Past exam of the topology course of the University of Oxford / 2014 / 2 / a

The product topology has basisOpen subsets of the product are arbitrary unions of these rectangles. The rectangles cover the product and their finite intersections are rectangles, so they satisfy the basis axioms. Equivalently, this is the smallest topology making both coordinate projections continuous.

First assume both factors are nonempty. If the product is compact, its continuous coordinate projections are onto, so each factor is compact by the continuous image of a compact space theorem.

For the converse, let $X,Y$ be compact and let $(W_{\lambda })$ be an open cover of their product. Fix $x\in X$. For every $y\in Y$, choose a basis rectangle $U_{x,y}\times V_{x,y}$ containing $(x,y)$ and lying in a cover member $W_{\lambda (x,y)}$. Compactness of $Y$ gives finitely many $V_{x,y_j}$ covering $Y$. Thenis an open neighbourhood of $x$, and $U_x\times Y$ is covered by the corresponding finitely many $W_{\lambda (x,y_j)}$.

The sets $U_x$ cover $X$. Compactness gives finitely many of them covering $X$, and the finitely many cover members associated to these choices now cover $X\times Y$. Hence the product is compact. This is the finite-product Tychonoff theorem proof, with the finite intersection creating a uniform neighbourhood along each fibre.

As printed, the assertion omits the nonempty hypothesis. If $X=\varnothing$ and $Y=\mathbb{R}$, the product is compact while $Y$ is not. The fully correct version without any nonemptiness assumption is