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I am learning set theory and have been working on this problem for a while.

I have read Mark's answer which uses forcing(is it?) to prove that a maximal chain of S is the desired total order.

But I think I also read somewhere that we can prove use this proposition using lexicographic order. First we have pair each element x up with set X=$\{y|y \preceq x\}$. Next arrange a well order for all X, which are subsets of S. Every subset is uniquely represented by a series of 0,1(of length |S|), and $x \preceq y$ iff $x$'s initial segment X is placed before $y$'s.

My question is: Is this second argument valid? What if |S| is uncountable?

More generally, assuming Well Ordering Theorem, if every set A is order isomorphic to an ordinal, why is not every set A "countable"? In other words, can we enumerate any set A by mapping each element of A to an ordinal, and since ordinals are well ordered, put the smaller one before the larger one? This argument implies all partially ordered sets can be extended to a well ordered set, which is even stronger than a total order.

John
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1 Answers1

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Proof in the link

Proof in your link uses Zorn's lemma, not forcing. Everything needed is in that link, so there is no point to reproduce it here. Just give it some time to understand.

Your efforts

By technical details of set theory, it is known, that any attempt to prove "every partial order can be extended to total" is obligated to use statement equivalent to Zorn's lemma. This is, why your attempts have to be unsuccessful. Let's take closer look: $\{y|y\preceq x\}$ is special case of downward directed set. These sets with subset relation as order are order-isomorphic to your starting order, not well ordered, not even totally ordered in general. Written explicitly: let $(S,\leq_S)$ be partial order, that is: $\ \leq_S\ \subset\ S\times S$ is reflexive, antisymmetric and transitive. Let $f:S\rightarrow\mathcal{P}(S)$, where $f(x)=\{y|y\leq_Sx\}$. Then

  • $f$ is injective
  • for each $x,y\in S:x\leq_Sy\iff f(x)\subseteq f(y)$

This implies, that image of f: $Img (f)=\{f(x)|x\in S\}$ with set inclusion is order isomorphic to $(S,\leq_S)$, so we gained nothing.

In the case of lexicographic order; it is only defined with respect to total order - won't work here.

Series series of $0$'s and $1$ you are referring to are called indicator function. These are just other way of representing subset, and are inducing the same order as set inclusion, namely: for $A,B\in X$, and $\chi_A,\chi_B:X\rightarrow \{0,1\}$ it is the case that: $A\subset B\iff \forall_{x\in X} \chi_A(x)\leq\chi_B(x)$

General case

There are uncountable ordinals. Well ordering Theorem states, that every set can be well ordered. It is not true, that every partial order extends to wel-order. For example set of rational numbers has order which is total - so cannot be extended, but it's not well-ordering.

Sidenote: case when $S$ is finite is much simpler, so it could be helpful to consider it first. In particular: every total order over finite set is well-ordering, and every partial order can be extended without Zorn's lemma. Such procedure is dubbed topological sort.

Przemek
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