From rapaport@buffalo.edu  Wed Apr  1 18:31:36 2009
Date: Wed, 1 Apr 2009 18:31:22 -0400 (EDT)
From: "William J. Rapaport" <rapaport@buffalo.edu>
Subject: HW9, #5 -- p 149 #74a

------------------------------------------------------------------------
Subject:  HW9, #5 -- p 149 #74a
------------------------------------------------------------------------

A student writes:
"I'm having trouble with HW9#5.  I might be misreading the question in
the book.  When I first read it, I was thinking that u was not a member
of the codomain which was giving me trouble because that didn't fall in
line with the definition of f being a subset of A x B...."

Co-domain of which function?  There are 2 functions in this problem:  f
and f*.  f is indeed a subset of AxB. However, u is not an element of B
(so it is not in the co-domain of f).

Continuing, the student writes:
"But it looks like f is mapping A to (B U {u})...."

Look again:  It is f* that maps A to (B U {u}).

Student:
"So does that really mean f is a subset of A x (B U {u})?"

Nope:  f* is a subset of that binary relation.

Think of it this way:  f maps A to B, but not all of A, which is why
it's a partial function.  To turn it into a total function, we need to
find something to map the missing elements (so to speak) into.  Let
a \in A be an element that is not in f's domain.  Let u be an element
not in B.  Create a new function f* that behaves just like f on all
elements of A that f is defined on, but that also maps lonely elements
like a into u.

That's the meaning of the definition of f*, which I'll repeat here in a
slightly different form (more like you might write it in a computer
program):

========================================================================
for any a \in A, 
     if f(a) \in B, then f*(a) = f(a) 
                 [i.e., if a belongs to the domain of definition of f,
                        then f* will map a into whatever f maps it into]
     else f*(a) = u
                 [i.e., if a doesn't belong to the dom of def of f,
                        then f* will map a into u]

What you have to prove is that f* is a function (i.e., that it
satisfies the definition of a function considered as a set of ordered
pairs)