Cartesian product of two collections of sets?

Solution 1:

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 $G$ as the set of the $\sigma$-algebra $\mathcal{G}$, i.e. $\mathcal{G} = (G,\mathcal{G})$. Further set $\mathcal{F} := \mathcal{G} \times \dots \times \mathcal{G}$. If you think of $\mathcal{F}$ as being a $\sigma$-Algebra on the power set of the Cartesian Product of $G$ 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 $\mathcal{P}(G^{k})$ (where $G$ is the set of the $\sigma$-algebra $\mathcal{G}$) and hence every member $B$ of it is a subset of the Cartesian Product $G \times \dots \times G$.

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.

Solution 2:

$\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

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

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 $B$.