. I should be more clear; $B(x) = \left\{ b^t \mid t \in \Bbb Q, t \le x \right\}$. – Emily Aug 14 '12 at 16:04
oh, okay thanks – funktor Aug 14 '12 at 16:26
I'm not sure you've asked a clear question yet. In other words, what mathematical statement(s) are you trying to prove that would justify the sentence "it makes sense to define ..."? Are you trying to prove that this definition of $b^x$ results in a strictly increasing function of $x$? Or that $b^{x+y} = b^x b^y$? Or...? – Greg Martin Aug 14 '12 at 16:48
I'm asking why it makes sense to define $b^x$ in such a way -- what justification is there in such a definition? (Also, this question leads into showing $b^{x+y} = b^xb^y$, but I have not yet investigated that). – Emily Aug 14 '12 at 16:52

Probably it is easier to accept the definition of $b^x$ as the unique element of separation of the two sets $$B(x)=\{b^t|t\in\mathbb{Q},t\leq x\}\\ C(x)=\{b^t|t\in\mathbb{Q},t\geq x\}$$ which implies $$b^{t_1}\leq b^x\leq b^{t_2}\quad\forall\;t_1,t_2\in\mathbb{Q}\;|\;t_1\leq x\leq t_2\quad .$$ As a consequence you can set indifferently $$b^x=\sup B(x)$$ or $$b^x=\inf C(x)$$ (some assumption about $b>1$ and $t>0$ should obviously be done).
What is the justification for the statement $b^{t_1} \le b^x \le b^{t_2}$? It is easy enough to see that $b^{t_1} \le b^{t_2}$, but how can we justify $b^{t_1} \le b^x$? – Emily Aug 14 '12 at 17:15
Wait, let me see... we can say that B(x) has the least upper bound property, and C(x) has the greatest lower bound property, and we can establish that B(x) and C(x) are ordered sets, so I see where you're going. $b^{t_1} \le b^x$ is a consequence of this definition. This makes sense. – Emily Aug 14 '12 at 17:22