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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.

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$\{0\}$ 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 $0<\dfrac{1}{N}<\dfrac{1}{|x|}$.

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 $x=0$ leads to a contradiction $x\neq 0$.

This shows that 0 is the only element of $\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)$ and $\bigcap_{n=1}^\infty\left(−\dfrac{1}{n}, \dfrac{1}{n}\right)\subseteq{0}$.

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$\cap_{n=1}(-\frac{1}{n}, \frac{1}{n}) = {0}$ 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

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    $\begingroup$ Those two statements are not equivalent, but the first implies the second. $\endgroup$ – Cameron Williams Jan 21 '16 at 3:28

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