A Note About the Base of a Product Topology A question came up in class about the base of a product topology. This document hopefully addresses it after first establishing a few things...

Consider two topological spaces (X, TX ) and (Y, TY ) and the product space

Z = X × Y.

A collection of sets BX ⊆ TX is a base (basis) for TX if every G ∈ TX can be written as a union of sets in BX . For example, [ G = Bα α∈I where I is some arbitrary, and possibly uncountable, index set, and {Bα}α∈I is a collection of sets in BX .

The product space Z can be endowed with the product topology which we will denote here by TZ . It is defined as the topology whose base is the collection of sets

{GX × GY : GX ∈ TX ,GY ∈ TY }.

Clearly there are other possible topologies for Z. For example, we could take the trivial topology {∅, Z}. However, the product topology will be an important one for us!

The question was:

Can we just take the base for TZ to be sets of the form

{BX × BY : BX ∈ BX ,BY ∈ BY }?

The answer is yes and we will prove it here shortly.

Forget product spaces for a few moments. Suppose that (X, TX ) is a topological space. It will be useful to use an alternate definition of a base for T . For clarity, we restate our first definition and label it as “Definition 1”.

Definition 1: A collection of sets B ⊆ T is a base for T if every G ∈ T can be written as a union of sets in B.

Definition 2: A collection of sets B ⊆ T is a base for T if for any G ∈ T and for any x ∈ G, there exists a B ∈ B such that x ∈ B ⊆ G. Proof that Def 1 ⇒ Def 2:

• Take any G ∈ T and any x ∈ G.

S • By Definition 1, we can write G = α∈I Bα for some sets Bα ∈ B and some index set I.

• So, [ x ∈ G = Bα ⇒ x ∈ Bα∗ α∈I ∗ for some Bα∗ ∈ B with α ∈ I. [ • But Bα∗ ⊆ Bα = G, so we have α∈I

x ∈ Bα∗ ⊆ G

for some Bα∗ ∈ B. ut

Proof that Def 2 ⇒ Def 1:

• Take any G ∈ T and any x ∈ G.

• By Definition 2, there exists a B ∈ B such that x ∈ B ⊆ G.

• Denote B by Bx and find such base sets for all x ∈ G. • Then clearly [ G ⊆ Bx x∈G

since every x ∈ G is contained in Bx. • On the other hand, [ Bx ⊆ G x∈G

since each Bx ⊆ G. [ [ • Thus, we have G = Bx, so we have written G in the form Bα. ut x∈G α

We are ready to prove the following Theorem. Theorem:

Let (X, TX ) and (Y, TY ) be two topological spaces with respective bases BX and BY . Let Z = X × Y. Then BX × BY := {BX × BY : BX ∈ BX ,BY ∈ BY }? is a base for the product topology TZ .

Proof:

Using Definition 2 of a topology base, We will show that, for any G ∈ TZ and for any z = (x, y) ∈ G, we can find BX ∈ BX and BY ∈ BY such that z ∈ BX × BY ⊆ G.

• Take any G ∈ TZ and any z = (x, y) ∈ G.

• By Definition 2 of the product topology, there exists a GX × GY with GX ∈ TX and GY ∈ TY such that (x, y) ∈ GX × GY ⊆ G.

• Since BX and BY are respective bases for TX and TY , there exist sets C ∈ BX and D ∈ BY such that x ∈ C ⊆ BX and y ∈ D ⊆ BY .

• Since we found C × D in BX × BY such that z = (x, y) ∈ C × D ⊆ G, we have, by Definition 2, that BX × BY is a base for the product topology TZ . ut