TITLE: Find a contradiction to: If $t^n < (\sup \{t \in \mathbb{R}_{>0}: t^n < x\})^n$ then $t^n < x$ when this supremum is greater than $x$.
QUESTION [0 upvotes]: 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 $t^n < x$.

REPLY [2 votes]: 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.
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