- #1

sujoykroy

- 18

- 0

I was reading "Introduction To Set Theory" by Karel Hrbacek and Thomas Jeck and stuck with some logical trap in the proposition that "([tex]\textbf{N}[/tex],[tex]\prec[/tex]) is a well ordered set" where [tex]\textbf{N}[/tex] 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[tex]\subseteq[/tex]A, then b[tex]\in[/tex]B is called least element of B in the ordering R if bRx for all x[tex]\in[/tex]B

Second,

The relation [tex]\prec[/tex] on [tex]\textbf{N}[/tex] is defined as follows,

m[tex]\prec[/tex]n if and only if m[tex]\in[/tex]n

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

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 ([tex]\textbf{N}[/tex],[tex]\prec[/tex]) is a well ordered set, then every non empty subset of [tex]\textbf{N}[/tex] will have least element in [tex]\prec[/tex].

Suppose, B={n} for some n[tex]\in[/tex][tex]\textbf{N}[/tex]

So, B is a subset of [tex]\textbf{N}[/tex] and if we say that S has least element, b, then

b[tex]\prec[/tex]x for all x[tex]\in[/tex]B

Since B is singleton, it implies from above assumption that n[tex]\prec[/tex]n or n[tex]\in[/tex]n , 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