$x \neq y\in$ metric space $M$, prove $\exists$ open sets $U,V$ s.t. $x\in U,\ y\in V$ and $\bar{U} \cap \bar{V} = \emptyset$

Question

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?

Answer 1

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.

Answer 2

Big Hint

Take $$U=\left\{t\in M\mid d(t,x)<\frac{d(x,y) }{4}\right\}\quad \text{and}\quad V=\left\{t\in M\mid d(t,y)<\frac{d(x,y)}{4}\right\}.$$

Answer 3

I'll call an open ball $B$ and a closed ball $D$.

Let $4r=d(x,y)$. So we define $U=B_r(x), V=B_r(y)$. Now, the closed balls $D_r(x)$ and $D_r(y)$ are closed sets (why?) which contain $U$ and $V$ respectively, hence they contain their closures. Also they are disjoint. So $\bar{U}\subset D_r(x)$, $\bar{V}\subset D_r(y)$ and hence $\bar{U}\cap\bar{V}$ is the empty set.

Answer 4

Consider $B_1 = \{ w : d(x,w) < r \}, B_2 = \{ u : d(u,y) < r \}$ , with $ r = \dfrac {d(x,y)} {3}$. These balls are the sets $U, V$ you need.

Answer 5

Definitions mean something. Do not think of the closure of a set $A$ simply as "the intersection of all closed sets containing $A$", but instead as "the smallest closed set containing $A$". Indeed, the closure $\overline{A}$ of $A$ is the only closed subset of $M$ satisfying the following two properties:

1. $A\subseteq\overline{A}$;
2. If $C$ is any other closed set containing $A$, then $\overline{A}\subseteq C$.

Your idea of taking balls of small radius is good. However, taking balls of radius $d(x,y)/2$ does not work even on euclidean space! It is always good to try to think of how your approach would work on some canonical example. So we need to take smaller radi.

Now, you have probably seen the following: If $B_r(x)=\left\{y:d(x,y)<r\right\}$ denotes the open ball of radius $r$ around $x$, and $C_r(x)=\left\{y:d(x,y)\leq r\right\}$ denotes the closed ball of radius $r$ around $x$, then $B_r(x)$ is open and $C_r(x)$ is closed. Clearly, $B_r(x)\subseteq C_r(x)$, so the description above of closures implies that $\overline{B_r(x)}\subseteq C_r(x)$.

(See A closed ball in a metric space is a closed set)

Moreover, note that if $r<R$, then $C_r(x)\subseteq B_R(x)$. These inclusions are useful because they allow us to include open balls, their closures, and closed balls in each other.

So instead take any radius strictly smaller than $d(x,y)/2$, for example $d(x,y)/3$. Then in fact we will have $$\overline{B_{d(x,y)/3}(x)}\cap\overline{B_{d(x,y)/3}(y)}\subseteq C_{d(x,y)/3}\cap C_{d(x,y)/3}\subseteq B_{d(x,y)/2}\cap B_{d(x,y)/2}=\varnothing,$$ and you probably have already seen the last equality.

Answer 6

Pretty straight forward. Let $r = d(x,y) > 0.$ Take open balls $U=B_{\frac {r} {2}} (x)$ and $V=B_{\frac r 2} (y)$ surrounding $x$ and $y$ respectively of radius $\frac r 2$ each. Then observe that $B_{\frac r 2} [x] \cap B_{\frac r 2} [y] = \emptyset.$ Since $\overline U \subseteq B_{\frac r 2} [x]$ and $\overline V \subseteq B_{\frac r 2} [y]$ it follows that $\overline U \cap \overline V = \emptyset,$ as claimed.