Product Topology - Topology, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

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

Obvious method

Call a subset of X x Y open if it is of the form A x 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 B is a basis of a topology J if every open set in J is a union of sets of B.

Example

In any metric space the set V of all ε-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 x Y is the topology whose basis is {A x B | A ∈ J_X , B ∈ J_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 ε-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 x 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 x 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 x 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 ε-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 J = {φ, {a, b}, {a} } on X = {a, b}.
Then the product topology on X x X is {φ, 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 x Y, there are projection maps p_X and p_Y from X x Y to X and to Y given by (x, y) → x and (x, y) → y.
The product topology on X x Y is the weakest topology (fewest open sets) for which both these maps are continuous.