The product topology

Given topological spaces X and Y we want to get an appropriate topology on the Cartesian product X Y.

Obvious method
Call a subset of X Y open if it is of the form A B with A open in X and B open in Y.

Difficulty
Taking X = Y = R would give the "open rectangles" in R² as the open sets. These subsets are open, but unfortunately there are lots of other sets which are open too.
We are therefore forced to work a bit differently.

Definition
A set of subsets is a basis of a topology if every open set in is a union of sets of .

Example
In any metric space the set of all -neighbourhoods (for all different values of ) is a basis for the topology.

Remark
This is a very helpful concept. For example, to check that a function is continuous you need only verify that f^-1(B) is open for all sets B in a basis -- usually much smaller than the whole collection of open sets.

We can now define the topology on the product.

Definition
If X and Y are topological spaces, the product topology on X Y is the topology whose basis is {A B | A _X , B _Y}.

Examples

1. The topology on R² as a product of the usual topologies on the copies of R is the usual topology (obtained from, say, the metric d₂).

   Proof
   The sets of the basis are open rectangles, and an -neighbouhood U in the metric d₂ is a disc. It is easy to see that every point of U can be contained in a small open rectangle lying inside the disc. Hence U is a union of (infinitely many!) of these rectangles and hence is in the product topology.
   Since every open set in the d₂ metric is a union of -neighbourhoods, every open set can be written as a union of the open rectangles.
   
   
   

2. A torus is the surface in R³:
   It can also be regarded as the product S¹ S¹ where S¹ is a circle (the curve, not the interior) in R². In this way it can be thought of as a subset of R⁴.
   The topology on S¹ is the subspace topology as a subset of R² and so we get the product topology on S¹ S¹.
   Fortunately this is the same as the topology on the torus thought of as a subset of R³.

   Proof
   A basis for the subspace topology on S1 is the set of "arcs"
   Hence a basis for the product topology on S¹ S¹ is sets of the form:
   A basis for the subspace topology on the torus as a subset of R³ is the intersection of the torus with -neighbourhoods of R³ (which are "small balls") and hence are sets of the form:
   As before, one can get these "ovals" as unions of the small "bent rectangles".
   
   

3. Take the topology = {, {a, b}, {a} } on X = {a, b}.
   Then the product topology on X X is {, X X, {(a, a)}, {(a, a), (a, b)}, {(a, a), (b, a)}, {(a, a), (a, b), (b, a)} } where the last open set in the list is not in the basis.
   

Remark
Given any product of sets X Y, there are projection maps p_X and p_Y from X Y to X and to Y given by (x, y) x and (x, y) y.
The product topology on X Y is the weakest topology (fewest open sets) for which both these maps are continuous.