Find the $\bigcap_{n = 1}^{\infty} (-\frac{1}{n}, \frac{2}{n})$ Find the $\bigcap_{n = 1}^{\infty} (-\frac{1}{n}, \frac{2}{n})$
So the way I understand it is that I'm trying to find 
$(\frac{-1}{1}, \frac{2}{1}) \bigcap (\frac{-1}{2}, \frac{2}{2}) \bigcap (\frac{-1}{3}, \frac{2}{3})\bigcap ...$ and so forth. Then the intersection would be $\varnothing$ right? Can I prove this by just writing out the first few elements and seeing that they do not intersect?
Or is the question asking me to find:
$\frac{-1}{1} \bigcap \frac{2}{2} \bigcap \frac{-1}{3} \bigcap...$? in this case the intersection would still be $\varnothing$, right?
 A: There are three things to prove:


*

*$0$ is in every interval, hence in the intersection of all of them.

*If $x>0$, then there is at least one interval that does not contain $x$, so $x$ is not in the mutual intersection.  (choose $n$ so that $2/n<x$).

*If $x<0$, then there is at least one interval that does not contain $x$, so $x$ is not in the mutual intersection.  (choose $n$ so that $-1/n>x$).
Combining, the intersection is $\{0\}$, as tetori points out.
A: Here is a longer and more formal version of vadim123's answer, with some more detail about where the case split is coming from.

$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Tag}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$Using a slightly different notation, and letting $\;n\;$ implicitly range over $\;\mathbb N^+\;$, we can simply start to calculate the elements $\;x\;$ of this set:
$$\calc
x \in \langle \cap n :: (-\tfrac{1}{n}, \tfrac{2}{n}) \rangle
\calcop\equiv{definition of $\;\cap\;$}
\langle \forall n :: x \in (-\tfrac{1}{n}, \tfrac{2}{n}) \rangle
\calcop\equiv{definition of interval; multiply by $\;n\;$ -- to try and isolate $\;n\;$}
\langle \forall n :: -1 < n \times x < 2 \rangle
\endcalc$$
Our strategy is to isolate $\;n\;$, and so we want to divide by $\;x\;$, and therefore we need to split into three different cases: for $\;x = 0\;$ we get
$$\calc
\tag 1
\langle \forall n :: -1 < n \times x < 2 \rangle
\calcop\equiv{substitute $\;x = 0\;$}
\langle \forall n :: -1 < 0 < 2 \rangle
\calcop\equiv{simplify}
\true
\endcalc$$
for $\;x > 0\;$ we get
$$\calc
\tag 2
\langle \forall n :: -1 < n \times x < 2 \rangle
\calcop\equiv{divide by $\;x\;$, using $\;x > 0\;$ so no sign flip}
\langle \forall n :: -\tfrac 1 x < n < \tfrac 2 x \rangle
\calcop\Rightarrow{choose any $\;n \geq \tfrac 2 x\;$, possible since $\;\mathbb N^+\;$ is unbounded upwards}
\false
\endcalc$$
and finally for $\;x < 0\;$ we get
$$\calc
\tag 3
\langle \forall n :: -1 < n \times x < 2 \rangle
\calcop\equiv{divide by $\;x\;$, using $\;x < 0\;$ so the signs flip}
\langle \forall n :: \tfrac 2 x < n < -\tfrac 1 x \rangle
\calcop\Rightarrow{choose any $\;n \geq -\tfrac 1 x\;$, possible since $\;\mathbb N^+\;$ is unbounded upwards}
\false
\endcalc$$
Summing up, we have proven that for all $\;x\;$, $\;x \in \langle \cap n :: (-\tfrac{1}{n}, \tfrac{2}{n}) \rangle \;\equiv\; x = 0\;$, in other words, the set in question is equal to $\;\{0\}\;$.

Note how the crucial property that we used was that $\;\mathbb N^+\;$ is unbounded.
