# Can the successor function be applied to N itself? ( With ' N' denoting the set of natural numbers)

I don't think my question leads to anything, but does it have an answer?

My question is: I am allowed to form the set

N U {N} ,

that is the set :

{    0, 1, 2 ,3 ...............................{ 0,1,2,3 .......}   } ,

applying the successor function S to N itself, where S(x) = x U {x} ?

Remark: I'm not asking whether N is itself a natural number, that would be forbidden, I think, by the rule according to which no set can be a member of itself.

• In set theory yes. See Von Neumann definition of ordinal. – Mauro ALLEGRANZA Mar 30 at 18:06
• Thanks. May I ask you whether this set is of any interest? Could it qualify as a number of some sort? – Ray LittleRock Mar 30 at 18:08
• @ Mauro Allegranza. Thanks for the link. – Ray LittleRock Mar 30 at 18:12
• Yes, it can qualify as a number... it's the ordinal number $\omega + 1$. – mjqxxxx Mar 30 at 18:14

For any set $$X,$$ the axiom of pairing implies $$\{X\}$$ is a set, and then applied again, says $$\{X,\{X\}\}$$ is a set, and then the axiom of union says $$\cup\{X,\{X\}\}=X\cup\{X\}$$ is a set. (This is only one way to construct it... there are lots of ways.) So the successor operation is defined for any set (although it's really only called the successor operation when it's used on ordinals).
So if $$\mathbb N$$ is a set, then we can form the successor set. Almost always, we define $$\mathbb N$$ to be $$\omega,$$ the first infinite ordinal, which is obtained as the closure of the empty set under the successor operation (this closure exists by the axiom of infinity). The successor of $$\omega,$$ $$\omega\cup\{\omega\},$$ is the next ordinal, denoted $$\omega+1.$$ Applying it again we get $$\omega+2,$$ $$\omega + 3$$ and so on.
• Is $\omega-1$ a well defined element of $\mathbb{N}$? – Count Iblis Mar 30 at 20:30