> > It states:  If X is a partially ordered set such that every chain in X
> > has an upper bound, then X contains a maximal element.  I'm trying to
> > understand Halmos' proof.  To summarize, instead of dealing with X, he
> > deals with Y, which is the set of all chains in X, and, after a long
> > series of contortions,  proves the existence of a maximal element  in
> > Y.
> >  Can someone please help explain:

> > 1.- How is this procedure equivalent to proving the existence of a
> > maximal element in X?

> To be explicit:

> Given x in X, the "weak initial segment of x", ws(x), is

> ws(x) = {y in X : y<=x}.

> You can consider ws as a function from X to P(X), the power set of X;
> the power set of X is ordered by inclusion, and we have that ws(x) is
> contained in ws(y) if and only if x<=y, so ws can be viewed as an
> order-preserving function between the partially ordered set X and the
> partially ordered set P(X). Morevoer, the function ws is one-to-one:
> if ws(x) = ws(y), then since x is in ws(x) and y is in ws(y), it
> follows that x<=y and y<=x, so x=y.

> So, Halmost says: "finding a maximal element in X is the same as
> finding a maximal element in" the range of ws. Because you have an
> order isomorphism between X and its image under ws. And since X
> satisfies that every chain in X has an upper bound in X, then every
> chain in the range of ws has an upper bound in the range of ws.

> Now let Y be the set of all chains in X, partially ordered by
> inclusion. Note that if A is a chain, then A has an upper bound x in
> X, so therefore, A is contained in ws(x) [here we are using the
> hypothesis on X]. Thus, every element of Y is dominated by an element
> of the range of ws.

> This means that if y is a maximal element in Y (with respect to the
> inclusion order of Y), then y has to be a maximal element in ws(x) as
> well. For if y is contained in ws(x), then y\/{x} would be a chain,
> hence an element of Y; since it contains the maximal element y of Y,
> we must have y = y\/{x}, so x lies in y. Thus, y is in fact in the
> range of ws, and is also maximal in the range of ws. Thus, finding an
> maximal element in Y gives a maximal element of the range of ws, which
> in turn gives a maximal element of X.

> > 2.- How or where is the original hypothesis invoked?

> We invoke the original hypothesis to get that every element of Y is
> contained in an element of the range of ws; this is used in order to
> justify that a maximal element of Y must in fact be of the form ws(x)
> for some x, which you need in order to obtain the corresponding
> maximal element of X.

> > In the proof, the only basic principle invoked is the Axiom of
> > Choice.  Any help appreciated.

> As with most proofs in Halmos's book, there is a fair amount unsaid;
> in this case, it is in the phrase "Since each set in Y is dominated by
> some set in S, the passage from S to X cannot introduce new maximal
> elements". The premise invokes the condition on X, the conclusion is
> the argument above.

> --
> Arturo Magidin