cardinal numbers 2

The cardinal number of an infinite set is very interesting. We are able to find some examples.

(1)Suppose $A_1=\left\{1,3,5,\cdots \right\}$ and $A_2=\left\{2,4,6,\cdots \right\}$. Then, obviously, card$A_1$=card$A_2$=$\aleph_0$. But, since $A_1\cup A_2=\mathbb{N}$, card$(A_1\cup A_2)=\aleph_0$. In the generality, for the direct sum of two infinite countable sets,
$$\aleph_0+\aleph_0=\aleph_0$$
(2)Similarly, in the cardinality of the continuum like the cardinal number of real numbers,
$$\aleph +\aleph =\aleph$$
It can be understood by following facts. Let two sets be interval $A_1=(0,1]$ and $A_2=(1,2)$. The cardinal number card$A_1$=card$A_2$=$\aleph$. But, card$(A_1\cup A_2)=$card$(0,2)=\aleph$.

(3)We have already seen the open interval $(0,1)$ is one to one corresponding to $\mathbb{R}.$Let a set $S$ be constructed by all points in a rectangle of the length $a>0$ and width $b>0$. Hence, $S$ is written by
$$S=\left\{ (x,y)|x\in [0,a], y\in [0,b], a,b\in \mathbb{R}, a,b>0  \right\}$$
Do not confuse the direct product $(x,y)$ with an open interval.

We are able to arrange a one to one function $(x,y)\in S\rightarrow z\in\mathbb{R}$. Supposing $a=b=1$ and $z\in [0,1]$, we shall simplify the problem to understand. Let $c_i$ be a single digit figure from $0$ to $9$. If we define $x,y,z$ as follow,

$x=0.c_1c_3c_5\cdots$
$y=0.c_2c_4c_6\cdots$
$z=0.c_1c_2c_3c_4c_5c_6\cdots$

this function satisfies the conditions. That is, the function $(x,y)\rightarrow z$ is a one to one correspondence. Therefore, the cardinal number of $S$ is $\aleph$. It means in a direct product set that the following equation is true.
$$\aleph^2=\aleph$$
Furthermore, in n-dimensions,
$$\aleph^n=\aleph$$
You should note that a deciding factor of the cardinal number of a set is the existence of a one to one function. Will your intuition say yes?