Two spaces can be combined into one whose points are pairs. The set is the Cartesian product, familiar from coordinates in the plane, and it comes with two maps that read off a coordinate. What it does not come with is a topology: that has to be chosen, and the choice is the subject of this chapter.

A seat in a theatre is a row and a number. The seating plan is the product of the two lists, and the two ushers who read off row and number are the projections.

A point of $X\times Y$ is a pair, and the two projections read off its coordinates: $\pi_1(p)$ is where it stands on the first axis and $\pi_2(p)$ where it stands on the second. The set is settled; which subsets count as open is not.

For sets $X$ and $Y$, the Cartesian product $X\times Y$ is the set of ordered pairs $(x,y)$ with $x\in X$ and $y\in Y$. It carries two projection maps, $\pi_1(x,y)=x$ and $\pi_2(x,y)=y$. A subset of the form $A\times B$ is called a box; not every subset of a product is one.

• $\mathbb{R}\times\mathbb{R}$ is the plane, and $\mathbb{R}^n$ is a product of $n$ copies of the line. The product of an interval with itself is a square.
• Most subsets of a product are not boxes: the diagonal $\{(x,x)\}$ and the disc $x^2+y^2<1$ are not products of two sets. A box is a very special shape.
• Boxes behave well under intersection: $(A\times B)\cap(C\times D)=(A\cap C)\times(B\cap D)$. Under unions they do not: a union of two boxes is usually not a box.
• The projections are onto whenever the other factor is non-empty, and they are almost never injective.
• Products of more than two factors are defined the same way, and a product of finitely many spaces can be built two at a time without changing anything.

Both sides are sets of pairs, so the proof reads each side at a single pair. A pair lies in the left-hand side when it lies in both boxes, which says its first coordinate is in A and in C and its second is in B and in D. That is exactly the condition for lying in the right-hand side. Nothing is used beyond the definition of a box and of an intersection, which is why this identity holds for any sets at all.

Take a pair in the intersection of the two boxes.

Membership of a box is a condition on each coordinate, so there are four conditions.

Group the two conditions on the first coordinate and the two on the second.

That is membership of the box on the right, and every step is reversible.

The product is the set of pairs; the projections read off a coordinate.

• $X\times Y$ is the set of ordered pairs, with projections $\pi_1$ and $\pi_2$.
• A box is a subset of the form $A\times B$; most subsets are not boxes.
• Boxes intersect in a box; they do not unite in one.
• The set is forced; the topology on it is still to be chosen.

Great work! You now know how to:

• form a product of sets and its projections;
• recognise which subsets are boxes;
• intersect boxes coordinate by coordinate;
• see that the topology on a product still has to be chosen.