Without using the fact that each positive real has a positive $n$-th root (which is what I'm trying to prove, but I'm stuck).

Basically I have the following situation:

Let $n$ be a positive integer and let $E(z) = \{t \in \mathbb{R_{>0}}: t^n < z\}$. Along the proof, which I started by assuming by contradiction that $x < (\sup E(x))^n$, I came to the conclusion that $E(x) = E(p)$ where $p$ is any real number satisfying $x < p < (\sup E(x))^n$. Now, if $p$ satisfies this, we have that $t^n < p$ implies $t^n < x$. Consequently, we alse have that $t^n < (\sup E(x))^n$ implies $t^n < x$.

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    $\begingroup$ $0 < 1 < 41$; $3^2 < 41$. But $3^2$ is not less than $1$. $\endgroup$ – John Hughes Feb 16 '18 at 20:47
  • $\begingroup$ @JohnHughes Sorry, I edited the question. $\endgroup$ – Doughnut Pump Feb 16 '18 at 21:07
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    $\begingroup$ Your title is clearly wrong. Please state clearly your question. $\endgroup$ – user99914 Feb 16 '18 at 21:35
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    $\begingroup$ @JohnMa Edited. $\endgroup$ – Doughnut Pump Feb 16 '18 at 21:51
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    $\begingroup$ I think what you are asking is that if $0<x<p$ and $n$ is a positive integer, then there is a $t$ such that $x<t^n<p$. (Which is essentially what you had said at the beginning, and implies what you want about suprema.) $\endgroup$ – Andrés E. Caicedo Feb 16 '18 at 22:15

I will show that for any positive integers $n,\ell,k$ there is an $M$ so large that for all positive integers $i$, if $i/M\le \ell$, then the difference $$ \left(\frac iM\right)^n-\left(\frac{i-1}M\right)^n $$ is less than $1/k$.

Let's prove this first, and then argue that the result follows from it.

Note that $$ (i+1)^n-i^n=\sum_{k=0}^{n-1}\binom nk i^k\le \sum_{k=0}^{n-1}\binom nk i^{n-1}<2^n i^{n-1}. $$ It follows that $$\left(\frac{i+1}M\right)^n-\left(\frac iM\right)^n<2^n\left(\frac iM\right)^{n-1}\frac1M\le 2^n\ell^{n-1}\frac1M<\frac1k$$ provided that $M$ is chosen larger than, say, $(2\ell k)^n$, with room to spare.

Ok. Now, we are given $0<x<p$ and we need to find some $t$ such that $x\le t^n<p$. Begin by choosing $k$ such that $p-x>1/k$ and a positive integer $\ell$ such that $p\le \ell^n$. With $M$ as above, we have that for all $i$ with $1\le i\le M\ell$, $$ \left(\frac iM\right)^n-\left(\frac{i-1}M\right)^n<\frac1k. $$ In particular, there should be at least one number $(j/M)^n$ with $0\le j\le M\ell$ with $x<(j/M)^n<p$, and we are done.

  • $\begingroup$ Does your last statement rely on the fact that $x - \left(\frac{j-1}{M}\right)^n \le 0$ for some $j$ with $0 < j \le M\ell$? How do you know such $j$ exists? $\endgroup$ – Doughnut Pump Feb 16 '18 at 22:40
  • $\begingroup$ Nevermind. We can choose $\ell$ large enough and then choose $j = \lceil \ell M / 2\rceil + 1$, so that $\left(\frac{j-1}{M}\right)^n \ge \left(\frac{\ell}{2}\right)^n \ge x$. Thanks, much appreciated. I really liked this strategy. $\endgroup$ – Doughnut Pump Feb 16 '18 at 23:02
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    $\begingroup$ The way I thought of it was to note that since $p\le\ell^n=(M\ell/\ell)^n$, there is a least $i$ in the required range such that $p\le(i/M)^n$. This implies that $\left(\frac{i-1}M\right)^n <p $, and therefore also $x <\left (\frac {i-1}M\right)^n $. $\endgroup$ – Andrés E. Caicedo Feb 16 '18 at 23:17

This might be useful:

If $0 < a < 1$ then $a = \dfrac1{1+b}$ where $b = \dfrac1{a}-1 \gt 0$.

Then, by Bernoulli's inequality, $(1+b)^n \ge 1+nb =1+n( \dfrac1{a}-1) \gt n( \dfrac1{a}-1) $ so $a^n = \dfrac1{(1+b)^n} \lt \dfrac1{n( \dfrac1{a}-1)} = \dfrac{a}{n( 1-a)} $.


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