Product Topology

Suppose that

X

and

Y

are topological spaces, then the collection

C

of all sets of the form

U \times V

with

U

open in

X

and

V

open in

Y

forms a basis

Let $(x, y) \in X \times Y$ , then the basis element $B = X \times Y$ works since $X, Y$ are open in themselves.

Suppose that $x \in B_{1} \cap B_{2}$ where $B_{1} = U_{1} \times V_{1}$ and $B_{2} = U_{2} \times V_{2}$ , then $B_{1} \cap B_{2} = (U_{1} \times V_{1}) \cap (U_{2} \times V_{2})$ but since intersection and cartesian product commute, then $B_{1} \cap B_{2} = (U_{1} \cap U_{2}) \times (V_{1} \cap V_{2})$ , but since $U_{1}, U_{2}, V_{1}, V_{2}$ are open in $X$ and $Y$ respectively, then so is their finite intersection so therefore $B_{1} \cap B_{2}$ is already a basis element so we can take $B_{3} = B_{1} \cap B_{2}$ in the definition of a basis.

product topology

Suppose that

X

and

Y

are topological spaces, then the product topology on

X \times Y

is the toplogy having the basis

B

of sets of the form

U \times V

where

U

is open in

X

and

V

is open in

Y

Suppose that

B

is a basis for the topology of

X

and

C

for

Y

. Then the collection

D = {B \times C : B \in B and C \in C}

then

D

is a basis for the topology of

X \times Y

To show it's a basis we will use the basis criterion. So let $W$ be an open set of $X \times Y$ , and let $(x, y) \in W$ , since the product topology is generated by the basis ${U \times V : X \in T_{X} and Y \in T_{Y}}$ , then we know that by the definition of a topology generated by a basis that there is an element $U \times V$ such that $(x, y) \in U \times V \subseteq W$ .

Since we assumed that $B$ and $C$ were bases for $X$ and $Y$ respectively then, we know that there is some element $B \in B$ such that $x \in B \subseteq U$ and there is some $C \in C$ such that $y \in C \subseteq V$ , thus we have found an element $B \times C \in D$ such that $(x, y) \in B \times C \subseteq U \times V = W$ which proves that $D$ is a basis and that it generates the toplogy of $X \times Y$

projections

We define

π_{1} : X \times Y \to X

to satisfy

π_{1} (x, y) = x

, and

π_{2} : X \times Y \to Y

with

π_{2} (x, y) = y

and say that

π_{1}

π_{2}

are projections of

X \times Y

into it's first and second factors.

inverse of a projection

Suppose that

U \subseteq X

, then

π_{1}^{- 1} (U) = U \times Y

, similarly if

V \subseteq Y

, then

π_{2}^{- 1} (V) = X \times V

We'll show that $π_{1}^{- 1} (U) = U \times Y$ first, so let $p \in π_{1}^{- 1} (U)$ $=$ ${(x, y) \in X \times Y : π_{1} (x, y) \in U}$ , but $π_{1} (x, y) = x$ , so this set is simply ${(x, y) \in X \times Y : x \in U} = U \times Y$

A symmetrical proof for $π_{2}$ can be obtained similarly.

subbasis for the product topology

The collection

S = {π_{1}^{- 1} (U) : U open in X} \cup {π_{2}^{- 1} (V) : V open in Y}

is a subbasis for the product topology on

X \times Y

Given this subbasis $S$ , and supposing that $T$ is the product topology on $X \times Y$ , our goal is to show that $T_{B_{S}} = T$ .

We'll first show that $T_{B_{S}} \subseteq T$ , an arbitrary element of $T_{B_{S}}$ is $M = ⋃_{E \in B} E$ where $B \subseteq B_{S}$ , since $E$ is a finite intersection of elements of $S$ (which we know are open in $X \times Y$ ), then $M \in T$ , showing that $T_{B_{S}} \subseteq T$

For the other inclusion, let $U \times V$ be a basis element for $T$ , but note that $U \times V = U \times Y \cap X \times V = π_{1}^{- 1} (U) \cap π_{2}^{- 1} (V)$ , thus by definition an element of $B_{S}$ , which shows that $T \subseteq T_{B_{S}}$