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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 **R**^{2} 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**

- The topology on
**R**^{2}as a product of the usual topologies on the copies of**R**is the usual topology (obtained from, say, the metric*d*_{2}).**Proof**

The sets of the basis are open rectangles, and an -neighbouhood*U*in the metric*d*_{2}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*d*_{2}metric is a union of -neighbourhoods, every open set can be written as a union of the open rectangles.

- A torus is the surface in
**R**^{3}:

It can also be regarded as the product*S*^{1}*S*^{1}where*S*^{1}is a circle (the curve, not the interior) in**R**^{2}. In this way it can be thought of as a subset of**R**^{4}.

The topology on*S*^{1}is the subspace topology as a subset of**R**^{2}and so we get the product topology on*S*^{1}*S*^{1}.

Fortunately this is the same as the topology on the torus thought of as a subset of**R**^{3}.**Proof**

A basis for the subspace topology on S1 is the set of "arcs"

Hence a basis for the product topology on*S*^{1}*S*^{1}is sets of the form:

A basis for the subspace topology on the torus as a subset of**R**^{3}is the intersection of the torus with -neighbourhoods of**R**^{3}(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".

- 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.

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