## The dreadful product topology (Pt. II)

In the previous post we gave the definitions of the product topology and some nice little tools about it. Here, we give some theorems regarding what properties of topological spaces are invariant under the formation of the product topology. Let us start out nice and easy

**Theorem:** Let be a finite number of topological spaces with corresponding open bases . Then, is an open base for under the product topology.

**Proof: **Let be open in , and let . There exists some (the defining open base) such that . But, by theorem 1 in the last post we know that is an open set containing for and since is an open base for we know there exists some such that . Clearly then and since and were arbitrary the conclusion follows.

We know use this to prove a nice little theorem.

**Theorem:** Let be a finite number of second countable spaces, then under the product topology is second countable.

**Proof:** Let be as above. Since is an open base for is remains to prove that it is countable. Let be given by . It is clear that this is a bijection. Recalling that the finite product of countable sets is countable finishes the argument.

It turns out that the above is a little more restrictive than is necessary as the following theorem illustrates.

**Theorem:** Let be a sequence of second countable spaces, then is second countable under the product topology.

**Proof:** Let be the corresponding countable open bases and define where and .

It is relatively easy to prove that given by is a bijection. It follows then, since the finite product of countable sets is countable, that is countable.

So, now define . It is clear that since is the countable union of countable sets that it is countable. So now, it remains to prove that is in fact an open base for

So, let be open in and let . There exists some in the defining open base such that . Considering the form in which basic open sets in the product topology take we know that there is a finite set of indices such that . So, let . Clearly we have that is a neighborhood of for and since is an open base there exists some such that . So, let . Then, it is evident that . Noting that finishes the argument. .

Note the necessity for finiteness again.

**Theorem:** Let be a finite number of toplogical spaces and let be a number of corresponding dense sets. Then, is dense in under the product topology.

**Proof:** Let be arbitrary and let be any neighborhood of . We then have by theorem 1 in the last post that is a neighborhood of for and since is dense in there exists some such that . Clearly then , and clearly . Noting the arbitrariness of finishes the argument.

From this we can derive the obvious consequence.

**Theorem:** Let be finite number of separable spaces and let under the product topology, then is separable.

**Proof:** Since each is separable it has a countable dense subset and by the above theorem we see that is dense in , but since the finite product of countable sets is countable the conclusion follows.

As in the case of second countable spaces we may extend this to the countable case

**Theorem:** Let be sequence of separable spaces, then is separable.

**Proof:** Since each is separable it has a countable dense subset we may extract an arbitrary but fixed sequence from them as where . We now define where

Clearly we have that by is a bijection and since the finite product of countable sets is countable it follows that is countable. So, let being the countable union of countable sets is countable.

From the previous paragraph it is clear that we must merely show that is dense in . To do this let be arbitrary and any neighborhood of . There exists some (the defining open base) such that . But, we know that for finitely man indices, say . Let . We know then that is a neighborhood of for every and since is dense in there exists some such that . Clearly then, we have that where is in and thus . But, and the conclusion follows.

We now discuss the product of Hausdorff spaces.

**Theorem:** be an arbitrary collection of Hausdorff spaces, then under the product topology is Hausdorff.

**Proof: **Let such that then for some and since is Hausdorff we know that there exists open sets such latex and . So define where and define where clearly we have that and and since we must have . The conclusion follows.

We can extend the last theorem even to c (continuum) many factors, but not to more. For a proof, see the Ask a Topologist forum of this year..

Comment by Henno | March 27, 2010 |

Thank you very much Henno. I was aware of this, but I only post proofs that I am capable of doing by myself. Thus, while I have never tried the proof for the product of copies I’m not sure I could manage.

That aside, I have seen you on Ask a Topologist. Great work!

Comment by drexel28 | March 28, 2010 |

It’s not that hard, really. But there is one “trick” to it that most wouldn’t think of by themselves.

Nice idea, this series! I’ll link to it on the forum!

Comment by Henno | March 29, 2010 |

Really? That would be great! I would love to hear feed-back/suggestions from more people!

Comment by drexel28 | March 29, 2010 |