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I'm working through Rudin's "Principles of Mathematical Analysis" on my own, so I don't want the full answer. I'm only looking for a hint on this problem.

As a follow-up to this question, Rudin asks me to prove that

If $b > 1$ and $B(r)$ is the set of all numbers $b^t$, where $t$ is rational and $t \leq r$, prove that $b^r = \sup B(r)$ where $r$ is rational.

I know that $B(r) \subset \mathbb{R}$ and that $\mathbb{R}$ has the least upper bound property, so to prove the existence of $\sup B(r)$, I just need to show that $B(r) \neq \emptyset$ and $B(r)$ is bounded above.

  1. $B(r)$ is obviously nonempty because $\forall r \in \mathbb{Q}$, $ r \leq r \Rightarrow b^r \in B(r)$.

  2. $b > 1$, so $b^t \leq b^r \Rightarrow t \leq r$, so $b^r$ is an upper bound of $B(r)$ because if $b^y > b^r$, for some $y \in \mathbb{Q}$, then $y > r$, so $b^y \notin B(r)$.

$B(r)$ is nonempty and bounded above, so $\sup B(r)$ exists.

I got stuck here proving that $b^r = \sup B(r)$. I'm thinking about this approach to prove that if $x < b^r$, then $x$ isn't an upper bound of $B(r)$.

  1. Pick some $x \in \mathbb{R}, x > 0, x < b^r$. I know that (I'm not sure how I know...) that $\exists s \in \mathbb{R}$ such that $b^s = x \Rightarrow x = b^s < b^r$. Using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, I know that there is some $p \in \mathbb{Q}$ such that $s < p < r$.

  2. Therefore $x = b^s < b^p < b^r$, so $x$ isn't an upper bound of $B(r)$, which implies that $b^r = \sup B(r)$.

I'm stuck here because I don't know how to do this without assuming that if $x \in \mathbb{R}, x > 0$, then $\exists s \in \mathbb{R}$ such that $b^s = x$.

If this proof isn't correct or assumes too much, can someone give me a hint in the right direction?

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  • Can anyone give me a hint in the right direction? I realize it's an old question, but I've updated it significantly with a new approach to the proof. – M T Nov 13 '14 at 21:51

1 Answers1

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For a given rational $r$, to prove $b^r=supB(r)$, you first need to pick $x \in B(r)$ and show $x \leq b^r$, so that $b^r$ is an upper bound according to Defintion 1.7. You know $x=b^t$ for some rational $t \leq r$, you know how rational exponents are defined in terms of integer exponents because of the work done in part (a) of this problem, and you know how order works when the exponents are integers. You need to put these things together to have (i) in Definition 1.8. After that, you pick any real $x$ with $x<b^r$ and you need to show there is some $y \in B(r)$ such that $x<y$ (so that ''$x \geq z$ for every $z \in B(r)$'' is a false statement, and $x$ is not an upper bound, according to Definition 1.7). This shows (ii) in Definition 1.8, which will mean that $b^r$ is the least upper bound. If you think about it correctly, you won't need hints for this part.

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    Thank you! I figured out the first part, but I must not be doing the last part right or "thinking about it correctly" because I'm still stuck. I know that $0 < x < b^r = b^{m/n}$, so $0 < x^n < b^m$ but then like always, I run into a wall. This is extremely discouraging and I'm only one chapter 1. I'll keep thinking about it. I need to show that if $0 < x < b^r$, there is some $b^{p/q}$ s.t. $0 < x < b^{p/q} < b^r$, right? I started working with this expression to see if I could "work backwards" and learn a bit about how to tackle it, but I'm still going. – M T Nov 14 '14 at 15:25
  • I wouldn't worry too much about it and get discouraged, I think this is one of those many things that are not obvious until you see it (and then go "aahhh, that WAS obvious"). Anyway, maybe there is a hint for the second part (and I'm not sure how much to say here): you don't need to construct the $y$ you need, it's out there already. –  Nov 14 '14 at 15:44
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    I don't know, I'll have to take a break at look at this again later. I guess I don't understand why I can't just use $b^r$ as $y$, because if $x < b^r$, then $x$ can't be an upper bound of $B(r)$ because $b^r \in B(r)$, so the statements a) $x$ is an upper bound of $B(r)$ and b) $x < b^r$ are contradictory, so $b^r$ must be the least upper bound. – M T Nov 14 '14 at 19:06
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    That's exactly what you should argue. If $x<b^r$ then $x$ is not an upper bound exactly because $b^r \in B(r)$ (since $r$ is a rational and $r \leq r$). This gives (II). –  Nov 14 '14 at 19:07
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    Hmm, notice that the $x$ in the part (i) and the $x$ in the part (ii) are not the same thing. First you pick some $x$ that satisfies $x \in B(r)$ (and this is the only requirement, we are not asuming $x<b^r$ here) and show $x \leq b^r$. Then on a different argument you pick some $x$ that satisfies $x<b^r$ (and this is the only requirement, we are not assuming $x \in B(r)$ here) and show $x<y$ for some $y \in B(r)$ (namely, take $y=b^r$, where $b^r$ is indeed in $B(r)$). –  Nov 14 '14 at 19:13
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    Right, because $x$ is a different quantity depending on the needs in each part of the problem. In the second part, there's no guarantee that $x = b^t$, where $t \in \mathbb{Q}$, because we only chose $x \in \mathbb{R}$. Thank you! – M T Nov 14 '14 at 19:26