# intersections of sets in metric spaces

(a) Let $A_{n}=B_{1/n}((0,0))$ in $\mathbb{R}^{2}$ with the usual metric. Show that $\bigcap_{n=1}^{\infty}A_{n}$ is not open.
(b) Find an infinite collection of distinct open sets in $\mathbb{R}^{2}$ with the usual metric whose intersection is a nonempty open set.

Attempt at (a): Clearly $(0,0)\in A_{n}$, $\forall n\in\mathbb{N}$ so $0\in\bigcap_{n=1}^{\infty}A_{n}$. Let $(a,b)\not=(0,0)$ s.t. $(a,b)\in\bigcap_{n=1}^{\infty}A_{n}$.
Since $d((a,b),(0,0))= \sqrt{a^2+b^2}\in\mathbb{R^+}>0$, can we say that by the archimidean property of $\mathbb{R}$ there exists an $n∈\mathbb{N}$ st. $\frac{1}{n}\lt d((a,b),(0,0))$ which means $(a,b)=(0,0)$ or $(a,b)\notin A_n$?

Attempt at (b): I need a collection of sets ${A_1,A_2,...}$ s.t. $\bigcap^{\infty}_{n=1} A_n$ is open. I'm using to finding the opposite -- an infinite collection of open sets whose intersection is closed as in (a). I'm having difficulty finding an open set $A$ s.t. $A\subseteq A_n$ $,\forall n$.

Edit: Had an idea for (b): The set $A_n=B_{1+n}((0,0)) \forall n\in\mathbb{N}$. The infinite intersection should be the open set $(-1,1)$, correct?

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For (b) you want the intersection to be a non-empty open set but your attempt has been to find a one-point set that is closed? I don't understand. – user38268 Jan 31 '12 at 16:43
Sorry Benjamin, Zev pointed out that error in (b) and I forgot to take it into account in my edits. It's now changed. – Emir Jan 31 '12 at 16:59
+1 for showing your efforts. – user38268 Jan 31 '12 at 17:06
a) is fine to me – leo Jan 31 '12 at 17:19
any hints for the set $A\subseteq A_n, \forall n$ for (b)? – Emir Jan 31 '12 at 17:50

You're on the right track for (a). Here is a hint for how to proceed: Let $(a,b)\in\mathbb{R}^2$ be distinct from $(0,0)$, i.e. $(a,b)\neq(0,0)$. What can you say about $d((a,b),(0,0))$? Can you show that, for any $(a,b)\neq(0,0)$, there is always an $n\in\mathbb{N}$ such that $$\frac{1}{n}<d((a,b),(0,0))\quad?$$ What does that say about the relationship between $(a,b)$ and $B_{1/n}((0,0))$?

You will then have to show that $\{(0,0)\}$ is not open. Recall that a set $S\subseteq\mathbb{R}^2$ is open when, for any $s\in S$, there is some $\epsilon>0$ such that $B_\epsilon(s)\subseteq S$. The negation of this (i.e., the conditions under which a set $S\subseteq\mathbb{R}^2$ is not open) is that there exists some $s\in S$ for which, given any $\epsilon>0$, we have $B_\epsilon(s)\not\subseteq S$.

Our set $S$ is $\{(0,0)\}$, so the only choice for $s\in S$ is $s=(0,0)$. Can you show that, for any $\epsilon>0$, the set $B_{\epsilon}((0,0))$ will contain points other than $(0,0)$, and therefore not be contained in $S=\{(0,0)\}$?

For (b), you are supposed to produce a open set in $\mathbb{R}^2$ - the set $\{(0,0)\}$ is not open (you should also be precise with your notation - it looks like you are working in $\mathbb{R}$??).

Here is a hint for how to proceed on (b): choose a non-empty open set $U\subseteq\mathbb{R}^2$. Find an infinite collection $\mathcal{V}=\{V_1,V_2,\ldots\}$ of distinct open sets of $\mathbb{R}^2$ such that $U\subset V_i$ for all $i$. Then let your collection be $\mathcal{V}\cup\{U\}$. Do you see why this works?

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Since $d((a,b),(0,0))=\sqrt{a^2+b^2}\in\mathbb{R^+}>0$, can we say that by the archimidean property of $\mathbb{R}$ there exists an $n\in\mathbb{N}$ st. $$\frac{1}{n}<d((a,b),(0,0))\quad?$$ which means $(a,b)=(0,0)$ or $(a,b)\not\in A_n$? – Emir Jan 31 '12 at 6:23
@Emir If you know about topological spaces then since a metric space is Haussdorff, it is also $T_1$ so that every finite point set is closed. – user38268 Jan 31 '12 at 17:06