Separating two points with open sets whose closure is also disjoint

Question

Let $x, y \in M$ be two different points in a metric space $M$, show that there are two open disjoint sets $U, V$ with $x \in U$ and $y \in V$. Moreover, show that $U$ and $V$ can be chosen such that $cl(U)$ and $cl(V)$ are disjoint.

If $r=d(x,y)$ then the open balls $B_{r/2}(x)$ and $B_{r/2}(y)$ are disjoint (actually any open ball with radius $r' < r/2$), but I'm having trouble finding the open sets whose closure is also disjoint.

Answer 1

A solution from first principles: If $x \neq y$, let $r = d(x,y) > 0$. Define $s = { r\over 4}$. (with room to spare..)

Then $\overline{B(x, s)} \cap \overline{B(y,s)} = \emptyset$ and we have the required open balls.

Proof: suppose for a contradiction that $z \in \overline{B(x, s)} \cap \overline{B(y,s)}$.

Then as $z \in \overline{B(x, s)}$ we pick $x' \in B(x,s) \cap B(z,s)$ so that $d(x',x) < s$ and $d(x',z) < s$.

Then as $z \in \overline{B(y, s)}$ we pick $y' \in B(y,s) \cap B(z,s)$ so that $d(y',y) < s$ and $d(y',z) < s$.

But then by repeatedly applying the triangle inequality, plus what we know about all the distances, we have: $$r= d(x,y) \le d(x,x') + d(x',z) + d(z,y') + d(y',y) < 4s = r$$

which is a contradiction. So the closures of the balls are disjoint.

Answer 2

First show B[r,a] = { x : d(x,a) <= r } is closed.
Then use B(r',x) and B(r',y) with r' < d(x,y)/2.