Prove that a metric space \(X\) is normal, i.e. if \(A, B \subset X\) are closed and disjoint then there exist open sets \(A \subset U \subset X, ~B \subset V \subset X\) such that \(U \cap V = \emptyset\).
Solution
Let \(d\) be the metric on \(X\), and let \(A,B\subseteq X\) be disjoint closed sets.
1. If \(A=\emptyset\) or \(B=\emptyset\), the required open neighborhoods exist.
If \(A=\emptyset\), take \[ U=\emptyset, \qquad V=X. \] If \(B=\emptyset\), interchange the two choices. Hence assume below that both \(A\) and \(B\) are nonempty.
2. For every \(a\in A\), choose \(\varepsilon_a>0\) such that\[ B_d(a,\varepsilon_a)\cap B=\emptyset, \]and for every \(b\in B\), choose \(\delta_b>0\) such that\[ B_d(b,\delta_b)\cap A=\emptyset. \]
Because \(B\) is closed, \(X\setminus B\) is open and contains every \(a\in A\). Thus some metric ball centered at \(a\) lies in \(X\setminus B\). The same argument with \(A\) and \(B\) interchanged gives the radii \(\delta_b\).
3. Define\[ U=\bigcup_{a\in A}B_d\left(a,\frac{\varepsilon_a}{2}\right), \qquad V=\bigcup_{b\in B}B_d\left(b,\frac{\delta_b}{2}\right). \]Then \(U\) and \(V\) are open, with\[ A\subseteq U, \qquad B\subseteq V. \]
Each metric ball is open, so both unions are open. Every \(a\in A\) belongs to its ball \(B_d(a,\varepsilon_a/2)\), and every \(b\in B\) belongs to \(B_d(b,\delta_b/2)\).
4. The open sets \(U\) and \(V\) are disjoint.
Suppose instead that \(z\in U\cap V\). Then for some \(a\in A\) and \(b\in B\), \[ d(a,z)<\frac{\varepsilon_a}{2}, \qquad d(z,b)<\frac{\delta_b}{2}. \] The triangle inequality gives \[ d(a,b) \le d(a,z)+d(z,b) <\frac{\varepsilon_a+\delta_b}{2}. \]
2.If \(\delta_b\le\varepsilon_a\), then\[ d(a,b)<\varepsilon_a, \]so \(b\in B_d(a,\varepsilon_a)\), contradicting the choice of \(\varepsilon_a\).Thus \(U\cap V=\emptyset\).
5. Therefore every metric space is normal.
For arbitrary disjoint closed subsets \(A,B\), steps <1>1–<1>4 produce disjoint open neighborhoods \(U\supseteq A\) and \(V\supseteq B\), which is the stated definition of normality.