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).

  • $\begingroup$ "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.) $\endgroup$ – 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.


$\newcommand{\set}[1]{\left\{ #1 \right\}} \newcommand{\G}{\mathcal{G}} $It might be easier to process if you name every set with a letter, say...

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

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

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


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

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



$\emptyset$ 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:

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




Don't overthink it.


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


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

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


  • $\begingroup$ Thanks for confirming correctness of the cartesian product. I will add a bit more detail in the question; please check that. $\endgroup$ – 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 $\mathcal G \times \mathcal G$.

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

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