Prove that for $\forall x \in (0,1]$, $\exists n \in \mathbb{N}$ such that $\frac{1}{n+1} \leq x \leq \frac{1}{n}$ 
Prove that $\forall x \in (0,1]$, $\exists n \in \mathbb{N}$ such that $\frac{1}{n+1} \leq x \leq \frac{1}{n}$

This is a completely obvious statement (it was taken for granted in another proof) and yet I cannot seem to come up with a strictly formal proof. Of course, if we take $A$ to be the set of $n$ such that $f(n)=\frac{1}{n}-x \geq 0$ and show that this set is bounded above, then $\sup(A)$ (or $\max(A)$ since $A \in \mathbb{N}$) will be the desired $n$. We could show that $A$ is bounded using the fact that $\lim_{n \to \infty} f(n) = -x < 0$, so there is an $N$ such that all $n>N \notin A$. Is this the right direction or am I just overthinking it?
 A: Hint: Can you prove that for all $y\geq 1$ there is an $n\in\mathbb N$ with $n\leq y\leq n+1$?
A: What about $n=\mathrm{floor}(\frac{1}{x})$ ?
A: You can get quite directly from the Archimedean Property of the Real Numbers that there exists some $n$ such that $x \leq \frac 1 n$. In particular, suppose that this $n$ is the natural number giving the closest $\frac 1 n$ to $x$, i.e. that there is no $m \in \Bbb N$ such that $x \leq \frac 1 m < \frac 1n$. (There must be such a 'closest' $n$ -- suppose there is only one $n$, then it's that. Suppose there are several -- then one of them will be closest because the real numbers are well-ordered.)
Then we can get the conclusion we want by contradiction. Take $\frac 1 {n+1}$. It is clear that $\frac 1 {n+1} < \frac 1 n$. Then suppose $x \leq \frac 1 {n+1} < \frac 1 n$. That's impossible, because we supposed that $n$ was the closest, above. Then we must have that $\frac 1 {n+1} \leq x \leq \frac 1 n$.
A: Set :
$$X:=\{n\in\mathbb{N}^*| \frac{1}{n}\geq x\} $$
This set is non empty because $1\in X$. Furthermore it is clear that if $n\in X$ then any $l\leq n$ will be in $X$ as well : $l\in X$. We would like to show that $X$ is bounded above. Hence (because of what I have written before) if we find $a>0$ such that $a\notin X$ then $X$ is bounded above by $a$.
Now to show that there exist an integer $a$ such that $a\notin X$ i.e. :
$$x>\frac{1}{a}\Leftrightarrow a>\frac{1}{x} $$
And this is the archimedean property of $\mathbb{R}$ that allows us to do this (one of the fundamental axiom of $\mathbb{R}$). Hence $X$ is bounded above, and as a non-empty set of integers it must have a maximum $n$. The maximum of $X$ clearly verifies (because $n+1\notin X$) :
$$\frac{1}{n+1}<x\leq \frac{1}{n} $$ 
