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\}
cardC 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 cardC 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?
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