# Cross product of two sets with vectors as elements.

If $$X \in \mathcal{X} = \{x_1,x_2\}$$, for some $$x_1,x_2 \in \mathbb{R}^{n \times 1}$$ and $$Y \in \mathcal{Y} = \{y_1,y_2\}$$, for some $$y_1,y_2 \in \mathbb{R}^{m \times 1}$$, can I say that the stacked vector $$Z = (X^T,Y^T)^T$$ belongs to $$\mathcal{X} \times \mathcal{Y}$$ ?

I think not, because $$Z = (X^T,Y^T)^T \in \mathcal{Z} = \left \{\ \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}, \begin{pmatrix} x_1 \\ y_2 \end{pmatrix}, \begin{pmatrix} x_2 \\ y_1 \end{pmatrix}, \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} \right\}$$, and

$$\mathcal{X} \times \mathcal{Y} = \{ (x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2) \}.$$

Then, is there a way to write $$\mathcal{Z}$$ in terms if $$\mathcal{X}$$ and $$\mathcal{Y}$$ ?

Firstly, it seems that you've already defined the transpose $$(\cdot)^T : \bigcup_{n=1}^\infty (\mathbb{R}^{n}\cup\mathbb{R}^{n \times 1}) \rightarrow \bigcup_{n=1}^\infty (\mathbb{R}^{n}\cup \mathbb{R}^{n \times 1})$$

for all row and column vectors (I'm just using plain $$\mathbb{R}^n$$ for the "rows" rather than $$\mathbb{R}^{1 \times n}$$).

So the image of subsets $$A \subseteq \bigcup_{n=1}^\infty (\mathbb{R}^{n}\cup \mathbb{R}^{n \times 1})$$ under $$(\cdot)^T$$ are defined already as

$$(\cdot)^T(A) = \{x^T \mid x \in A\}$$

so for readability you might as well define

$$A^T = (\cdot)^T(A)$$

This means that you can define $$\mathcal{Z}$$ as

$$\mathcal{Z} = (\mathcal{X}^T \times \mathcal{Y}^T)^T$$