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).

 A: $\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 $B$.
A: 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

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

?
A: 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.
A: $(a,b)$ 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 $\mathcal G \times \mathcal G$.
