axiomatic sets 8 (subsets)

By axiom of extensionality, we defined two same sets.

Its condition is all elements of each sets is same.
If $a=\left\{ x,y  \right\}$  and $b=\left\{ x,y  \right\}$ , then $a=b$ .

If there are different elements in two sets,
or there are not some elements in each sets,
then two sets are not same.

Given two sets $a,b$ , if
$$\forall x[x\in a\rightarrow x\in b] ,$$
then the set $a$  is called the subset of the set $b$ .
We will write it $a\subset b$ .

Usually $a\subset b$  accepts the case of $a=b$ .

Hence, in addition if
$$\exists y[y\notin a\wedge y\in b ] ,$$
then $a$  is called the proper subset of $b$ .

The proper subset means the elements of the subset is always in the set ,
and there are some elements of the set not in the subset . Therefore, $a\neq b$ .

Please understand the difference of $\in$  and $\subset$ .

Although both are the binary relationship, $\in$  is the relation of a element and a set,
and $\subset$  is the relation of a set and a set.

Given two sets $a=\left\{ x,y  \right\},b=\left\{ x,y,z  \right\}$ .
As $a\subset b$ ($a$  is the proper subset of $b$ ) , $a\notin b$ .

As $x$ (and $y$) is in both $a$ and $b$ , $x\in a$ and $x\in b$ .
However, $x$ are not the subset of $a$  and $b$ ,
because the elements of $x$ are not in $a$ and $b$ .

If $a=\left\{ x,y  \right\},b=\left\{ x,y,\left\{ x,y  \right\}  \right\}$ , then $a\in b$  and $a\subset b$ .

Please note again the difference of $\left\{ x  \right\}$  and $\left\{ \left\{ x  \right\}  \right\}$ .


(This is not axiom of ZFC. We will use in axiom of power set. )

axiomatic sets 7 (union of sets)

By using axiom of pairing, we can define the union of sets.

$$\forall x\exists z\forall w[w\in z\leftrightarrow \exists v[w\in v\wedge v\in x]]$$

It is axiom of union.
The collection of all the elements $w$ of the elements $x,y$ of a set $a$ becomes a set.
We will write the set $\cup a$ .

If you know the naive set theory, you might feel the axiom of union strange.
Why the elements of the elements of a set are needed?
Because $a=\left\{ x,y  \right\}$  is needed.

Given a set (non ordered pair) $a=\left\{ x,y  \right\}$ and
$x=\left\{ x_1,x_2  \right\},y=\left\{ y_1,x_2  \right\}$ .

Axiom of union asserts all elements of the elements $x,y$ of a set $a$ is a set.
$$\cup a=\left\{ x_1,x_2,y_1  \right\} .$$
We will define $\cup a=\cup\left\{ x,y  \right\}=x\cup y$ .

We will also write $\cup a=\left\{ x_1,x_2  \right\}\cup \left\{ y_1,x_2  \right\}$ .
In the same way,
$$\left\{ x_1,\cdots ,x_n  \right\}=\left\{ x_1\right\}\cup \left\{ x_2,\cdots ,x_n  \right\}$$
will be defined.

By Aixiom of union, we are able to use a set whose elements are three or more.

Using the symbol $x\cup y$ , the union of sets is stated by
$$\forall w[w\in x\cup y\leftrightarrow w\in x\vee w\in y] .$$

If a union of three sets $x_1,x_2,x_3$ is needed, then put $a=\left\{ x_1,\left\{ x_2,x_3 \right\}  \right\}$  and
$$\cup a=\cup \left\{ x_1, \cup\left\{ x_2,x_3  \right\}  \right\}=x_1\cup x_2\cup x_3 .$$
Continuing this operation, the union of $x_1,\cdots x_n$ will be gotten by
$$x_1\cup x_2\cup \cdots \cup x_n .$$

For concrete example, the union of $a=\left\{ 0,1  \right\}$  is
$$\cup a=0\cup 1=\left\{ \right\}\cup \left\{ \left\{ \right\} \right\} =\left\{ \left\{ \right\} \right\}=1 .$$
This is not able to be gotten in the naive set theory.