To derive this we need to use the completeness of $\mathbb{R}$:
Every nonempty set of real numbers that is bounded above has a supremum in $\mathbb{R}$.
There are two approaches to this, the axiomatic approach (where it is an axiom) and the constructive approach (where it is a theorem). More information can be found in analysis texts.
Now we prove the Archimedean property of $\mathbb{R}$:
$\mathbb{N}$ is not bounded above in $x\in\mathbb{R}$. That is, for each $x\in\mathbb{R}$ there exists $n\in\mathbb{N}$ such that $n>x$.
For $x<0$ it is trivial. Suppose $x\geq0$. Let $A=\{n\in\mathbb{N}\ ;n\leq x\}$. By completeness we have $\sup A\in\mathbb{R}$. Obviously, there is $a\in A$ such that $\sup A-1/2<a$ (Why?), then let $n=a+1$ and $n>x$ and the theorem is proved.
Finally, we turn to the question:
For any real number $a>0$, there exists a natural number $n$ such that $0<1/n<a$.
We proceed using proof by contradiction. Let $0<1/n<a$ for all $n\in\mathbb{N}^{\times}$, then $n\leq 1/a$ for all $n\in\mathbb{N}^{\times}$. Thus $\mathbb{N}$ is bounded in $x\in\mathbb{R}$, contradicting the Archimedean property.
HINT: Given any real number, say $x$, you can a positive integer $N$ so that $N>x$. Now what should you let $x$ be so that when you `flip both sides upside-down', you get the desired inequality?
Best Answer
$\cos$ is a continous function, the set $\{0\}$ is closed, and therefore $\cos^{-1}(\{0\})$ is a closed set.
We conclude that $\cos^{-1}(\{0\})\cap[0,\infty)$ is closed. This set is equal to $\cos^{-1}(\{0\})\cap (0,\infty)$ (because $\cos(0)\neq 0$).
This set is bounded from below and must therefore have a greatest lower bound $\alpha$. Recall that the greatest lower bound of a set is always an adherent point for the set. Since $\cos^{-1}(\{0\})\cap(0,\infty)$ is closed we conclude $\alpha\in \cos^{-1}(\{0\})\cap(0,\infty)$. Therefore the set $\cos^{-1}(\{0\})\cap (0,\infty)$ has a minimum as desired.