cardinal numbers 3

We will see a representative relationship lying between the cardinal number $\aleph_0$ of a countable set and $\aleph$ of an uncountable set.

Suppose a set $B$ be $\left\{0,1 \right\}$. For example, the cardinal number of the direct product set $B\times B\times B$ is card$(B\times B\times B)=2^3=8$, i.e.,
\[ 000, 001, 010, 011, 100, 101, 110, 111 \]
Hence, the cardinal number of the finite direct product set $B_1\times \cdots \times B_n$ is $2^n$.

We put a set $C$ as the infinite direct product set of $B$.
\[ C=\left\{ B\times B\times B\times \cdots   \right\} \]
card$C$ is expressed by $2^{\aleph_0}$ if $n\rightarrow \infty$. The set $C$ has all sequences of a infinite combinations of $0$ and $1$. In other words, $C$ contains all results of having thrown a coin with no end. We can show that $C$ is uncountable.

If $C$ is countable, all elements of $C$ can be listed in any order. However, after all elements were listed, by Cantor's diagonal argument, we can find a new sequence which should be in $C$. It is a contradiction. Therefore, $C$ must be uncountable. (refer to "countable sets 2")

This conclusion will be accepted by the simple fact that the binary number system is equivalent to the decimal in a scale of infinite digits.

As $C$ is uncountable, the cardinal number of $C$ becomes $\aleph$. However, since card$C$ is $2^{\aleph_0}$,
\[  2^{\aleph_0}=\aleph\]
It also means that
\[  \aleph_0 < 2^{\aleph_0} \]
Will you be convinced that it is true?