Prove Theorem 19.4: if each space \(X_\alpha\) is a Hausdorff space, then \(\prod X_\alpha\) is a Hausdorff space in both the box and product topologies.
Solution
Let \(x=(x_\alpha)\) and \(y=(y_\alpha)\) be distinct points of \(\prod X_\alpha\). Then for some index \(\beta\), \[ x_\beta\ne y_\beta. \] Since \(X_\beta\) is Hausdorff, choose disjoint open neighborhoods \(U_\beta,V_\beta\) of \(x_\beta,y_\beta\).
In the product topology, set \[ U=\pi_\beta^{-1}(U_\beta),\qquad V=\pi_\beta^{-1}(V_\beta). \] These are disjoint open neighborhoods of \(x,y\). Hence the product topology is Hausdorff.
The box topology is finer than the product topology, so the same two sets are box-open. Thus the box product is Hausdorff as well.