Prove that every metrizable space is normal Hausdorff (aka \(T_4\)).
Solution
Let \(d\) be a metric inducing the topology of \(X\).
1. \(X\) is Hausdorff.
If \(x\neq y\), let \(r=d(x,y)/3\). A point \(z\in B(x,r)\cap B(y,r)\) would give \(d(x,y)\le d(x,z)+d(z,y)<2r=\frac23d(x,y)\), a contradiction; so \(B(x,r)\) and \(B(y,r)\) are disjoint.
2. Let \(A,B\subseteq X\) be disjoint nonempty closed sets. Then \(d(a,B)>0\) for every \(a\in A\) and \(d(b,A)>0\) for every \(b\in B\).
Because \(B\) is closed and \(a\notin B\), the open set \(X\setminus B\) contains a ball \(B(a,\varepsilon)\), so \(d(a,B)\ge\varepsilon>0\). The same argument with \(A\) and \(B\) interchanged gives \(d(b,A)>0\).
3. With \(A\) and \(B\) as in step 2, the sets \[U=\bigcup_{a\in A}B\!\left(a,\frac{d(a,B)}3\right), \qquad V=\bigcup_{b\in B}B\!\left(b,\frac{d(b,A)}3\right)\] are disjoint open sets with \(A\subseteq U\) and \(B\subseteq V\).
They are unions of open balls, and each center lies in its ball because the radius is positive by step 2. Suppose \(z\in U\cap V\), and choose \(a\in A\) and \(b\in B\) with \(d(z,a)<d(a,B)/3\) and \(d(z,b)<d(b,A)/3\). Since \(d(a,B)\le d(a,b)\) and \(d(b,A)\le d(a,b)\), \[d(a,b)\le d(a,z)+d(z,b)<\frac{d(a,B)+d(b,A)}3\le\frac23d(a,b),\] which is impossible because \(d(a,b)>0\).