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**octa01** I try to understand the following proof of the well ordering theorem, by Halmos:

Given the set X, consider the collection W of all well ordered subsets of X.

Explicitly: an element of W is a subset A of X together with a well ordering of A.

We partially order W by continuation.

The collection W is not empty, because, for instance, ø is an element of W. If X unequals ø, less annoying elements of W can be exhibited; one such is {(x,x)}, for any particular element x of X. If C is a chain in W, then the union U of the sets in C has a unique well ordering that makes U "larger" than (or equal to) each set in C. This means that the principal hypothesis of Zorn's lemma has been verified; the conclusion

is that there exist a maximal well ordered set, say M, in W. The set M must be equal to the entire set X.

Reason: if x is an element of X not in M, then M can be enlarged by putting x after all the elements of M.

Let A and B be two well ordered subsets of X. In my observation B is a continuation of A if and only if A is a subset of B and A is the initial segment of some element of B.

The collection W could have well ordered subsets with different orders? The statement about the maximal well ordered set follows from Zorn's lemma. The point what is ambiguous for me is that M equals X.

If we have found a maximal well ordered set with a given order, how can we say that adjoining a x (doesn't belong to M) to M leads to a well ordered set?

Because "clearly" (can you actually see why?) every subset of $\displaystyle M\cup\{x\}$ has a first element after "putting" x "after" any other element of M...

Tonio

Why is this element related (the given order) with all the elements of M?