Let $x,y\in M, \ x\neq y,\ M \ \text{being a metric space}$.
I'm asked to prove that there exists open sets $U,V\subset M$ such that $x\in U,\ y\in V \ \text{and} \ \bar{U}\cap\bar{V} = \emptyset$ where $\bar{U}, \ \bar{V}$ defines the closure of the respective set.
My understanding of closure is that the closure $\bar{U}$ of $U$ is the intersection of all closed sets containing $U$, like
$$\bar{U} = U_1 \cap U_2 \cap \dots \cap U_n$$
where $U_i$ is a closed set containing $U$.
A closed set in turn is a set where every converging sequence of that set converges to a point of the set.
My attempt so far is to construct open balls with the radius of half the distance between the points $x,y$ but I can't satisfy the last condition. How can I construct $U,V$ such that their closure does not intersect? How do I know what closed set will contain $U$ or $V$ without more information?
Best Answer
Constructing a pair of small balls is exactly the right idea. In order to understand the closure of a ball $B(x,r)$, you would just need to prove that that the closed ball $\bar{B}(x,r)=\{y \in M| d(x,y) \leq r \}$ is a closed set containing the open ball. Therefore the closure of the open ball is contained in the closed ball.
Once you've made your balls small enough (you might be able to see why half the distance isn't small enough, now) that will give you what you want.