Please help make my infinite intersection proof more rigorous! I am working on problems from my textbook for self-study and would like to know how I can make my proof more rigorous - I am having trouble expressing what I am thinking mathematically. I want to show that the infinite intersection of open sets may not be open. To demonstrate this, I am using the sequence $A_n$ = $B_{1/n}((0,0))$, which is the ball with radius $\frac 1 n$ around the origin. Below is what I have:
Denote the infinite intersection of $A_n$ as $U$. We know that $(0,0) \in U$ and thus $(0,0) \in A_n$ for all $n$. Since for all $n$, $A_n$ is open, we also know that $B_{1/n}((0,0)) \subset U$. Let $r$ be the smallest radius of $A_n$ for all $n$. Then $B_r((0,0)) \subset B_{1/n}((0,0)) \subset U$. But since $\lim \frac 1 n \to 0$, $r=0$. Thus $B_r ((0,0)) = (0,0)$, which is closed as its complement is open.
 A: You’ve said a couple of things exactly backwards. 

We know that $\langle 0,0\rangle\in U$ and thus $\langle 0,0\rangle\in A_n$ for all $n$.

No, it’s the other way around: we know that $\langle 0,0\rangle\in U$ because we know that $\langle 0,0\rangle\in A_n$ for each $n\in\Bbb Z^+$
It’s also not true that $B_{1/n}(\langle 0,0\rangle)\subseteq U$: the inclusion goes the other way. $U=\bigcap_{n\in\Bbb Z^+}A_n$, so $U\subseteq A_n$ for each $n\in\Bbb Z^+$, which of course means that $U\subseteq B_{1/n}(\langle 0,0\rangle)$ for each $n\in\Bbb Z^+$, by the definition of the sets $A_n$.
There is no smallest radius of the sets $A_n$: no matter what $r>0$ you choose, there is an $n\in\Bbb Z^+$ such that $\frac1n<r$, and the radius of $A_n$ is then less than $r$. In fact this is what you really want; you’re just not using it quite right. You want to use it to prove that $\langle 0,0\rangle$ is the only point of $U$, so suppose that $\langle x,y\rangle\in\Bbb R^2\setminus\{\langle 0,0\rangle\}$. Let $r=\sqrt{x^2+y^2}$; then $r>0$, so there is an $n\in\Bbb Z^+$ such that $\frac1n<r$, and you can easily check that $\langle x,y\rangle\notin A_n$ and hence $\langle x,y\rangle\notin U$. This shows that $U=\{\langle 0,0\rangle\}$, which, as you say, is closed.
However, a set can be both closed and open; $\varnothing$ is such a set, for example. Thus, it’s not enough to say that $\{\langle 0,0\rangle\}$ is closed: you must actually show that it’s not open. This is easy: if it were open, it would contain $B_\epsilon(\langle 0,0\rangle)$ for some $\epsilon>0$, and it clearly does not, since it does not contain (for instance) $\left\langle\frac{\epsilon}2,0\right\rangle$.
