Every metric space is completely normal.

Question

Every metric space is completely normal.

Let $(X,d)$ be a metric space. Our aim is to prove, For any subsets $A$ and $B$ of $X$ such that $\overline A\cap B=B\cap\overline A=\emptyset,$ We need to find two disjoint non-empty open subsets of $X$ such that $A\subseteq U$ and $B\subseteq V.$ Since, $A$ and $B$ are non-empty subset of $X$. we can find $\text{dist}(A,\overline{B})>0$ and $\text{dist}(B,\overline{A})>0$. Choose $d=\text{dist}(A,\overline{B})$. Let $U=\cup_{x\in U}B_d(x,d)$ and $U=\cup_{x\in A}B_d(x,d)$ and $V=\cup_{x\in \overline B}B_d(x,d): A\subseteq U$ and $B\subseteq V$. (Where $\text{dist}(A,\overline{B})=\inf\{d(x,y):x\in A \wedge x\in \overline{B}\} $). Is my proof correct?

Answer 1

A better idea is to adapt the distance per point instead:

For each $a \in A$ pick $r_a >0$ such that $B(a,r_a) \cap B = \emptyset$, and also for each $b \in B$ pick $s_b >0$ such that $B(b, s_b) \cap A = \emptyset$.

Then use $U = \bigcup\{B(a,\frac{r_a}{3}): a \in A\}$ and $V=\bigcup \{ B(b, \frac{s_b}{3}): b \in B\}$ as the required open sets (open balls are open, and so are all unions of them, $A \subseteq U$ and $B \subseteq V$ are obvious, and use the triangle inequality to show that $U$ and $V$ are disjoint.

There need not be a positive distance between two separated sets: consider $A=(0,1), B=(1,2)$ in the reals e.g.