Problem with a limit and the infinite intersection of open sets This is probably a really stupid question, but it's been annoying me for a while and I still can't find an answer that convinces me.
We know that $\lim_{x \rightarrow \infty} \frac{1}{x} = 0$.
But, in my lecture, we saw that $\{0\} = \bigcap_{n=1}^{\infty} \left(-\frac{1}{n} , \frac{1}{n} \right) \subset \mathbb{R}$.
What I don't get is, why isn't the above intersection equal to $(0,0)$, i.e. to the empty set? Again, sorry if this is really stupid. Someone told me it's because $\frac{1}{n}$ approaches 0 but is never equal to $0$, but then why would $\lim_{x \rightarrow \infty} \frac{1}{n}$ be equal to 0?
Thanks in advance for any help!
 A: The intersection of an infinite amount of sets is defined to be the elements present in every set.
$0$ is in every set of the form $\left( -\dfrac1n, \dfrac1n \right)$, so it is also in their intersection.
There is no rules saying that arbitrary intersections of open set must be open, and this is a counter-example.
A: Do not apply limit to the numbers, what  $ \bigcap_{n=1}^{\infty} \left(-\frac{1}{n} , \frac{1}{n} \right) \subset \mathbb{R}$ means is that it is a set of all element that is contained by all sets of $(-1/n, 1/n) $ for all $n \in \mathbb N$.
Thus only $0$ satisfies it.
Notice we could naturally define arbitrarily number of set intersections and unions, even uncountably many is fine, thus no limit is needed -  the $\infty$ in your formula is just a notation saying that we take $n$ for all natural numbers.
A: Because the limit is the value where there are an infinity of intervals "around". In this case you can choose any $\epsilon>0$ in which $1/x_{n}$ stays in $(0-\epsilon,0+\epsilon)$ for all $n> N(\epsilon)$ (when $x_{n}$ goes to $\infty$).
And so, the intersection always contain $0$ because it is an open interval (neighbourhood of $0$).
