Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 1 / 12A (Metric and Topological Spaces)

The product topology on $X\times Y$ has as a basis the rectangles $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Equivalently, it is the coarsest topology making both coordinate projections continuous.

To show that the product of two Hausdorff spaces is Hausdorff, take distinct points $(x_1,y_1)$ and $(x_2,y_2)$. If $x_1\ne x_2$, choose disjoint open neighborhoods $U_1,U_2$ in $X$; the rectangles $U_1\times Y$ and $U_2\times Y$ separate the product points. If the first coordinates agree, the second coordinates differ, and the same argument uses disjoint open neighborhoods in $Y$. Thus every pair of distinct product points has disjoint open neighborhoods.

The three implications below prove the requested equivalence without assuming metrizability.

Assume $X$ is Hausdorff. For any $(x_1,x_2)$ outside the diagonal, the points differ, so choose disjoint open neighborhoods $U_1,U_2$. Their rectangle contains $(x_1,x_2)$ and misses the diagonal. Thus the complement of the diagonal is open in the product topology, proving (i) implies (ii). This is one direction of the closed diagonal theorem.

Assume the diagonal $\Delta \subset X\times X$ is closed. For continuous $f,g:Y\to X$, the map $h:Y\to X\times X$, $h(y)=(f(y),g(y))$, is continuous: the inverse image of a basic rectangle $U\times V$ is $f^{-1}(U)\cap g^{-1}(V)$. Thereforeis closed as the inverse image of a closed set under a continuous map. This proves (ii) implies (iii).

Assume (iii). Take $Y=X\times X$ and let $f,g$ be the two coordinate projections, continuous by the definition of the product topology. Their equality set is the diagonal, so it is closed. For distinct $x_1,x_2$, the point $(x_1,x_2)$ lies in its open complement. Choose a basic rectangle $U_1\times U_2$ containing it and contained in that complement. If $U_1\cap U_2$ were nonempty, a point $u$ in their intersection would give $(u,u)$ in the rectangle, impossible. Thus $U_1,U_2$ are disjoint open neighborhoods of $x_1,x_2$, proving that $X$ is Hausdorff. Hence (iii) implies (i), completing all three equivalences.