# Having trouble understanding this example in my textbook about Cartesian products of sigma algebras

I am reading a chapter on probability in my textbook where they give a counterexample to show that Cartesian products of sigma algebras are not necessarily sigma algebras:

It is not true that $$\mathcal F_1 \times \mathcal F_2 = \sigma( \mathcal F_1 \times \mathcal F_2)$$. Take, for example, $$\mathcal F_1 = \mathcal F_2 = 2^{\{1,2\}}$$. Then, $$|\mathcal F_1 \times \mathcal F_2| = 1+3\times 3 =10$$ (because $$\emptyset \times X = \emptyset$$), while, since $$\mathcal F_1 \times \mathcal F_2$$ includes the singletons of $$2^{\{1,2\} \times \{1,2\}}$$, $$\sigma( \mathcal F_1 \times \mathcal F_2) = 2^{\{1,2\} \times \{1,2\}}$$. Hence, six sets are missing from $$\mathcal F_1 \times \mathcal F_2$$. For example, $$\{(1,1), (2,2)\} \in \sigma( \mathcal F_1 \times \mathcal F_2) \setminus \mathcal F_1 \times \mathcal F_2$$.

From what I understand, the set $$\mathcal{F}_1 \times \mathcal{F}_2 = \{\varnothing, (\{1\}, \{1\}), (\{1\}, \{2\}), (\{1\}, \{1,2\}), (\{2\}, \{1\}), (\{2\}, \{2\}), (\{2\}, \{1,2\}), (\{1, 2\}, \{1\}), (\{1, 2\}, \{2\}), (\{1, 2\}, \{1,2\})\}$$.
The ‘missing’ elements in the above set would be those with $$\varnothing$$ but in the text they say these are set to $$\varnothing$$ so I don’t understand what elements are missing in order to make this a sigma algebra.
The Cartesian Product of sigma-algebras isn't literally the cartesian product (the set of all pairs). Instead, $$\mathcal F_1 \times \mathcal F_2 := \{A\times B \mid A\in \mathcal F_1, B\in \mathcal F_2\},$$ so in the example, for the space $$X=\{1,2\}$$ and sigma-algebras $$\mathcal F_1 = \mathcal F_2 = 2^{\{1,2\}}$$, we get $$\mathcal F_1 \times \mathcal F_2 = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}\times \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$$ $$=\left\{ \begin{array}{c} \emptyset \times \emptyset, \emptyset \times \{1\}, \emptyset \times \{2\}, \emptyset \times \{1,2\},\\ \{1\} \times \emptyset, \{1\}\times \{1\}, \{1\}\times \{2\}, \{1\}\times \{1,2\},\\ \{2\} \times \emptyset, \{2\}\times \{1\}, \{2\}\times \{2\}, \{2\}\times \{1,2\},\\ \{1,2\} \times \emptyset, \{1,2\}\times \{1\}, \{1,2\}\times \{2\}, \{1,2\}\times \{1,2\}\\ \end{array} \right\}$$ $$=\Big\{\emptyset, \{(1,1)\}, \{(1,2)\}, \{(1,1),(1,2)\}, \{(2,1)\}, \{(2,2)\}, \{(2,1),(2,2)\}, \{(1,1),(2,1)\}, \{(1,2),(2,2)\}, \{(1,1),(1,2),(2,1),(2,2)\}\Big\}.$$ Then, we can see that this set is missing the subset $$\{(1,1),(2,2)\}$$, which should be in there for it to be a sigma-algebra, since sigma-algebras are closed under countable unions, and this contains $$\{(1,1)\}, \{(2,2)\} \in\mathcal F_1 \times \mathcal F_2.$$
Pictorially, $$\mathcal F_1 \times \mathcal F_2$$ is a collection of subsets of $$\{1,2\}\times \{1,2\}$$, a $$2\times 2$$ grid, and this missing element is a diagonal on the grid.