axiomatic sets 14 (a natural number)

Natural number is an ordinal.

You must know well,
the first=1,
the second=1+1=2,
the third=2+1=3,
the fourth=3+1=4,
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In addition, '0' is also a natural number.

In axiomatic set theory, those must be sets.

We have already defined
\[ 0=\phi=\left\{ \right\} . \]

As, if 'n' is a natural number,　then 'n+1' is also a natural number and an ordinal,
we will define
\[ n+1=n\cup\left\{ n\right\}=\left\{ n, \left\{ n\right\} \right\}  . \]

That is to say,
$1=0\cup\left\{ 0\right\}=\phi\cup\left\{ 0\right\}=\left\{ 0\right\}=\left\{ \phi \right\}=\left\{ \left\{ \right\}\right\}$ ,
$2=1\cup\left\{ 1\right\}=\left\{ 0\right\}\cup\left\{ \left\{ 0\right\} \right\}=\left\{ 0,\left\{  0\right\}\right\}$ ,
$3=2\cup\left\{ 2\right\}=\left\{ 0,\left\{  0\right\},\left\{ 0,\left\{  0\right\}\right\}  \right\}$ ,
$4=3\cup\left\{ 3\right\}=\left\{ 0,\left\{  0\right\},\left\{ 0,\left\{  0\right\}\right\},\left\{ 0,\left\{  0\right\},\left\{ 0,\left\{  0\right\}\right\}   \right\}  \right\}$ ,
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You will understand
$1=\left\{0 \right\}$ ,
$2=\left\{0,1 \right\}$ ,
$3=\left\{0,1,2  \right\}$ ,
$4=\left\{0,1,2,3  \right\}$ ,
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We are able to get all natural numbers.

axiomatic sets 13 (functions)

We can get a definition of a function,
adding conditons to the definition of a relation.

As a relation is a set, a function is a subset of a relation.

'$F$' is a function if and only if '$F$' is a relation and
\[ \forall x\forall y\forall z[xFy\wedge xFz\rightarrow y=z] \]

It means there exists only one $y$ in a function $xFy$  for giving $x$.

This definition of a function is equivalent to
\[ \forall z[z\in F \rightarrow E!(y)[z=<x,y>\rightarrow xFy]]  \]
,where $<x,y>$  is an ordered pair.

In general, the realtion $\lt(\gt)$ are not a function
because when $x\lt(\gt) y$ and $x\lt(\gt) z$ , we are not able to say $y=z$ .

You must know that in standard notations we write $xFy$  $y=F(x)$ ,
namely $y$  is $F$ of $x$ .

We are able to define a funtion as
\[ F(x)=y\leftrightarrow [E!z[xFz]\wedge xFy]\vee [\neg(E!z[xFz])\wedge y=0] .   \]

In this case, for example, if
\[ F=\left\{<x,y> ; <1,2>,<1,3>,<3,4>,<4,4>    \right\}  \]
,then
\[ F(1)=0,\qquad F(2)=0,\qquad F(3)=4,\qquad F(4)=4 .  \]

You know $\left\{ \right\}=0 $ .