# Well-ordered set of Natural Numbers

1. Apr 20, 2008

### sujoykroy

Hi,
I was reading "Introduction To Set Theory" by Karel Hrbacek and Thomas Jeck and stuck with some logical trap in the proposition that "($$\textbf{N}$$,$$\prec$$) is a well ordered set" where $$\textbf{N}$$ is set of all natural numbers. I will try to present the argument briefly to clarify the subjective trap that i am facing.

First, the least element is defined as,
If R is an ordering of set A and B$$\subseteq$$A, then b$$\in$$B is called least element of B in the ordering R if bRx for all x$$\in$$B

Second,
The relation $$\prec$$ on $$\textbf{N}$$ is defined as follows,
m$$\prec$$n if and only if m$$\in$$n

Third, ($$\textbf{N}$$,$$\prec$$) is a linearly ordered set because $$\prec$$ is a strict ordering on $$\textbf{N}$$ and every two elements of the $$\textbf{N}$$ is comparable in $$\prec$$

Fourth, (A,R) will be called well-ordered if every non-empty subset of A has least element in the linear ordering R.

Now, the the the trap is,
if ($$\textbf{N}$$,$$\prec$$) is a well ordered set, then every non empty subset of $$\textbf{N}$$ will have least element in $$\prec$$.
Suppose, B={n} for some n$$\in$$$$\textbf{N}$$
So, B is a subset of $$\textbf{N}$$ and if we say that S has least element, b, then
b$$\prec$$x for all x$$\in$$B
Since B is singleton, it implies from above assumption that n$$\prec$$n or n$$\in$$n , which i guess violates Axiom of Choice.

I may have misinterpreted some (or all) of the definition, thats why i am here asking for help. I am not a Math Professional or Math Student but pursue Math for personal interest of theory, so if i have asked very stupid question i do apologize for that.

Regards
SR

2. Apr 20, 2008

### HallsofIvy

Staff Emeritus
Okay, so, since you are talking about m being a member= of n, this is the model for the natural numbers in which 0= empty set, 1= set whose only member is the empty set, 2 is the set whose only members are 1 and 2, etc.

You mean that B is the set containing the single natural number "n"? But what is S here? I think you meant B itself. Obviously, if B contains only the natural number n, then its least element is n. Oh, I see your difficulty. This definition is strict inequality. Clearly, it is impossible for any member of a set B to be strictly less than every member of B- it can't be strictly less than itself! You need to say "b is the least element of B if and only if it is less than every other member of B". That is $b\prec a$ for every $a\in B$ and $a\ne b$.

3. Apr 20, 2008

### sujoykroy

Then set of all natural numbers is not well-ordered, because each subset of N is supposed have least element in $$\prec$$ in order to be called as well-ordered.