# Cartesian product of two collections of sets?

If we have $$\mathcal{G} = \{\phi, \{0\}, \{1\}, \{1,2\} \},$$ would $$\mathcal{G} \times \mathcal{G}$$ have the form $$\{\phi, (\{0\}, \{0\}), (\{0\}, \{1\}), (\{0\}, \{0, 1\}), \dots \}$$?

In that case, what would an example of $$(w_1, w_2) \in B \in \mathcal{G} \times \mathcal{G}$$ be?

My working: Take for example $$B = (\{0\}, \{0, 1\})$$. This is not a set; so, I don't understand how one could take an element $$(w_1, w_2) \in B$$.

P.S.: I am taking a Probability Theory class now and even though I did a Measure Theory class last term, I am finding things very difficult. But I really want to get good at this stuff. Would there be any reading material you'd recommend? (I find the lectures hard to follow.)

EDIT: I encountered this statement in the book Theory of Probability and Random Processes by Koralov. In Definition 2.1, he states:

A finite-dimenstional cylinder is a set of the form $$A = \{\omega: (\omega_{t_1}, \dots, \omega_{t_k}) \in B \},$$ where $$t_1, \dots, t_k \geq 1$$, and $$B \in \mathcal{G} \times \dots \times \mathcal{G}$$ (k times).

• "Take for example B=({0},{0,1}). This is not a set; so, I don't understand how one could take an element (w1,w2)∈B." You are correct. "$(w_1, w_2) \in B \in \mathcal{G} \times \mathcal{G}$" doesn't actually make any sense. Pretty much for the reason you gave. If $B\in \mathcal G \times \mathcal G$ then $B$ is an ordered par of sets and $B$ itself would not have an ordered pair as an element... (Although an ordered pair is a sort of set.. a set where order matters and a set with exactly two elements. It's elements would be.. the first item in the pair and.. the second item in the pair.) – fleablood Jan 28 at 7:33

The context of the definition that you give in your edit is quite important. The 'Cartesian Product' can then be interpreted in a different way.

First let's denote $$\mathcal{F} := \mathcal{G} \times \dots \times \mathcal{G}$$. If you think of $$\mathcal{F}$$ as being a $$\sigma$$-Algebra on the power set of the Cartesian Product instead of the Cartesian Product itself, then the expression

$$(\omega_{t_1}, \dots, \omega_{t_k}) \in B \in \mathcal{F} \ \big(=\mathcal{G} \times \dots \times \mathcal{G}\big)$$ makes much more sense.

Because then $$\mathcal{F}$$ is a subset of the power set $$\mathcal{P}(\mathcal{G} \times \dots \times \mathcal{G})$$ and hence every member $$B$$ of it is a subset of the Cartesian Product $$\mathcal{G} \times \dots \times \mathcal{G}$$.

In fact, in the context of your definition in the book this "$$\mathcal{G} \times \dots \times \mathcal{G}$$" is - by abuse of notation - a shorthand for the Product $$\sigma$$-algebra of $$k$$-times the $$\sigma$$-algebra $$\mathcal{G}$$.

Sometimes this is denoted $$\mathcal{G} \otimes \dots \otimes \mathcal{G}$$ to prevent confusion with the Cartesian Product.

(And the book then goes on to make a connection between the $$k$$-times Product $$\sigma$$-algebra and the $$n$$-times Product $$\sigma$$-algebra for $$k \leq n$$. As seen here on page 25.)

About your P.S.: I usually recommend Klenke's Probability Theory which starts with some detailed measure theory then covers the basic theory, stochastic processes, martingales up to stochastic differential calculus. (Plus I got the notation of the Product $$\sigma$$-algebra from there.)

But you should have a look at many books to find one that might suit you and your class best.


$$A = \varnothing \quad B = \set{0} \quad C = \set{1} \quad D = \set{1,2}$$

Then $$\G = \set{A,B,C,D}$$ and $$\G^2$$ is given by

\begin{align*} \G \times \G = \Big\{ &(A,A),(A,B),(A,C),(A,D),\\ &(B,A),(B,B),(B,C),(B,D),\\ &(C,A),(C,B),(C,C),(C,D),\\ &(D,A),(D,B),(D,C),(D,D)\Big\} \end{align*}

You can then replace $$A,B,C,D$$ with their explicit definitions:

\begin{align*} \G \times \G = \Big\{ &\Big(\varnothing,\varnothing\Big) \; , \;\Big(\varnothing,\set{0}\Big) \; , \;\Big(\varnothing,\set{1}\Big) \; , \;\Big(\varnothing,\set{1,2}\Big),\\ &\Big(\set{0},\varnothing\Big) \; , \;\Big(\set{0},\set{0}\Big) \; , \;\Big(\set{0},\set{1}\Big) \; , \;\Big(\set{0},\set{1,2}\Big),\\ &\Big(\set{1},\varnothing\Big) \; , \;\Big(\set{1},\set{0}\Big) \; , \;\Big(\set{1},\set{1}\Big) \; , \;\Big(\set{1},\set{1,2}\Big),\\ &\Big(\set{1,2},\varnothing\Big) \; , \;\Big(\set{1,2},\set{0}\Big) \; , \;\Big(\set{1,2},\set{1}\Big) \; , \;\Big(\set{1,2},\set{1,2}\Big)\Big\} \end{align*}

More loosely,

$$\G \times \G = \Big\{ \text{all possible pairs } (X,Y) \text{ where } X,Y \in \G \Big\}$$

What you end up with, then, is ordered pairs of sets. Be careful with the ordered pair $$(\varnothing,\varnothing)$$; you seem to think that might be equal to $$\varnothing$$, but recall that ordered pairs have a formal definition:

$$(x,y) = \Big\{ \set{x},\set{x,y} \Big\}$$

You should see why, then, $$(\varnothing,\varnothing) \ne \varnothing$$.

Of course, $$\G^2$$ is still a set, so you can take elements from it, and those elements are ordered pairs. If we say $$(w_1,w_2) \in \G \times \G$$, then we are just saying $$(w_1,w_2)$$ is some ordered pair in $$\G \times \G$$ (and, moreover, that means $$w_1 \in \G$$ and $$w_2 \in \G$$ from definitions).

But can we take elements of a $$B \in \G \times \G$$ instead? Like you saw: can we take $$(w_1,w_2) \in B$$ for $$B \in \G \times \G$$? (To avoid confusion, this need not be the same $$B$$ as earlier.)

Note that $$B \in \G \times \G$$ may be characterized as set in terms of ordered pairs. It might be best to work with an example, say $$B = (\set{1,2},\set{1,2})$$. Then

$$B = \Big( \set{1,2},\set{1,2} \Big) = \Big\{ \set{1,2} \; , \; \big\{ \set{1,2},\set{1,2} \big\} \Big\}$$

but

$$(w_1,w_2) = \Big\{ w_1, \set{w_1,w_2} \Big\}$$

If $$(w_1,w_2) \in B$$, then it means that $$(w_1,w_2)$$ is represented by the set, larger set in $$B$$.

Almost.

$$\emptyset$$ wouldn't be in the cartesian product but $$(\emptyset,\emptyset)$$ will be.

Don't overthink it. $$A\times B$$ is nothing more than the set of ordered pairs and an ordered pair is just a pair of an element from the first set and an element for the second.... in order.

So if $$\color{blue}{\mathcal{G} = \{\phi, \{0\}, \{1\}, \{1,2\} \}}$$ is a set with four elements and $$\color{red}{\mathcal{G} = \{\phi, \{0\}, \{1\}, \{1,2\} \}}$$ is the same set of four elementts then $$\color{blue}{\mathcal{G}}\times \color{red}{\mathcal{G}}$$ will be a set of the sixteen pairs that can be made:

$$\color{blue}{\mathcal{G}}\times \color{red}{\mathcal{G}}= \{(\color{blue}{\emptyset},\color{red}{\emptyset}),(\color{blue}{\emptyset},\color{red}{\{0\}}),(\color{blue}{\emptyset},\color{red}{\{1\}}),(\color{blue}{\emptyset},\color{red}{\{1,2\}}),$$

$$\{(\color{blue}{\{0\}},\color{red}{\emptyset}),(\color{blue}{\{0\}},\color{red}{\{0\}}),(\color{blue}{\{0\}},\color{red}{\{1\}}),(\color{blue}{\{0\}},\color{red}{\{1,2\}}),$$

$$\{(\color{blue}{\{1\}},\color{red}{\emptyset}),(\color{blue}{\{1\}},\color{red}{\{0\}}),(\color{blue}{\{1\}},\color{red}{\{1\}}),(\color{blue}{\{1\}},\color{red}{\{1,2\}}),$$

$$\{(\color{blue}{\{1,2\}},\color{red}{\emptyset}),(\color{blue}{\{1,2\}},\color{red}{\{0\}}),(\color{blue}{\{1,2\}},\color{red}{\{1\}}),(\color{blue}{\{1,2\}},\color{red}{\{1,2\}})\}$$

Don't overthink it.

......

In that case, what would an example of (w1,w2)∈B∈G×G be?

Oh.....

Now I see why you were overthinking it. That statement makes no sense.

$$B\in \mathcal G \times \mathcal G$$ would mean that $$B$$ is an ordered pair $$(J,K)$$ where $$J$$ is one of the four sets and $$K$$ is one of the four sets: $$\emptyset, \{0\}, \{1\}, \{1,2\}$$.

Then, if I were to say $$w \in B = (J,K)$$ then that means $$w$$ is one of either $$J$$ or $$K$$.

So that makes no sense. Are you sure the statement wasn't:

$$(w_1, w_2) = B \in \mathcal G \times \mathcal G$$

or maybe

$$(w_1, w_2) \in B \subset \mathcal G \times \mathcal G$$

?

• Thanks for confirming correctness of the cartesian product. I will add a bit more detail in the question; please check that. – spideyonthego Jan 28 at 7:35

$$(a,b)$$ is a way to denote an element of the Cartesian product $$A \times B$$ where $$a \in A$$ and $$b \in B$$.

There are several ways is set theory to define the ordered pair. A usual one is

$$(a,b)= \{\{a\}, \{a,b \}\}.$$

Based on that you can derive $$\mathcal G \times \mathcal G$$.

• OP's Cartesian product is not correct, since he has $\varnothing \in \mathcal{G}^2$ which is not an ordered pair. – Eevee Trainer Jan 28 at 7:12
• @EeveeTrainer You’re right. I read it too quickly. – mathcounterexamples.net Jan 28 at 7:32