According to Tao $$x^{1/n} :=sup\{y\in \mathbb{R}: y\geq 0\; and \; y^n \leq x \}$$
Tao asks the reader to prove that if $z = x^{1/n}$ then $z^n = x$. Tao gives hint that proof by contradiction might be a useful tool here
With some effort i was able to show that if the following implication
$$\forall\; \epsilon > 0\;\exists\; y \geq 0\;(z-\epsilon <y \leq z) \implies \forall \;\delta >0 \;\exists \;y^n (z^n-\delta <y^n \leq z^n)$$ is true then surely Proposition is true. But then i was unable to prove the implication.
Proving proposition 5.6.6 from Terence Tao Analysis I
real-analysis
Best Answer
Left hand side of your arrow is automatically true for $y\in A$ (I call right hand side of your first equation $A$) by the definition of least upper bound. Therefore you just need to show the right hand side is true.
For any given $\delta>0$ you want to find $y\in A$ such that $z^n-\delta< y^n\leq z^n$.
What you already have is for any $\epsilon>0$ there is a $y\in A$ such that $z-\epsilon< y\leq z$. For this $y$ on our choice of $\epsilon$, we already have $(z-\epsilon)^n< y^n\leq z^n$.
Therefore you just need to find an $\epsilon$ such that $(z-\epsilon)^n>z^n-\delta$ for a given $\delta$. By Bernoulli's inequality (if $z>\epsilon>0$, $n\geq 2$) we have $(z-\epsilon)^n=z^n(1-\epsilon/z)^n\geq z^n(1-n\epsilon /z)=z^n-n\epsilon z^{n-1}$.
Therefore you just need to pick an $\epsilon$ such that $n\epsilon z^{n-1}<\delta$ or equivalently $\epsilon<\frac{\delta}{nz^{n-1}}$.