Past exam of the options I for mathematics and philosophy course of the University of Oxford / 2013 / E / 1

A basis of a topology $\mathcal{T}$ is a collection $\mathcal{B}\subseteq \mathcal{T}$ such that every open set is a union of members of $\mathcal{B}$. Equivalently, for each open $U$ and every $x\in U$, there exists $B\in \mathcal{B}$ with $x\in B\subseteq U$. The empty union represents the empty set. In particular, the basis covers $X$, and any intersection of two basis elements is itself a union of basis elements.

The following open rectangles form a basis of a topology for the product topology:These open rectangles in a finite product cover $X\times Y$, and intersections remain rectangles since $(U_1\times V_1)\cap (U_2\times V_2)=(U_1\cap U_2)\times (V_1\cap V_2)$. One may equivalently take products of elements from a chosen basis of a topology in each factor.

Choose $v_0\in V$, using its nonemptiness. For each $u\in U$, the openness of $U\times V$ in the product topology supplies open $O_u\subseteq X$, $W_u\subseteq Y$ such thatSince $v_0\in W_u$, every point of $O_u$ belongs to $U$. Thus $U={\bigcup }_{u\in U}{O}_u$ is open. Interchanging the factors, using a fixed $u_0\in U$, proves $V$ open. Therefore $\boxed{U\in {\mathcal{T}}_X,V\in {\mathcal{T}}_Y}$. Nonemptiness matters: an empty product is open whatever the other factor is.

Yes: at least one factor topology must be indiscrete. Suppose instead that both have a nonempty proper open set, say $\varnothing \ne U⊊X$ and $\varnothing \ne V⊊Y$. Thenis open in the product topology. Both coordinate projections of $W$ are the full factors, because $U,V$ are nonempty. If $W$ were a rectangle $A\times B$, this would force $A=X$, $B=Y$, hence $W=X\times Y$. But choosing $x\notin U$ and $y\notin V$ gives $(x,y)\notin W$, a contradiction. Thus both factors cannot have nontrivial open sets. This is rectangularity of every product-open set.