In this article, we consider the product of two topological spaces. To motivate our definition, we first begin with metric spaces (*X*, *d _{X}*) and (

*Y*,

*d*). Letting

_{Y}*Z*=

*X*×

*Y*be the set-theoretic product, we wish to define a metric on

*Z*from

*d*and

_{X}*d*. Here’re some possibilities, all of which seem rather reasonable. For any , let:

_{Y}- ;
- ;
- .

It’s not hard to prove that all the above are metrics, but which one to choose? It turns out all these are topologically equivalent anyway. E.g. to show equivalence of *d* and *d*_{1}, note that for any ,

and

Since this is not the main point, we’ll leave the details of the proof and the remaining cases to the reader. Instead, the point we wish to make is the following:

Theorem. The above metrics all define the topology T with basis given by

**Proof**.

Since we can pick any metric and use the corresponding open balls as a basis, let’s pick:

Let *B’* denote this collection of open balls. Clearly, *B’* is contained in *B*. Conversely, each element of *B* is obviously a union of elements of *B’*. Thus, they generate the same topology. ♦

So let’s generalise the definition of product space to arbitrary topological spaces.

Definition. If X and Y are topological spaces, then the corresponding topology on X × Y is defined by the basis

Note that in any topological space, our *B* is truly a basis since (i) the union of all elements of *B* is *X* × *Y* (in fact, *X* × *Y* lies in *B*), and (ii) the intersection of (*U* × *V*) and (*U’* × *V’*) is (*U* ∩ *U’*) × (V ∩ V’) which is in *B*. So the definition is valid.

**Note**. If are closed subsets, then *C* × *D* is closed in *X* × *Y*, because its complement is , which is a union of two open subsets.

## Relationship with Bases and Subbases

If we’re given bases or subbases of *X* and *Y*, then these can be used to define a corresponding basis or subbasis of *X* × *Y*.

Theorem. Let X and Y be topological spaces.

- If B
_{X}and B_{Y}are given bases of X and Y respectively, then is a basis of X × Y.- If S
_{X}and S_{Y}are given subbases of X and Y respectively, then is a subbasis of X × Y.

**Proof.**

For the first statement, we already saw that is a basis of *X* × *Y*. But since *B _{X}* and

*B*are bases of

_{Y}*X*and

*Y*, we can write , for some . This gives so every element of

*B’*is expressible as a union of elements of

*B*.

For the second, suppose finite intersections of *S _{X}* (resp.

*S*) give the basis

_{Y}*B*(resp.

_{X}*B*). Hence, if , we can write and for some , . This gives Thus every element of the collection is a finite intersection of elements of

_{Y}*S*; by the first statement,

*S*is a subbasis for

*X*×

*Y*. ♦

## Consistency of Successive Products

Next, let’s consider some basic properties on commutativity and associativity of products. Recall that we mentioned two topological spaces *X* and *Y* are homeomorphic if there’s a bijective map *f* : *X* → *Y*, such that a subset *U* of *X* is open if and only if *f*(*U*) is open in *Y*.

Proposition. Let X, Y, Z be topological spaces. Then the following spaces are homeomorphic.

- , via ;
- , via .

**Proof**.

The first property is obvious since the map takes *U* × *V* to *V* × *U*, for open subsets , . For the second property, since *B* = {*U* × *V* : *U* open in *X*, *V* open in *Y*} is a basis for *X* × *Y*, the above theorem tells us that:

{(*U* × *V*) × *W* : *U* open in *X*, *V* open in *Y*, *W* open in *Z*}

is a basis for (*X* × *Y*) × *Z*. Likewise, the collection of *U* × (*V* × *W*) is a basis for *X* × (*Y* × *Z*). Since the above map takes (*U* × *V*) × *W* to *U* × (*V* × *W*), this gives a homeomorphism. ♦

From the proof, it follows that for the topology on *X* × *Y* × *Z*, one can take a basis comprising of *U* × *V* × *W*, for open subsets Also, given a finite number of topological spaces , one can unreservedly take their product since product of topological spaces is commutative and associative.

## Consistency with Subspace

Finally, we have the following.

Proposition. Let X, Y be topological spaces and , be subsets. One can define a topology on X’ × Y’ in two distinct ways.

- Take subspace topologies on X’ and Y’, then form the product topology X’ × Y’.
- Take the product topology on X × Y, then the subspace topology on X’ × Y’.
These two are identical.

**Proof**.

For the first case, the subspace topology on *X’* is given by {*U* ∩ *X’* : *U* open in *X*} and that on *Y’* is given by {*V* ∩ *Y’* : *V* open in *Y*}. Hence, the product topology on *X’* × *Y’* is given by the basis {(*U* ∩ *X’*) × (*V* ∩ *Y’*)}, for open subsets *U* of *X* and *V* of *Y*.

For the second case, the product topology on *X* × *Y* has basis *U* × *V* for open subsets *U* of *X* and *V* of *Y*. From our previous article, a basis on *X’* × *Y’* can thus be given by the collection of:

(*U* × *V*) ∩ (*X’* × *Y’*) = (*U* ∩ *X’*) × (*V* ∩ *Y’*)

for open subsets *U* of *X* and *V* of *Y*. ♦

**Examples**

- The product of two (or finitely many) discrete topological spaces is still discrete. We’ll see later that this is not true for an infinite product of discrete spaces.
- The product of
**R**^{n}and**R**^{m}, with topology given by the usual Euclidean metric, is**R**^{n+m}with the same topology. In particular, each**R**^{n}has the product topology of*n*copies of**R**. - If
*X*and*Y*are metrisable, then the initial paragraphs of this article show that*X*×*Y*is metrisable. - Suppose
*X*is any topological space and*Y*= {1, 2} with the discrete topology. Then the picture of*X*×*Y*is that of two identical copies of*X*. We’ll expound on disjoint unions in the next article. - Let
*X*=*S*^{1}, the set of points (*x*,*y*) in**R**^{2}satisfying*x*^{2}+*y*^{2}= 1. Then the space*X*×*X*is called a**torus**.