Suppose that \( X \) and \( Y \) are topological spaces, then the collection \( \mathcal{C} \) of all sets of the form \( U \times V \) with \( U \) open in \( X \) and \( V \) open in \( Y \) forms a basis

Let \( \left ( x , y \right ) \in X \times Y \), 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 \( B_{1} \cap B_{2} = \left ( U_{1} \times V_{1} \right ) \cap \left ( U_{2} \times V_{2} \right ) \) but since intersection and cartesian product commute, then \( B_{1} \cap B_{2} = \left ( U_{1} \cap U_{2} \right ) \times \left ( V_{1} \cap V_{2} \right ) \), 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 \( \mathcal{B} \) of sets of the form \( U \times V \) where \( U \) is open in \( X \) and \( V \) is open in \( Y \)

Suppose that \( \mathcal{B} \) is a basis for the topology of \( X \) and \( \mathcal{C} \) for \( Y \). Then the collection \( \mathcal{D} = \left \lbrace B \times C : B \in \mathcal{B} \text{ and } C \in \mathcal{C} \right \rbrace \) then \( \mathcal{D} \) 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 \( \left ( x , y \right ) \in W \), since the product topology is generated by the basis \( \left \lbrace U \times V : X \in \mathcal{T}_{X} \text{ and } Y \in \mathcal{T}_{Y} \right \rbrace \), then we know that by the definition of a topology generated by a basis that there is an element \( U \times V \) such that \( \left ( x , y \right ) \in U \times V \subseteq W \).

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

projections

We define \( \pi_{1} : X \times Y \to X \) to satisfy \( \pi_{1} \left ( x , y \right ) = x \), and \( \pi_{2} : X \times Y \to Y \) with \( \pi_{2} \left ( x , y \right ) = y \) and say that \( \pi_{1} \), \( \pi_{2} \) are projections of \( X \times Y \) into it's first and second factors.

inverse of a projection

Suppose that \( U \subseteq X \), then \( \pi_{1}^{- 1} \left ( U \right ) = U \times Y \), similarly if \( V \subseteq Y \), then \( \pi_{2}^{- 1} \left ( V \right ) = X \times V \)

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

A symmetrical proof for \( \pi_{2} \) can be obtained similarly.

subbasis for the product topology

The collection \( S = \left \lbrace \pi_{1}^{- 1} \left ( U \right ) : U \text{ open in } X \right \rbrace \cup \left \lbrace \pi_{2}^{- 1} \left ( V \right ) : V \text{ open in } Y \right \rbrace \) is a subbasis for the product topology on \( X \times Y \)

Given this subbasis \( S \), and supposing that \( \mathcal{T} \) is the product topology on \( X \times Y \), our goal is to show that \( \mathcal{T}_{\mathcal{B}_{S}} = \mathcal{T} \).

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

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