Proving one infinite intersection of sets I've a doubt on how do we prove this kind of stuff involving infinite intersections. My point is: the book I'm working with gives the following example to prove that the intersection of infinite sets may not be open: let $a \in \mathbb{R}^n$, given the familly of balls $B(a; 1/k)=\left\{x \in \mathbb{R}^n \mid d(x,a)<1/k\right\}$ we have:
$$\bigcap_{k=1}^\infty{B(a; 1/k)}=\left\{a\right\}$$
However I can't see this intuitively, nor how to prove it formally. How do we deal with this kind of problem involving infinite intersections of sets? I tried to understand intuitively what this means: we are enclosing the point with balls on which as $k$ grows the "size" of the neighborhood reduces, so that if we make $k$ really big the only point in common on all of these balls will be the point at it's center. I don't know if my intuition is correct, but even if it is, how do we make this formal ?
Thanks in advance for the help.
 A: Your intuition is sound. Pick any other point $x\in\mathbb R^n$.  Then we have $d(a,x)=\varepsilon>0$.  We may find $N\in\mathbb N$ such that $\varepsilon>1/N>0$.  Thus $x\notin B(a;1/N)$, and so it cannot be in the infinite intersection either.  Does this help?
A: Formally, if we have some collection $\mathcal{C}$ of sets, then
$a\in\bigcap_{S\in\mathcal{C}}S$ if and only if $a\in S$ for every $S\in\mathcal{C}$.
Thus, in your example, clearly $a\in B(a,\frac{1}{k})$ for every $k$, hence $a\in\bigcap_{k=1}^\infty B(a,\frac{1}{k})$. Furthermore, for any $b\neq a$, we have $d(b,a)>0$ so we can pick some $k$ such that $\frac{1}{k}<d(b,a)$ and hence $b\notin B(a,\frac{1}{k})$. Thus $b\notin \bigcap_{k=1}^\infty B(a,\frac{1}{k})$ and we have proven $\bigcap_{k=1}^\infty B(a,\frac{1}{k}) = \{a\}$.
A: Your intuition is perfectly fine. Formally, I will call the intersection $I$ and suppose that there were some element $b\in I$ s.t. $b\neq a$. Since $d$ is a metric function $a\neq b$ implies $d(a,b)>0$ Let this distance be $\epsilon$. Since the sequence $x_k=\frac{1}{k}$ converges to 0 for large $k$, there will exist $K$ s.t. $x_K<\epsilon$. Since $x_K=\frac{1}{K}<\epsilon$ you have that $b\notin B(a;x_K)$ . But $b\in\cap_{k=1}^\infty B(a;\frac{1}{k})$ implies $b\in B(a;x_K)$. This is contradiction. This means our initial assumption was false. It follows that there doesnt exist $b$ such as assumed. 
