errata



Max Weiss wrote:
> It seems to imply AC since, once again,
> let X be a set; let S={X};

> Then S has a choice function, namely {<{X},X>};
No. "A choice-function for a set" is NOT a function
that takes the set as argument and returns an
element of the set as result. A choice-function for
a set is a function whose DOMAIN IS the set and that
therefore takes each ELEMENT of the set as an argument,
and returns, for THAT element, an element OF THAT element.
A choice function for S={X} looks like {<X,x>}, with xeX.
But you don't necessarily HAVE an {<X,x>} unless you already
had a choice-function to compute x from X.

> but since U(S)=X,
> it follows
> by (*), that X has a choice function.

To do this the way you wanted, you would have to
let S={{X}}. THEN, S has a choice-function, namely,
{<{X},X>}. Since U(S) now = {X}, then the theorem
yields that {X} has a choice-function.
But then you could just apply the theorem AGAIN
(since U({X})=X) to get that X has a choice-function...

> Maybe it's the authors' mistake...

I was really hoping so, since there are a lot of errors
in this book. A pdf-list of them is at
http://comet.lehman.cuny.edu/fitting/errata/book_errors/setbookerrors/seterrata.pdf
and it is only 4 months old. Maybe you could direct your
inquiry to some of the other people who are listed as having
found errors.

.