The natural numbers are defined as $\displaystyle \mathbb{N}=\left \{ 0,1,2,3,...\right \}$, the integers are defined as $\displaystyle \mathbb{Z}=\left \{ ...-3,-2,-1,0,1,2,3,...\right \}$, and the real numbers ($\displaystyle \mathbb{R}$) are defined as the set of all non-complex numbers. As such, $\displaystyle \mathbb{N}$ is a subset of $\displaystyle \mathbb{Z}$, and $\displaystyle \mathbb{Z}$ is a subset of $\displaystyle \mathbb{R}$.

This is a theorem for well ordered sets and the proof is as follows.

First identify what is given in the statement.

1. The sets are onto order isomorphisms

$\displaystyle f:A_1\rightarrow A_2$ and $\displaystyle g:B_1\rightarrow B_2$

2. The goal is to make an onto order isomorphism. Let us call it $\displaystyle t$.

$\displaystyle t:A_1\times B_1 \rightarrow A_2\times B_2$

Thus, $\displaystyle t$ can be defined as the function,

$\displaystyle \\t:A_1\times B_1\rightarrow A_2\times B_2 \\t(a,b)=(f(a),g(b))$

To show $\displaystyle t$ is a well defined, one-to-one,  function on $\displaystyle A_1\times B_1$ since $\displaystyle f$ and $\displaystyle g$ are one-to-one, the following is performed.

$\displaystyle t(a,b)=t(u,v)\Rightarrow(f(a),g(b))=(f(u),g(v))$

                                $\displaystyle \\\Rightarrow f(a)=f(u), g(b)=g(v) \\\Rightarrow a=u, b=v \\\Rightarrow (a,b)=(u,v)$

Thus, $\displaystyle t$ is one-to-one.

Now to prove $\displaystyle t$ in onto $\displaystyle A_2\times B_2$ if $\displaystyle (a,b)\ \epsilon\ A_2\times B_2$ 

$\displaystyle a=f(u), b=g(v)$

for some

$\displaystyle u\ \epsilon\ A_1, v\ \epsilon\ B_1$

thus

$\displaystyle t(u,v)=(a,b)$

proving $\displaystyle t$ is onto $\displaystyle A_2 \times B_2$.

Lastly prove ordering.

$\displaystyle (a_1,b_1)\leq_{A\times B}(a_2,b_2)\Leftrightarrow \left\{\begin{matrix} a_1< _aa_2\\ a_1=a_2, b+1\leq_{r}b_2 \end{matrix}\right.$

                                             $\displaystyle \Leftrightarrow\left\{\begin{matrix} f(a_1)< _{A}f(a_2)\\ f(a_1)=f(a_2), g(b_1\leq_{r}g(b_2)) \end{matrix}\right.$

                                             $\displaystyle \Leftrightarrow\left\{\begin{matrix} (f(a_1),g(b_1))< _{A\times B}(f(a_2),g(b_2))\\ (f(a_1),g(b_1))\leq(f(a_2), g(b_2)) \end{matrix}\right.$

Thus proving $\displaystyle t$ is isomorphic. Therefore the statement is true.

Given a set $\displaystyle W$, the cardinality is the number of elements in $\displaystyle W$. This is also written as $\displaystyle |W|$.

Looking at this particular problem, 

$\displaystyle A=\begin{Bmatrix} 3,7,5,4,9,11,24 \end{Bmatrix}$

Therefore the number of elements in a is,

$\displaystyle |A|=7$

Given a set $\displaystyle W$, the cardinality is the number of elements in $\displaystyle W$. This is also written as $\displaystyle |W|$.

Looking at this particular problem, 

$\displaystyle A=\begin{Bmatrix} a,a,b,c,d,e,e \end{Bmatrix}$

First, rewrite the set to depict a more accurate description of the space.

$\displaystyle A=\begin{Bmatrix} a,b,c,d,e \end{Bmatrix}$

Therefore, counting up the number of elements, results in the following cardinality,

$\displaystyle |A|=5$