Let $M$ be a metric space. Show that $M$ is normal. 
Let $M$ be a metric space. Show that $M$ is normal.

Let $F_1$ and $F_2$ be disjoint closed sets of $M$. Then there exists $r >0$ such that for all $x \in F_1$ and $y \in F_2$ we have  $d(x,y) >r>0$.
Now take $U=\bigcup_{x \in F_1} B(x, r/4)$ and $V=\bigcup_{y \in F_2}B(y,r/4)$. Then $U$ and $V$ are disjoint open neighborhoods of $F_1$ and $F_2$.
This was a proof given by a professor and I wonder where are the sets $U=\bigcup_{x \in F_1} B(x, r/4)$ and $V=\bigcup_{y \in F_2}B(y,r/4)$ coming from and what proves that they're disjoint?
 A: Ultimately, our objective is to find open neighborhoods containing $F_1$ and $F_2$ that are also disjoint. How do we do this? Well, we first define the distance function between a point and a set of points via $\text{dist}(x,S)=\inf_{s\in S} d(x,s)$.
It is not the case that $F_1$ and $F_2$ are separated by some distance $r>0$ as the original post suggests, but instead we can see that for every $x\in F_1$, there is some $r(x)>0$ such that $\text{dist}(x,F_2)\geq r(x)$, and we can similarly define $s(y)>0$ for $y\in F_2$. (Why is this true? Well, we can intersect $F_2$ with some closed ball of large radius to get a compact set and then apply the Extreme Value Theorem to $\text{dist}(\cdot,F_2)$ on this set (one can verify that it is continuous) to get a minimum value. This minimum value is nonzero as otherwise $x$ is a limit point of $F_2$ and would be in its closure.)
Now, we can take $U=\bigcup_{x\in F_1} B(x,\tfrac{r(x)}{4})$ and $V=\bigcup_{y\in F_2} B(y,\tfrac{s(y)}{4})$ and proceed as follows: Towards a contradiction, suppose there is some $z\in U\cap V$. Then there exist $x\in F_1,y\in F_2$ such that $z\in B(x,\tfrac{r(x)}{4})\cap B(y,\tfrac{s(y)}{4})$. Noting that $\tfrac{r(x)}{4}\leq \frac{d(x,y)}{4}$ and $\tfrac{s(x)}{4}\leq \frac{d(x,y)}{4}$ by definition, by the triangle inequality we have
$$d(x,y)\leq d(x,z)+d(z,y)\leq \frac{r(x)}{4}+\frac{s(x)}{4}\leq \frac{d(x,y)}{4}+\frac{d(x,y)}{4}=\frac{d(x,y)}{2},$$
which is nonsense since $x\neq y$.
A: I think you can avoid using the distance from a point to a closed set if you argue like this: since  $F_1\subseteq M\setminus F_2$ and $F_2\subseteq M\setminus F_1$ and these sets are open, for each $x\in F_1$ there is a ball $B(x,r_x)\subseteq M\setminus F_2.$ Similarly for $y\in F_2$.
Now let $U$ be the union of the $B(x,r_x/2)$ as $x$ ranges over $F_1$ and $V$ be the union of the $B(y,r_y/2)$ as $y$ ranges over $F_2.$
Suppose $z\in U\cap V$, then there are $x\in B(x,r_x/2)$ and $y\in B(y,r_y)$ such that $z\in B(x,r_x/2)\cap B(y,r_y/2).$  Without loss of generality, $r_x\le r_y.$ But now $d(x,y)\le d(x,z)+d(z,y)<r_y\Rightarrow x\in M\setminus F_1,$ which is a contradiction.
