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I am currently self studying Baby Rudin and I'm having some problems with his proof of Theorem 3.20(a) on page 58. I have read all previous chapters and I can't find any mention of real exponents besides its definition in the exercise section of Chapter 1 (but nothing about order/this identity). He defines $x^a$ for rational $a$ and $x>1$ then this is used to extend it to all real $a$ by $x^a = \sup \{x^t : t \leq a,~ t \in Q\}$.

Am I missing something or does he expect the reader to fill this huge gap in the proof?


Prefatory note on parts for Theorem 3.20: We shall now compute the limits of some sequences which occur frequently. The proofs will all be based on the following remark: If $0\leq x_n\leq s_n$ for $n\geq N$, where $N$ is some fixed number, and if $s_n\to 0$, then $x_n\to 0$.

Theorem 3.20 (a): If $p>0$, then $\lim_{n\to\infty}\frac{1}{n^p}=0$.

Proof. Take $n>(1/\varepsilon)^{1/p}$. (Note that the archimedean property of the real number system is used here.)

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  • $\begingroup$ What's his definition of $x^a$? $\endgroup$ – Simon S Jul 16 '15 at 22:21
  • $\begingroup$ Does he prove it for base $e$? $\endgroup$ – GFauxPas Jul 16 '15 at 22:23
  • $\begingroup$ No, he does not. If I remember correctly this is how Apostol does it but Rudin introduces $e$ a few pages later. I will update the op with the definition. $\endgroup$ – B. Freitas Jul 16 '15 at 22:26
  • $\begingroup$ If real exponents are defined by a limiting process with rational exponents, then you can prove this property pretty easily using the definition. You need to know that any real can be approximated by rationals, and that any sequence of rationals converging to a given real exponent yields the same value for $x^\text{power}$. (You need these things for exponentiation to be well-defined at all). $\endgroup$ – Bobby Grizzard Jul 16 '15 at 22:27
  • $\begingroup$ Which edition of Baby Rudin are you using? I have the second edition, and Theorem 3.20 is on page 50. $\endgroup$ – Rory Daulton Jul 16 '15 at 22:31
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"Am I missing something or he expects the reader to fill this huge gap in the proof?" It's not only a gap in the proof, Rudin hasn't even defined what $1/n^p$ means for real $p>0.$ In fact beyond integer powers, he has only defined $a^p$ for $a\ge 0$ and $p=1/n$ for some $n\in \mathbb {N}.$ Yes, there are the exercises in Chapter 1 concerning rational and real powers, but the whole lot of them make for a huge expenditure of effort at that point. I don't think this is one of Rudin's finest moments.

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    $\begingroup$ Not answer, more of a comment $\endgroup$ – Simon S Jul 16 '15 at 22:50
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    $\begingroup$ @SimonS I'm amazed at how many bad answers there were to this question. $\endgroup$ – Cameron Williams Jul 16 '15 at 22:52
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    $\begingroup$ Excuse me, I quoted the very question asked and then answered it! Read again. $\endgroup$ – zhw. Jul 16 '15 at 22:52
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    $\begingroup$ Valid answer IMO. Most of my problem with the proof was the somewhat hand-wavy use of real expoents and this confirms that Rudin was a bit hasty here. $\endgroup$ – B. Freitas Jul 16 '15 at 22:52
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    $\begingroup$ @CameronWilliams You too should read the real question posed and then my answer. $\endgroup$ – zhw. Jul 16 '15 at 22:54
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First show that it holds for $a, b \in \mathbb N$. Then show it holds for $a,b \in \mathbb Q$ (this is easy: since $x^{p/q} = x^p \cdot (x)^{1/q}$, and $p, q \in \mathbb N$)

Finally for reals $a, b$ you have that $x^a = \sup\{x^t, t\le a, t \in \mathbb Q\}$; hence you get

\begin{eqnarray*}(x^{a})^b &=& \sup\{\sup\{x^t, t\le a, t \in \mathbb Q\}^s, s\le b, s \in \mathbb Q\} \\[0.5em] &=& \sup\{\sup\{(x^t)^s, t\le a, t \in \mathbb Q\}, s\le b, s \in \mathbb Q\} \\[0.5em] &=& \sup\{\sup\{x^{ts}, t\le a, t \in \mathbb Q\}, s\le b, s \in \mathbb Q\} \\[0.5em] &=& \sup\{x^{ts}, t\le a, s\le b, s \in \mathbb Q, t \in \mathbb Q\} \\[0.5em] &=&\sup\{x^{k}, k\le ab, k \in \mathbb Q\} \\[0.5em] &=& x^{ab} \end{eqnarray*}

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  • $\begingroup$ Your argument is nearly how I'd do it as well, but just for readability's sake I would have just shown that $(x^a)^t = x^{at}$ for rational $t$ first (which is the meat of your argument). Then combined it at the end. $\endgroup$ – Cameron Williams Jul 16 '15 at 22:46
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    $\begingroup$ You would still have to prove $\sup\{x\}^{a} = \sup\{x^{a}\}$. $\endgroup$ – layman Jul 17 '15 at 0:12
  • $\begingroup$ @user46944 right! But I'll leave that to the reader :) $\endgroup$ – Ant Jul 17 '15 at 8:26
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Well, when we consider integers...

Note that $$ (x^a)^1 = x^a = x^{a \cdot 1} $$

Use induction:

IF $(x^a)^{n} = x^{an}$, then also $(x^a)^{n+1} = x^{a(n+1)}$ , as $$ (x^a)^{n + 1} = (x^a)^n x^a = x^{an} x^a = x^{a(n+1)} $$

It is true for $n=1$, and by induction it is true for any positive integer.

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    $\begingroup$ This works for integral $b$. You can extend this to rational $b$ in a fairly straightforward way. But to extend to real $b$, you have to invoke continuity or some other machinery. Your answer is not sufficient. $\endgroup$ – Cameron Williams Jul 16 '15 at 22:35
  • $\begingroup$ @CameronWilliams, you are correct, but when I read this post, I saw all kind of answers about integer, so I did it for integers... Sorry for the insufficient answer... $\endgroup$ – johannesvalks Jul 16 '15 at 22:50
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    $\begingroup$ No worries! It happens. It was a bit of a miscommunication by the OP and lots of people rushed to answer the question which caused confusion. $\endgroup$ – Cameron Williams Jul 16 '15 at 22:51

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