Well-ordered set of Natural Numbers

[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\subseteqA, then b\inB is called least element of B in the ordering R if bRx for all x\inB

Second,
The relation \prec on \textbf{N} is defined as follows,
m\precn if and only if m\inn

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\precx for all x\inB
Since B is singleton, it implies from above assumption that n\precn or n\inn , which i guess violates Axiom of Choice.I may have misinterpreted some (or all) of the definition, that's 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

 

[HallsofIvy]

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\subseteqA, then b\inB is called least element of B in the ordering R if bRx for all x\inB

Second,
The relation \prec on \textbf{N} is defined as follows,
m\precn if and only if m\inn

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.

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\precx for all x\inB
Since B is singleton, it implies from above assumption that n\precn or n\inn , which i guess violates Axiom of Choice.

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.

I may have misinterpreted some (or all) of the definition, that's 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

 

[sujoykroy]

Thanks for your reply.

You mean that B is the set containing the single natural number "n"? But what is S here? I think you meant B itself.

Oops, that's my mistake. Yes , it should be B instead of S.

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.

That means B does not have a least element as per as the above definition of least element in a strict ordering, since B={n} and n=n .

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.

Regards
SR