Prove that \(X, Y\) are Hausdorff iff \(X\cross Y\) is Hausdorff.
Solution
1. If \(X\) and \(Y\) are Hausdorff, then \(X\times Y\) is Hausdorff.
For distinct \((x_1,y_1),(x_2,y_2)\), at least one coordinate differs. Separate that coordinate by disjoint open neighborhoods and take the product with the whole other factor.
2. Conversely, if \(X\times Y\) is Hausdorff and both factors are nonempty, then \(X\) and \(Y\) are Hausdorff.
Choose \(y_0\in Y\). The slice \(X\times\{y_0\}\) is a subspace of the Hausdorff product and is homeomorphic to \(X\), hence \(X\) is Hausdorff. Similarly \(Y\) is Hausdorff using a point of \(X\).
3. Thus, for nonempty factors,\[\boxed{X\times Y\text{ is Hausdorff}\iff X\text{ and }Y\text{ are Hausdorff}.}\]
Combine <1>1–<1>2. If an empty factor is allowed, the converse needs the stated nonemptiness qualification because the empty product space is Hausdorff regardless of the other factor.