How to prove that 'every totally ordered set X contains a cofinal well-ordered subset'?(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# A question about ordered set and cofinality

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