## The dreaded product topology

**NOTE: **All sets and spaces in the following discussion will be assumed to be, without loss of generality, non-empty. Also, if not stated it is assumed that under the product topology where is a non-empty collection of topological spaces.

To continue any further we mus make an, admittedly not brief, sojourn into the concept of product topologies. Now, this isn’t a text-book so I’ll assume that the reader is familiar with the basic concepts of Cartesian products and projections (denote ). With this in mind we first make a definition for the arbitrary Cartesian product of sets:

**Arbitrary Cartesian Product:** Let be an arbitrary class of non-empty sets, then we define .

Now, if is a non-empty class of topological spaces we define the product topology on as the topology generated by the subbase (where is the topology on ,) in the sense that it is the set of all finite intersections of elements of form an open base for , and thus the topology on is made up of all arbitrary unions of elements of this open base. We call in the above discussion the *defining open subbase* for and the *defining open base* for under the product topology.

So what do elements of look like? Well, let be an open set in then is (in words) all such that . It is clear that the only limitation then that needs to be made on is that it’s th coordinate is in . From this it is clear that where . Thus, is made up of all sets of the form where for all but one

**Example:** If under the usual topology then whereas

So, this being said we see that the elements of (being the finite intersection of elements of ) are going to be the product of open sets such that for all but finitely many .

So, let’s start proving some theorems!

**Theorem 1: **All projections are open mappings.

**Proof: **Let be open then where , and so and since the th coordinate of every basic open set (whether equal to the full space or not) is an open set in and so is the union of open sets in and thus open.

**Corollary 1:** If is a neighborhood of then is a neighborhood of

*Remark:* Projections need not be closed mappings. Consider the set as a subset of with the product topology. It is relatively easy is closed, but which is open.

One might ask “Why not just make the open base for a topology on the set of all arbitrary products of open sets?”. The topology that this is describing is called the *box topology.* Namely, the box topology has the set of all arbitrary products of open open sets as an open base. It is clear that the box and product topologies coincide for finite cases but do not for infinite cases. So, why did we choose the box topology over the product topology? The answer comes from the following theorem.

**Theorem:** Let be a mapping of a topological space into the product space . Then, is continuous if and only if is continuous for all .

**Proof:**

:

Let be open in then and since is subbasic open in and continuous we see that is open in .

:

**Lemma:** A mapping (where and are arbitrary topological spaces) is continuous if and only if is open for every where is an open base for .

**Proof:**

: Let be open in then where . So then, which by assumption is the union of open sets in and thus open. The conclusion follows.

: This is obvious.

So, suppose that is basic open in , then where and is open in . It follows then that

and since each is continuous it follows that is the finite intersection of open sets in and thus open. The conclusion follows.

Note that the key point here was that was the *finite* intersection of inverse images of open sets. If we had the box topology we could have had that was the infinite intersection of inverse images of open sets we would have no guarantee that would be open since it would be the *infinite* intersection of open sets…which we all know need not be open.

Now, one may ask “Topologically does it matter which order we product the sets?”. The answer is “no…for finite cases”

**Theorem: **Let be a finite collection of topological spaces then where is an permutation of

**Proof: **It is readily verified that the canonical mapping given by is a homeomorphism.

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