# Union and successor

Definition : The successor of a set $x$ is the set $S(x)=x \cup \{ x \}$

This use of union seems to differ from normal usage.

Normal usage: $\{ 1,2,3 \} \cup \{ 4,5 \} = \{ 1,2,3,4,5 \}$ i.e., we do a union of the elements in the set.

We don't normally do $\{ 1,2,3 \} \cup \{ 4,5 \} = \{ \{ 1,2,3 \}, \{ 4,5 \} \}$ i.e., we don't usually consider each subset as an element.

But with successor, it seems we consider {$x$} a element.

Stated another way, if the universe $Z$is {$1,2,3,4,5$} then

with normal union $5 \cup Z = \{ 1,2,3,4,5 \}$ not { $5$, { $1,2,3,4,5$} }

Update: apparently { { 1,2,3 }, { 4,5 } } is more basic.

I found the following in internet doc "An Introduction to Axiomatic Set Theory, Joseph R. Mileti, February 6, 2007" but don't yet fully understand it.

Suppose that we have two sets A and B, say A = {u, v,w} and B = {x, z}. We want to be able to assert the existence of the union of A and B, which is {u, v,w, x, z}. First, by by the Axiom of Pairing, we may form the set F = {A,B}, which equals {{u, v,w}, {x, z}}. Now the union of A and B is the set of elements of elements of F. . . . in the presence of the Axiom of Separation, we state this axiom in the (apparently) weaker form that for any set F, there is set containing all elements of elements of F.

## 2 Answers

Comment to your updated question.

Suppose that we have two sets $A$ and $B$, say $A = \{ u, v, w \}$ and $B = \{ x, z \}$.

From Herbert Enderton, Elements of Set Theory (1977), page 3 :

The union of sets $A$ and $B$ is the set $A \cup B$ of all things that are members of $A$ or $B$ (or both).

Thus, in your example :

$A \cup B = \{ u, v, w, x, z \}$.

The existence of $A \cup B$ is licensed by the Union Axiom :

$\forall a \forall b \exists B \forall x(x \in B \leftrightarrow x \in a \lor x \in b)$.

see Enderton, page 18 or Axioms of Zermelo-Fraenkel Set Theory.

Thus, regarding your examples, you are right about : $\{ 1,2,3 \} \cup \{ 4,5 \} = \{ 1,2,3,4,5 \}$ but not about $Z = \{ 1,2,3,4,5 \}$ and $5$.

Writteh in that way, $5$ is not intended as a set; thus $5 \cup Z$ is not defined.

In set theory every object is a set; thus : $x \cup \{ x \}$ "makes sense" because is the set which has as elements the elements of $x$ and of $\{ x \}$. The elements of $\{ x \}$ are only one : $x$.

$\{ x \}$ is not considered as an element, but as a "normal" set, with $x$ as element.

Now, going on to the set-theoretic counterparts of natural numbers, we have that in set theory $5$ is "proxied" by $\{ 0,1,2,3,4 \}$; thus :

$6 = S(5) = 5 \cup \{ 5 \}$, i.e. $6 = \{ 0,1,2,3,4 \} \cup \{ 5 \} = \{ 0,1,2,3,4,5 \}$.

As you can see, the definition of "successor" make use of the "normal" union between sets.

In $S(x) = x \cup \{x\}$ we take a normal union of two sets, one of which is the set $x$ (with whatever elements it has) and the other is the set $\{x\}$, which has exactly one element, namely the set $x$ itself. In set theory it's perfectly fine for a set to be itself a member of another set; one could say that everything is a set.

• For example, if $x=\{1,2\}$ then $x\cup\{x\}=\{1,2,\{1,2\}\}$ – ajotatxe May 10 '14 at 12:29