Prove that $\bigcap\limits_{n \in \mathbb{N}} B_{\frac{1}{n}} (0) = \{0\}$ I'm trying to prove that the intersection of the open interval $\left(- \frac{1}{n}, \frac{1}{n}\right)$ is not open, but I'm not sure if my final step is fully valid. Here is my attempt.

We show that the arbitrary intersection of open sets is not necessarily open. Consider $\mathbb{R}$ with the Euclidean metric, and for each $n \geq 1$, let $U_n := B_{\frac{1}{n}} (0) = \left(- \frac{1}{n}, \frac{1}{n}\right)$, each of which is open by problem 1. I claim that $\bigcap\limits_{n=1}^{\infty} U_n = \{0\}$. Indeed, for every $n$, $0 \in \left(- \frac{1}{n}, \frac{1}{n}\right) = B_{\frac{1}{n}} (0)$. Furthermore, if $x \in \bigcap\limits_{n=1}^{\infty} U_n$, then for every $n$, we have $d(x,0) = |x| < \frac{1}{n}$. Taking $n \to \infty$, we find $|x| = 0$, so $x = 0$. But $\{0\}$ is not open since, for every $\epsilon > 0$, $B_{\epsilon} (0) = (- \epsilon, \epsilon) \not \subset \{0\}$, as required.

Is it fully rigorous to "take $n \to \infty$" here? I'm not sure if I'm treating this as a limit of a sequence or as a function or if the distinction even matters. I can proceed by contradiction as well. I think I can also argue that because $n$ is arbitrary, $x < \frac{1}{n}$ implies $x \leq 0$, and $x > - \frac{1}{n}$ implies $x \geq 0$, so $x = 0$. I thought I was combining those cases with the aid of the squeeze lemma.
Is this a correct argument?
 A: Let $X$ be the described intersection.  Let $x \in X$.  Then we see clearly that $|x| \leq \frac{1}{n}$ for all $n \in \mathbb{N}$.  But the only $x \in \mathbb{R}$ that satisfies $|x| \leq \frac{1}{n}$ for all $n \in \mathbb{N}$ is $x = 0$, so $X = \{0\}$.
Notice that by quantifying over all $n$ directly, we can avoid explicit discussion of limits altogether (at least for this step of the proof).  If you "unwrap" the limit notation, it will ultimately say the same thing.
A: This is fine. A more explicit way to proceed is to say that for any fixed $x \neq 0$, there is some $n(x)$ such that $x \not \in (-1/n(x),1/n(x))$. This $n(x)$ can be taken to be, say, $\left \lceil \frac{1}{|x|} \right \rceil$.
A: There is no real need to take $n \to \infty$ here.
First note that
$$
A_{N} := \bigcap_{n=1}^N \left(-\frac{1}{n}, \frac{1}{n} \right) = \left(-\frac{1}{N}, \frac{1}{N} \right)
$$
for each $N \in \mathbb N$ and let
$$
A := \bigcap_{n=1}^\infty \left(-\frac{1}{n}, \frac{1}{n} \right)
$$
Also note that $A_{N+1} \subset A_N$ and $A \subset A_N$ for each $n \in \mathbb N$.
You have already shown that $0 \in A$.
Now you can continue by contradiction: Assume that $x \in A$, $x \neq 0$.
If $x > 0$, we are able to find a $N \in \mathbb N$ such that
$$
\frac{1}{N} < x
$$
by the Archimedean property.
This means that $x \notin A_N$ which implies that $x \notin A$. A similar argument can be made if $x < 0$, so we conclude that $x=0$ which is a contradiction.
Finally we conclude that $A = \{0\}$.
