Intersection of all open intervals of the form $(-\frac{1}{n}, \frac{1}{n})$ is $\{0\}$? In Mathematical Analysis by Apostol he mentions that the "Intersection of all open intervals of the form $(-\frac{1}{n}, \frac{1}{n})$ is $\{0\}$"
Obviously this is a super basic question but I thought that an open interval does not include the endpoints, so from the limit as $n\to\infty$ on both sides we get (0,0), i.e. if $x$ belongs to the intersection then $0<x$ and $x<0$ which no real number satisfies. 
 A: $\{0\}$ is a subset of $\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$.
Since $0$ is an element of the open interval $\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$ for every positive integer $n$, therefore
$$0\in\left\{x\in\mathbb{R} \quad|\quad x\in\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)\quad\hbox{for every}\quad  n\in\mathbb{Z}\right\}=
\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$$
For $\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$ is a subset of $\{0\}$.
Let $x$ be an element of $\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$. By the definition of intersection this means that $x\in\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$ for every $n\in\mathbb{Z}$. Suppose that $x$ does not equal $0$.
Then $|x|>0$ and, by the Archimedean Principle, there is a positive integer $N$ such that $0<\dfrac{1}{N}<|x|$.
Therefore $x\notin\left(−\dfrac{1}{N}, \dfrac{1}{N}\right)$ which contradicts the assumption that $x\in\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$. Since $x\ne0$ leads to a contradiction $x=0$.
This shows that $0$ is the only element of $\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$ and $\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)\subseteq\{0\}$.
A: $\cap_{n=1}(-\frac{1}{n}, \frac{1}{n}) = {0}$ is equivalent to the statement that for all $n \in \mathbb{N}, 0 \in (-\frac{1}{n}, \frac{1}{n})$, which is obviously true.
As $n \rightarrow\infty$, $(-\frac{1}{n}, \frac{1}{n})$ approaches the zero, but never reaches the zero
