What is the set $ C:= \bigcap_{n \in \mathbb{N}} \left[0, {1\over n}\right[$ What is the set $$ C:= \bigcap_{n \in \mathbb{N}} \left[0, {1\over n}\right[$$
I guess the result will be that $C$ is an empty set, because the upper limit converges to $0$,
as the limit of ${1\over n}$ is $0$. 
$$\left[0, 0\right[ = \emptyset $$
I tried to prove this with the Archimedean property:
Assume $C$ is non-empty: 
$$\exists x\in \mathbb{R}: x \in \bigcap_{n \in \mathbb{N}} \left[0, {1\over n}\right[ $$ which is equal to $$ \forall n \in \mathbb{N}: \left[0, {1\over n}\right[ $$ concluding that $x$ must be positive, therefore $n > 0$ and $x > 0$.
By looking for solutions I found that the following completes the proof by contradiction. $$ \forall n \in \mathbb{N}: n < {1\over n} $$
I still don't quite get how the last statement was made and how this applies the Archimedean property.
Any help would be greatly appreciated!
 A: I claim that $$C: = \bigcap_{n \in \mathbb{N}} \left[0, \frac{1}{n}\right[ = \{0\}$$
Proof: Suppose, for the sake of reaching a contradiction, that there exists $\epsilon > 0$ such that $\epsilon \in \left[0, \frac{1}{n}\right[$ for all $n \in \mathbb{N}$. Then, $\epsilon < \frac{1}{n}$ for every $n \in \mathbb{N}$, contradicting the archimedian property of the real numbers. Hence $C$ can't contain any strictly positive real numbers. Clearly, $C$ can contain no negative real numbers (because $C$ is the intersection of positive intervals), and it is also obvious that $C$ contains $0$, since $0 \in \left[0, \frac{1}{n}\right[$ for all $n \in \mathbb{N}$ and we are done!
A: Here is a way to calculate the result.$%
\require{begingroup}
\begingroup
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\subcalch}[1]{\\ \quad & \quad #1 \\ \quad &}
\newcommand{\subcalc}{\quad \begin{aligned} \quad & \\ \bullet \quad & }
\newcommand{\endsubcalc}{\end{aligned} \\ \\ \cdot \quad &}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\false}{\text{false}}
%$
Which $\;x\;$ are in this set of yours?  Letting $\;n\;$ range over the positive integers, we have for all $\;x\;$
$$\calc
    x \in \bigcap_n \left[ 0 , \tfrac 1 n \right[
\op\equiv\hint{definition of $\;\bigcap\;$; definition of half-open interval}
    \langle \forall n :: x \in \mathbb R \;\land\; 0 \le x < \tfrac 1 n \rangle
\op\equiv\hints{rightmost part: multiply both sides by $\;n\;$}\hint{-- work towards isolating $\;n\;$}
    \langle \forall n :: x \in \mathbb R \;\land\; 0 \le x \;\land\; n\,x \lt 1 \rangle
\op\equiv\hints{rightmost part: divide both sides by $\;x\;$, special}\hint{case $\;x=0\;$ -- isolating $\;n\;$}
    \langle \forall n :: x \in \mathbb R \;\land\; (x = 0 \;\lor\; (x > 0 \land n \lt \tfrac 1 x)) \rangle
\op\equiv\hints{logic: move all parts that don't have $\;n\;$ out of $\;\forall n\;$}\hint{-- to simplify}
    x \in \mathbb R \;\land\; (x = 0 \;\lor\; (x > 0 \land \langle \forall n :: n \lt \tfrac 1 x \rangle))
\op\equiv\hint{choose, e.g., $\;n := \left\lceil \tfrac 1 x \right\rceil\;$ which is a positive integer}
    x \in \mathbb R \;\land\; (x = 0 \;\lor\; (x > 0 \land \false))
\op\equiv\hint{logic: simplify}
    x = 0
\endcalc$$
so we've proven
$$
\bigcap_n \left[ 0 , \tfrac 1 n \right[ \;=\; \{0\}
$$
$%
\endgroup
%$
