Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 3 / 13B / Solution

The product topology on $X\times Y$ consists of arbitrary unions of rectangles $U\times V$ with $U\subseteq X$, $V\subseteq Y$ open. Such rectangles form a basis of a topology: every point lies in one, and intersections of two rectangles are rectangles with open factors.

If $d_X,d_Y$ induce the factor topologies, setDefiniteness and symmetry are immediate. Each factor triangle inequality bounds its distance by the sum of the two relevant product distances; taking the maximum proves the product triangle inequality. A ball of radius $r$ is exactly $B_X(x,r)\times B_Y(y,r)$, so every metric-open set is product-open. Conversely an open rectangle containing $(x,y)$ contains a product ball with radius the minimum of two suitable factor radii. Hence the maximum product metric induces precisely the product topology.

A compact space is one for which every open cover has a finite subcover. Let $\mathcal{U}$ cover $X\times Y$, with both factors compact; if either is empty the result is immediate. For each fixed $x$ and each $y$, choose a rectangle $U_{x,y}\times V_{x,y}$ containing $(x,y)$ and lying in a member of $\mathcal{U}$. The sets $V_{x,y}$ cover $Y$, so finitely many, indexed by $y_1,\dots ,y_{m_x}$, suffice. Their $X$-neighbourhoods have open intersectioncontaining $x$. The finitely many selected members of $\mathcal{U}$ cover all of $U_x\times Y$. Now the sets $U_x$ cover $X$, so select finitely many of these by compactness of $X$. Taking the union of their finite selections from $\mathcal{U}$ gives a finite cover of the whole product. The product of two compact spaces is compact, without requiring the factors to be Hausdorff.