Product Topology

Cartesian Product of open sets is a basis for the Cartesian Product

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

Product Topology by Universal Properties

Given

(X, 𝒯_{X}), (Y, 𝒯_{Y})

, the product topology is the unique topology satisfying:

$π_{X} : X \times Y \to X$ and $π_{Y} : X \times Y \to Y$ are continuous
Suppose that $f : Z \to X$ and $g : Z \to Y$ are continuous then $f \times g : Z \to X \times Y$ defined as $(f \times g) (z) = (f (z), g (z))$ is continuous

Let $U, V \in 𝒯_{X}, 𝒯_{Y}$ respectively, and then note that $π_{X}^{- 1} (U) = U \times Y$ and that $π_{Y}^{- 1} (V) = X \times V$ since those are both basis elements, then they are open sets.

Now suppose that $f, g$ were continuous, so now we'll show that $f \times g$ is continuous, so we take a basis element $U \times V$ now we have $\begin{matrix} {(f \times g)}^{- 1} (U \times V) & = {z : (f (z), g (z)) \in U \times V} \\ = {z : f (z) \in U \land g (z) \in V} \\ = {z : f (z) \in U} \cap {z : g (z) \in V} \\ = f^{- 1} (U) \cap g^{- 1} (V) \end{matrix}$ thus, as an intersectino of two open sets, it is also open as needed.

We now show uniqueness, which we do by supposing that $𝒯_{1}$ and $𝒯_{2}$ are both topologies on $X \times Y$ which satisfy the properties.

Note that in the above if $h$ was continuous, then so are the compositions, so it's really if and only only if in the second bullet point.

The Basis for the product Topology of two Topologies generated by Bases are all the Cartesian Products of all the Basis Elements

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 $X \times Y$

The Dictionary Order Topology on Rxr Is the Same as the Product Topology of Rdxr

The dictionary order topology on

ℝ \times ℝ

is the same as the product topology on

ℝ_{d} \times ℝ

where

ℝ_{d}

is the discrete topology.

In order to prove that the two topologies are the same we will use this, so first we have to identify the basis for each topology, we recall the basis for $ℝ \times ℝ$ . Then looking at $ℝ_{d} \times ℝ$ we recall that, thus since we know that ${{x} : x \in ℝ}$ is a basis for $ℝ_{d}$ and that ${(a, b) : a < b \in ℝ}$ is a basis for $ℝ$ then a basis for their product is given by the set ${{x} \times (a, b) : x, a, b \in ℝ}$

Following the original corollary let $x \times y \in ℝ \times ℝ$ and let $(a \times b, c \times d)$ be a basis element of the dictionary order topology containing $x \times y$ , if $a = c$ then we must have $a = x = c$ and that $b < y < d$ in such a case the basis element ${x} \times (b, d)$ is contained within $(a \times b, c \times d)$ and contains $x \times y$ as needed. If it's the case that $a < c$ then it must be that $a < x < c$ with no restriction on $y$ as we are in the dictionary order, because for any $p \in ℝ$ we have that $a \times b < x \times p < c \times d$ therefore we can consider the basis element ${x} \times (y - 1, y + 1)$ , thus in either case we've found a basis element containing $x \times y$ contained within the original basis element.

So now suppose that we had a basis element of the form ${p} \times (a, b)$ containing $x \times y$ , this implies that $p = x$ then note that ${x} \times (a, b) = (x \times a, x \times b)$ which is already a basis element of the correct form, so we are done and the two topologies are equal.

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 $𝒯 \subseteq 𝒯_{ℬ_{S}}$