Finding an $n$ such that $x^{n}0$. let $(x,y)$ be such that $0<x<1$ and $y>0$. Your goal is to find such $n$ that $x^n<y$
 A: Let $x,y$, $0<x<1$ $y>0$


*

*If $x\le y $ then take $n = 2,3,4....$ since $$.....<x^3<x^2<x\le y$$

*if $x>y$ then 
$$A=\{p\in\Bbb N:x^p>y\}\neq\emptyset$$
is none empty and bounded subset of $\Bbb N$ then has a maximum.
indeed,
$1\in A$ and $$1>x>x^p>y >0\implies p\ln x>\ln y\implies 0< p< \frac{\ln y}{\ln x}$$


whrere this $ p\ln x>\ln y\implies 0< p< \frac{\ln y}{\ln x}$
holds true because $\ln x<0$ ,
$$\ln a<0~~~if~~0<a<1$$
Now let $$j=\max A\implies j\in A~~~and ~~j+1\not\in A$$ that is $$x^j>y\ge x^{j+1}>x^{j+2}$$ 
Then taking $n=j+2$ yields $x^n<y$

In each case there always exists $n$ such that $x^n<y$

Now Let Prove that 
$$\bigcap\limits_{n\in \mathbb N}^{\infty} T_{n} =\{(x,0)\in \mathbb R^{2}:0\le x\le 1\}\cup\{(1,y)\in \mathbb R^{2}:0\le y\le 1\}:=T  $$
where $T_{n}=\{(x,y)\in \mathbb R^{2}:0\le x\le 1,0\le y\le x^{n}\}$


*

*let $(x,y)\in\bigcap\limits_{n\in \mathbb N}^{\infty} T_{n}$


then $$0\le y\le x^n~~~\forall ~~n\in\Bbb N \implies0\le y\le \lim_{n\to \infty}x^n = \begin{cases}0&if ~~~~0\le x<1\\
1&if~~x=1\end{cases}$$
Hence $y= 0$ if $x\neq 1$ i.e $(x,y)= (x,0) \in T$ and $0\le y\le  1$ if $x= 1$ i.e $(x,y)= (1,y)\in T.$
Then we have $$\bigcap\limits_{n\in \mathbb N}^{\infty} T_{n}  \subset T$$


*The fact that, 


$$\bigcap\limits_{n\in \mathbb N}^{\infty} T_{n}  \supset T$$
is trivial since $y= 0$ implies $y \le x^n$ and  if $x = 1$ implies 
$0\le y\le 1^n$.
A: If $0<x<1$ and $y>0$, then $$x<1 \Rightarrow \frac{1}{x} >1$$ and we can find an $a\in \mathbb R>0$ such that $$\frac{1}{x}=1+a$$ holds. From Bernoulli's inequality we get that $$\frac{1}{x}^{n}\ge 1+na, \forall n\in \mathbb N.$$ Thus we get $$x^{n}\le \frac{1}{1+na}.$$ Now let $n$ be such that $\frac{1}{1+na}<y$.Then, $$x^{n}\le \frac{1}{1+na}< y.$$ QED
