Cartesian product of two collections of sets?

$\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 $B$.

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

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

?