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The role of $t$ at the end of of the proof of Theorem 1.21 of Rudin's Principles of Mathematical Analysis (page 10 in Ed. 3) is not clear to me and I am not sure how it is being used.

Rudin states in the proof

Assume $y^n > x$. Put

$$k = \frac{y^n-x}{ny^{n-1}}.$$

Then $0 < k < y$. If $ t \ge y - k$ we conclude that

$$ y^n - t^n \le y^n - (y-k)^n < kny^{n-1} = y^n - x.$$

Thus $t^n > x$, and $ t \notin E$. It follows that $y-k$ is an upper bound of $E$.

Question 1: Why can't we simplify the proof by not introducing $t$ at all, and directly claiming that

$$ y^n - (y-k)^n \overset{(1)}{<} kny^{n-1} \overset{(2)}{=} y^n - x,$$

where inequality $(1)$ follows from the given identity $b^n-a^n < (b-a)nb^{n-1}$, and the equality $(2)$ follows from the definition of $k$? From here we can conclude that $(y-k)^n > x $, and thus $(y-k)^n \notin E$ and is an upper bound of $E$.

Question 2: When Rudin says "It follows that $y-k$ is an upper bound of $E$", what does this have to do with the immediately preceding sentence "Thus $t^n > x$, and $ t \notin E$"?

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    It'd really help if you put the full proof in. At least *what* is being proved and what the definition of E is. – 2017-01-28
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    1) That proves $(y-k)^n$ is an upper bound, but not that $y-k$ is an upper bound. – 2017-01-28
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    You need the $t$ because to prove (y-k) is an upper bound you have to prove *any* t > (y-k) is not in E. It's not enough to prove $(y-k)^2 > x$ – 2017-01-28

1 Answers 1

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You really should have included more detail.

But going to my copy.

$E = \{w\in \mathbb R| w^n > x\}$

(well, Rudin used the variable $t$ but I'll use $w$ to avoid confusion.)

We are attempting to prove that $y = \sup E$ is such that $y^n = x$

So we are attempting to prove that be assuming if $y^n > x$ then $y-k$ is an upper bound of $E$ which contradicts $y = \sup E$.

okay.... now to you question.

Showing that $(y-k)^2 > x$ does prove that $y-k \not \in E$ but it does not prove that $y-k$ is an upper bound. There could very possibly be $t > y-k$ that are in $E$.

So this arbitrary $t \ge y-k$ is needed to prove that nothing larger or equal to $y-k$ is in $E$.

For any $t \ge y-k$ we see that $t \not \in E$ so that means if $w \in E$ then $w \not \ge y-k$ so $w < y-k$ so $y-k$ is an upper bound.

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    Thank you @fleablood -- I will include more context next time. – 2017-01-28
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    @fleablood +1 for the nice answer! I know I’m a bit late to the discussion, but I had two questions myself: (a) in the boxed proof above why is it true that $00$ is simple, as by hypothesis the numerator of the fraction for $k$ is strictly positive. But, if we assume the latter inequality, I can write $$y^n-xx.$$ But, if $n=1$ then $x<0$, a contradiction; (b) why do we have $t^n>x$? Shouldn’t it be $t^n\ge x$? Thanks so much! – 2018-05-17
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    $y = \frac {y^n}{y^{n-1}} > \frac {y^n -x}{y^{n-1}} > \frac {y^n - x}{ny^{n-1}}$ – 2018-05-17
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    @fleablood Also, why $t^n>x$? – 2018-05-17
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    $y^n - t^n \le y^n - (y-k)^n < kny^{n-1} = y^n - x$ – 2018-05-17