- #1
sujoykroy
- 18
- 0
Hi,
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
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