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A question about ordered set and cofinality

  1. Aug 11, 2011 #1
    How to prove that 'every totally ordered set X contains a cofinal well-ordered subset'?
    I'm reading Halmos's Navie set theory,this is an excersice in the chapter 'well-ordering'.
    According to Halmos( in the same chapter),if a set C cotains well-orderd sets,and it is a chain respect to continuation,then there exists an unique set U,which is the union of sets in C: U is well-orderd,and U is a contunuation of every set in C.
    this lemma seems like the tool needed to construct such a subset.but i met some problems here.
    The set A={s(x),x belong to X} is a chain respect to continuation ,but s(x) is not well ordered.
    the set Ax= {x,f(x),ff(x),...} (where f is a chioce function on X-s(x))is well ordered,but the family {Ax} seems not like a chain respect to continuation here.
    in those cases,Halmos's lemma could not be used directly both. I try to construct some kind of subset of s(x),keep well-orderd,and their famliy forms a chain.for example,take some 'a ' belong to X,just consider the set X-s(a),and let Ax={a,x1,x2,x3,....,x}where xn comes from the chioce function on the set {a<y<x},it's obvious Ax is well ordered,for z>x, Az={a,z1,z2,z3,...,z},but we could refomulate Az as the union of Ax and some set B,B contains the values of choice function on s(z)-s(x),it then seems that Az is a continuation of Ax.
    but between x and z there would be infinite many other elements of X.I'm not sure whether the construction on Az is suitable.If it makes sense,then we just take the union of all Ax,and get a well ordered set,this set is obvious cofinal(it's a continuation of all Ax)
    Maybe i'm totally wrong on the beginning,the real proof need not halmos' lemma or choice functions.But i havn't got other methods yet,i'm still not sure about this.
     
    Last edited: Aug 11, 2011
  2. jcsd
  3. Aug 11, 2011 #2
    i've solved this problem by apply zorn's lemma instead of use AC directly.consider the set of well ordered sets of X,by Halmos' lemma every chain (respect to continuation)in it has an upper bound,so it has an maximal element.this maximal element is cofinal.
    does anyone knows other methods?for example by AC.
     
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