Past exam of the options I course of the University of Oxford / 2012 / E / E2

A basis of a topology $\mathcal{B}$ is a family of open subsets such that every open set is a union of members of $\mathcal{B}$. Equivalently, for every $U\in \mathcal{T}$ and $x\in U$, some $B\in \mathcal{B}$ satisfiesThe empty open set is represented by the empty union.

Define $\mathcal{T}$ to be all unions of members of $\mathcal{B}$, including the empty union. Then $\varnothing \in \mathcal{T}$ and the covering assumption gives $X\in \mathcal{T}$. Arbitrary unions of such sets are again unions of basis members.

For two members $U={\bigcup }_i{B}_i$ and $V={\bigcup }_j{B}_j^{\prime }$, distribute intersection over union:Each intersection is a union of members of $\mathcal{B}$ by the second assumption, so $U\cap V\in \mathcal{T}$. Repeating gives closure under all finite intersections. These are precisely the topology axioms. Every $B\in \mathcal{B}$ belongs to $\mathcal{T}$, and every set in $\mathcal{T}$ is a union of such members by construction. Thus

The product topology on $X\times Y$ has as a basis all open rectanglesThus a subset $W\subseteq X\times Y$ is open precisely when, around every $(x,y)\in W$, there are open $U\ni x$ and $V\ni y$ with $U\times V\subseteq W$. These rectangles cover the product, and their pairwise intersections are rectangles because $(U\times V)\cap (U^{\prime }\times V^{\prime })=(U\cap U^{\prime })\times (V\cap V^{\prime })$. Hence they define a topology by part (b).

This is equivalently the coarsest topology making the two coordinate projection maps continuous: the preimages of factor-open sets are the open strips $U\times Y$ and $X\times V$, whose finite intersections are the basis rectangles.

Let $\{B_i:i\in I\}$ and $\{C_j:j\in J\}$ be countable bases of the two factors. The familyis countable, since the Cartesian product of two countable sets is countable. It is a basis of a topology for the product topology: given a product-open neighbourhood of $(x,y)$, choose a contained open rectangle $U\times V$, then basis members with $x\in B_i\subseteq U$ and $y\in C_j\subseteq V$. Their product contains $(x,y)$ and is contained in the original neighbourhood. ThereforeIn standard terminology, a finite product of second-countable spaces is second-countable.