1. ## Well Ordering Theorem

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

2. Originally Posted by 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 $M\cup\{x\}$ 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?
.

3. The phrase "putting" x "after" any other element of M, I don't understand. How do you know that adjoining a {x} to M is consistent with respect to the order of the maximal well ordered set? In other words: is the adjoining {x} bigger (with respect to the order of M) than all elements of M?

If this statement is clear, then I indeed see that {x} adjoined with M is well ordered.

4. Originally Posted by octa01
The phrase "putting" x "after" any other element of M, I don't understand. How do you know that adjoining a {x} to M is consistent with respect to the order of the maximal well ordered set? In other words: is the adjoining {x} bigger (with respect to the order of M) than all elements of M?

If this statement is clear, then I indeed see that {x} adjoined with M is well ordered.

I think "bigger" here means x is going to be a maximum element of $M\cup \{x\}$

Tonio

5. How is it possible to make a (arbitrary) x the maximum element of M U {x}?

6. Originally Posted by octa01
How is it possible to make a (arbitrary) x the maximum element of M U {x}?

Simple: determine that $m\leq x\,\,\,\forall m\in M$...this extends the well ordering of M, whatever it is.

Tonio