As a set is a gathering of some matters, the elements of a set have no orders.
Given a set having elements $x,y,z$ , then
$\left\{ x,y,z \right\}=\left\{ x,z,y \right\}=\left\{ y,x,z \right\}=\left\{ y,z,x \right\}=\left\{ z,x,y \right\}=\left\{ z,y,x \right\} $ .
We want to give an order for elements and
distinguish two sets having same elements of which the orders are not same.
At first we will define a set having only two elements.
\[ \forall x\forall y\exists a\forall w[w\in a\leftrightarrow w=x\vee w=y] \]
Namely, $a=\left\{ x,y \right\}$ or $\left\{ y,x \right\}$ . It is called axiom of pairing.
If we write $a=\left\{ x,y \right\}$ , then it means $x\neq y$ , and
if $x=y$ , then we will write $a=\left\{ x \right\}$ or $a=\left\{ y \right\}$ .
Therefore, $\left\{ x,x \right\}=\left\{ x \right\}$ .
Given two sets $a=\left\{ x,y \right\}$ and $b=\left\{ u,v \right\}$ ,
if $a=b$ , then $x=u$ and $y=v$ , or $x=v$ and $y=u$ by axiom of extensionality.
That is to say $\left\{ x,y \right\}=\left\{ u,v \right\}$ and $\left\{ x,y \right\}=\left\{ v,u \right\}$ .
However, if $a=b$ means only the case of $\left\{ x,y \right\}=\left\{ u,v \right\}$
and $\left\{ x,y \right\}\neq \left\{ v,u \right\}$ ,
then the only conditions of $x=u$ and $y=v$ will be required.
We will define an ordered pair for two elements by
\[ \left\{ \left\{ x \right\}, \left\{ x,y \right\} \right\} \]
and write it $(x,y)$ or $\lt x,y \gt$ . Hence, definitely $\lt x,y \gt\neq \lt y,x \gt$ .
With concrete description, $\lt 0,1 \gt\neq\lt 1,0 \gt$ . Because
$\lt 0,1 \gt=\left\{ \left\{ 0 \right\}, \left\{ 0,1 \right\} \right\}$ ,
$\lt 1,0 \gt=\left\{ \left\{ 1 \right\}, \left\{ 1,0 \right\} \right\}$ ,
although $\left\{ 0,1 \right\} =\left\{ 1,0 \right\}$ .
(and you might know $\left\{ \left\{ 0 \right\}, \left\{ 0,1 \right\} \right\}=\left\{ 1, \left\{ 1,0 \right\} \right\}$ . )
By this definition, an order of two elements in a set will be decided and
two same sets having two different elements become equivalent,
just only when the order of elements is equivalent.
If you want to set out three elements, you may put
\[ \lt a,b,c \gt=\lt a,\lt b,c \gt \gt=\left\{ \left\{ a \right\}, \left\{ a,\left\{ \left\{ b \right\}, \left\{ b,c \right\} \right\} \right\} \right\} \]
More elements if you need, then
\[ \lt a_1,a_2,a_3,\cdots,a_n \gt=\lt a_1,\lt a_2,a_3,\cdots,a_n \gt \gt \]
Axiom of paring means that there is a set such that has only one or two elements.
Using the axiom, we are able to get an order of elements.
For example, an arity of functions will be noted by a set.
2015/11/16
2015/11/05
axiomatic sets 5 (extensionality)
Second axiom of ZFC is the extensionality of a set.
By this axiom, we are able to distinguish sets by means of members.
It is stated by following formula;
\[ \forall x\forall y[\forall z[z\in x\leftrightarrow z\in y]\leftrightarrow x=y] \]
Using a familiar expression,
\[ x=y\leftrightarrow \forall z[z\in x\leftrightarrow x\in y] \]
This axiom means three things.
one is a set (except for empty set) has elements,
two is the difference of two sets is only dependent on each elements, and
three is a set can be extended by adding distinct elements.
If $x\neq y$ are two sets, then there are some elements (or a element)
that belong to one but not the other.
For example, if
$x=\left\{ x;1\leq x\leq \sqrt{2}, x=\mbox{ real number} \right\}$ , and
$y=\left\{ y;1\leq y\lt \sqrt{2}, y=\mbox{ real number} \right\}$ , then $x\neq y$ .
If
$x=\left\{ x;1\leq x\leq \sqrt{2}, x=\mbox{ rational number} \right\}$ , and
$y=\left\{ y;1\leq y\lt \sqrt{2}, y=\mbox{ rational number} \right\}$ , then $x= y$ .
The axiom of extensionality requires that a set will be only identified by the elements.
When we deal with a set, we have to observe the elements of a set carefully.
By contraries if the elements are not defined or are defined ambiguously,
the set can not exist.
(However, you will see the gathering of some matters not a set,
although the elements are defined with no doubt. )
As the elements create a set, empty set is always one in any spaces.
There do not exist two different empty sets.
It can be possible that empty set becomes a element.
\[ \left\{ \phi \right\} =\left\{ \left\{ \right\}\right\} \]
We will define this set $\left\{\phi\right\}$ as $1$ ;
\[ \left\{\phi\right\}=1 \]
because in the theory $1$ also has to be a set.
By this definition, you must understand that $\phi$ is not equivalent to $\left\{\phi\right\} $ .
In a addition, we will always be able to use the number $0$ and $1$ in any spaces.
By this axiom, we are able to distinguish sets by means of members.
It is stated by following formula;
\[ \forall x\forall y[\forall z[z\in x\leftrightarrow z\in y]\leftrightarrow x=y] \]
Using a familiar expression,
\[ x=y\leftrightarrow \forall z[z\in x\leftrightarrow x\in y] \]
This axiom means three things.
one is a set (except for empty set) has elements,
two is the difference of two sets is only dependent on each elements, and
three is a set can be extended by adding distinct elements.
If $x\neq y$ are two sets, then there are some elements (or a element)
that belong to one but not the other.
For example, if
$x=\left\{ x;1\leq x\leq \sqrt{2}, x=\mbox{ real number} \right\}$ , and
$y=\left\{ y;1\leq y\lt \sqrt{2}, y=\mbox{ real number} \right\}$ , then $x\neq y$ .
If
$x=\left\{ x;1\leq x\leq \sqrt{2}, x=\mbox{ rational number} \right\}$ , and
$y=\left\{ y;1\leq y\lt \sqrt{2}, y=\mbox{ rational number} \right\}$ , then $x= y$ .
The axiom of extensionality requires that a set will be only identified by the elements.
When we deal with a set, we have to observe the elements of a set carefully.
By contraries if the elements are not defined or are defined ambiguously,
the set can not exist.
(However, you will see the gathering of some matters not a set,
although the elements are defined with no doubt. )
As the elements create a set, empty set is always one in any spaces.
There do not exist two different empty sets.
It can be possible that empty set becomes a element.
\[ \left\{ \phi \right\} =\left\{ \left\{ \right\}\right\} \]
We will define this set $\left\{\phi\right\}$ as $1$ ;
\[ \left\{\phi\right\}=1 \]
because in the theory $1$ also has to be a set.
By this definition, you must understand that $\phi$ is not equivalent to $\left\{\phi\right\} $ .
In a addition, we will always be able to use the number $0$ and $1$ in any spaces.