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I am reading Walter Rudin's "Principles of Mathematical Analysis".

There is the following theorem in this book:

p.57
Theorem 3.20(a)
If $p > 0$, then $\lim_{n\to\infty} \frac{1}{n^p}=0$.

Take $n > (\frac{1}{\epsilon})^{\frac{1}{p}}$. Then $n^p > ((\frac{1}{\epsilon})^{\frac{1}{p}})^p = \frac{1}{\epsilon}$. So $\epsilon > \frac{1}{n^p}$. To prove $((\frac{1}{\epsilon})^{\frac{1}{p}})^p = \frac{1}{\epsilon}$, I think we need the property $(a^x)^y = a^{x y}$. And Rudin didn't write this property on p.22 ex6.

On p.22 Exercise 6, Rudin defined $b^x$ for $b \in \{y \in \mathbb{R} | y > 1\}, x\in\mathbb{R}$.
And the reader proves that $b^{x+y} = b^x b^y$ for all $x, y \in \mathbb{R}$.

But Rudin didn't define $b^x$ for $b \in \{y \in \mathbb{R} | 0 < y \leq 1\}, x\in\mathbb{R}$.

And Rudin didn't write other properties of $b^x$.
For example, Rudin didn't write $(b^x)^y = b^{xy}$ for all $x, y \in \mathbb{R}$.

I am disappointed and sad.
Walter Rudin's "Principles of Mathematical Analysis" isn't perfect.

I can guess $b^x$ is defined as $(\frac{1}{b})^{-x}$ for $b \in \{y \in \mathbb{R} | 0 < y \leq 1\}, x\in\mathbb{R}$.

Isn't Walter Rudin's "Principles of Mathematical Analysis" self-contained?

Is there a self-contained analysis book in the world?

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    $\begingroup$ You are making conclusion too hastily. $x^\alpha$ for positive $x$ and real $\alpha$ is defined on p.181. That said, most reviewers on the internet seem to agree that Rudin is not suitable for self studies (if one learns analysis for the first time) and it should be used as a textbook in a first course on analysis only if the students are guided by a good lecturer. $\endgroup$
    – user1551
    Jan 21, 2019 at 13:09
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    $\begingroup$ Even if you want to use $(a^x)^y=a^{xy}$, there still isn't a circular argument, because the general definition of exponential on p.181 does not depend on the result of the 3.20(a). And strictly speaking, you don't actually need $(a^x)^y=a^{xy}$, because there are other ways to prove the statement. E.g. just pick a sufficiently large $n$ such that $n>(\frac1\epsilon)^{1/q}$ for some rational number $0<q\le p$ instead. Then $\frac1{n^p}\le\frac1{n^q}<\epsilon$. $\endgroup$
    – user1551
    Jan 22, 2019 at 4:20
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    $\begingroup$ A simpler approach is to consider a positive integer $k>1/p$ and then $1/n^p<1/n^{1/k}$ and this is less than $\epsilon$ if $n>\epsilon^{-k}$. $\endgroup$
    – Paramanand Singh
    Jan 22, 2019 at 8:54
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    $\begingroup$ Also the approach by Rudin to define $b^x$ is somewhat complicated. A better approach is to develop logarithms first. Or if you want to avoid logarithm then better use limits instead of sup or inf. Limits obey nice algebraic properties which inf and sup may not. See this post for more details. $\endgroup$
    – Paramanand Singh
    Jan 22, 2019 at 8:57
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    $\begingroup$ @tchappyha: you have caught a minor (but worth noting) issue here! If the function $f$ is continuous and strictly monotone then we can exchange $f$ and limit operation without worrying about the existence of limit. See the theorem mentioned at the end of this answer: math.stackexchange.com/a/1073047/72031 I will update this in blog after some time. $\endgroup$
    – Paramanand Singh
    Jan 22, 2019 at 14:19

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You can easily extend the definition of $b^x$ for $0 < b \leq 1$ by defining $$b^x := \frac{1}{(1/b)^x}$$ and show the property $$b^{x+y} = b^x b^y$$ still holds.

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  • $\begingroup$ Is $(b^x)^y = b^{xy}$ easy? $\endgroup$
    – tchappy ha
    Jan 21, 2019 at 13:03

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