# Product Topology

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 $$\mathcal{T} \subseteq \mathcal{T}_{\mathcal{B}_{S}}$$