Show that a closed subspace of a normal space is normal.
Solution
Goal. Show a closed subspace of a normal space is normal.
1. Let \(X\) be normal and \(Y \subseteq X\) closed.
setup.
2. \(Y\) is \(T_1\).
a subspace of a \(T_1\) space is \(T_1\) (singletons are closed in \(X\), hence in \(Y\)).
3. \(Y\) is normal.
3.1. Let \(A, B \subseteq Y\) be disjoint closed subsets of \(Y\).
take arbitrary disjoint closed sets in \(Y\).
3.2. \(A\) and \(B\) are closed in \(X\).
\(A\) is closed in \(Y\) and \(Y\) is closed in \(X\), so \(A\) is closed in \(X\); same for \(B\).
3.3. By normality of \(X\), there are disjoint open \(U, V \subseteq X\) with \(A \subseteq U\) and \(B \subseteq V\).
\(A, B\) are disjoint closed sets in the normal space \(X\).
3.4. Then \(U \cap Y\) and \(V \cap Y\) are disjoint open sets in \(Y\) separating \(A\) and \(B\).
restrict the open sets to \(Y\).
4. Q.E.D.
3 shows \(Y\) is normal.