Show that if \(\prod X_\alpha\) is Hausdorff, or regular, or normal, then so is \(X_\alpha\). (Assume that each \(X_\alpha\) is nonempty.)
Solution
Fix an index \(\alpha\), and choose basepoints \(a_\beta\in X_\beta\) for every \(\beta\ne\alpha\). The map \[ i:X_\alpha\to\prod_\beta X_\beta, \qquad i(x)_\alpha=x,\quad i(x)_\beta=a_\beta\ (\beta\ne\alpha) \] is a homeomorphism from \(X_\alpha\) onto the corresponding slice \(S\).
Hausdorffness and regularity are hereditary to subspaces, so if the product is Hausdorff or regular, then \(S\), hence \(X_\alpha\), has the same property.
If the product is normal, then it is \(T_1\), so every singleton \(\{a_\beta\}\) is closed. Hence \[ S=\bigcap_{\beta\ne\alpha}\pi_\beta^{-1}(\{a_\beta\}) \] is closed in the product. Closed subspaces of normal spaces are normal. Therefore \(X_\alpha\) is normal as well.