You may remember the structure of natural number $\mathbb{N}$ ,
\[ 0=\left\{ \right\},1=\left\{0\right\},2=\left\{0,1\right\},3=\left\{0,1,2\right\},\cdots , \]
based on axiom of infinity ($n+1=n\cup\left\{n\right\}$) .
We will also define
\[ \mathbb{w}=\left\{0,1,2,\cdots \right\} \]
as all natural numbers $n$ .
If we go to the next number from $\mathbb{w}$, then
\[ \mathbb{w}+1=\left\{0,1,2,\cdots,\mathbb{w} \right\}, \]
Furthermore,
\[ \mathbb{w}+2=\left\{0,1,2,\cdots,\mathbb{w},\mathbb{w}+1 \right\} , \]
and $\cdots\cdots\cdots$
Then,
\[ \mathbb{w}+\mathbb{w}=\left\{0,1,2,\cdots,\mathbb{w},\mathbb{w}+1,\mathbb{w}+2,\cdots \right\} \]
and we naturally put
\[ 2\mathbb{w}=\mathbb{w}+\mathbb{w} . \]
Further and furthermore,
\[ 2\mathbb{w},3\mathbb{w},\cdots, \mathbb{w}\mathbb{w} \]
\[\mathbb{w}^2=\mathbb{w}\mathbb{w}=\mathbb{w}+\mathbb{w}+\cdots \]
\[ \mathbb{w}^3=\mathbb{w}^2+\mathbb{w}^2+\cdots \]
(you must also remember that $3^2=3\times 3=3+3+3,4^2=4\times 4=4+4+4+4$ and $3^3=3^2\times 3$)
Although we can not write anymore,
we are formally able to get $\mathbb{w}^{\mathbb{w}},\mathbb{w}^{\mathbb{w}^{\mathbb{w}}}\cdots$ .
These are called an Ordinal number.
It is the expansion of the property of Natural number.
When you find out eternal infinite repetitions of an infinite number,
could you feel the depth of real number $\mathbb{R}$ ?
2017/04/17
2017/04/04
axiomatic sets 34 (regularity 3)
We shall give the easy proof in which Axiom of regularity
\[ \forall x[x\ne \phi\rightarrow \exists y[y\in x\wedge y\cap x=\phi]] \]
means that there is not a infinite sequence $a_1\ni a_2\ni a_3\ni\cdots$ .
Suppose that there is a infinite sequence $a_1\ni a_2\ni a_3\ni\cdots$ .
Then, we are able to make the new set $A=\left\{ a_1,a_2,a_3,\cdots \right\}$ .
If we accept axiom of regularity,
then there must be a $b$ such that $A\cap b=\phi$ in $a_1,a_2,a_3,\cdots $ .
Let $b$ be $a_i$ . However, as $\cdots\ni a_{i-1}\ni a_i\ni a_{i+1}\ni \cdots$ ,
$a_{i+1}\in b(=a_i)$ and $a_{i+1}\in A$ are true.
It is a contradiction, because $a_{i+1}\in(A\cap b)$ .
Therefore, there is not a infinite sequence $a_1\ni a_2\ni a_3\ni\cdots$ .
Conversely, we accept axiom of regularity.
For two arbitrary sets $A,a$ , even if $A\cap a=a_1$ ,$A\cap a_1=a_2$ ,$\cdots$ continue,
by axiom of regularity, the sequence will stop somewhere.
If $A\cap a_i=\phi$ , then by $a_i=b$ , axiom of regularity will be satisfied.
\[ \forall x[x\ne \phi\rightarrow \exists y[y\in x\wedge y\cap x=\phi]] \]
means that there is not a infinite sequence $a_1\ni a_2\ni a_3\ni\cdots$ .
Suppose that there is a infinite sequence $a_1\ni a_2\ni a_3\ni\cdots$ .
Then, we are able to make the new set $A=\left\{ a_1,a_2,a_3,\cdots \right\}$ .
If we accept axiom of regularity,
then there must be a $b$ such that $A\cap b=\phi$ in $a_1,a_2,a_3,\cdots $ .
Let $b$ be $a_i$ . However, as $\cdots\ni a_{i-1}\ni a_i\ni a_{i+1}\ni \cdots$ ,
$a_{i+1}\in b(=a_i)$ and $a_{i+1}\in A$ are true.
It is a contradiction, because $a_{i+1}\in(A\cap b)$ .
Therefore, there is not a infinite sequence $a_1\ni a_2\ni a_3\ni\cdots$ .
Conversely, we accept axiom of regularity.
For two arbitrary sets $A,a$ , even if $A\cap a=a_1$ ,$A\cap a_1=a_2$ ,$\cdots$ continue,
by axiom of regularity, the sequence will stop somewhere.
If $A\cap a_i=\phi$ , then by $a_i=b$ , axiom of regularity will be satisfied.