Re: Review of Mueckenheims book.

MoeBlee wrote:
On Mar 8, 1:38 pm, Virgil <vir...@xxxxxxxxxxx> wrote:

Oh, by the way, for about the fifth time, please, Virgil, what is your
definition of 'codomain'?

The second factor of the Cartesian product of which the set of ordered
pairs of the function is necessarily a subset.

But there is NO SINGLE Cartesian product of which the function is
"necessarily" a subset EXCEPT dom(f)Xrange(f). The function is a
subset of MANY Cartesian products. IF you want to say that the
Cartesian product you're using for this definition is domXrange, then
'codomain' reduces to 'range'.

If you say f:D->C, so that you claim DxC is the Cartesian product of
which f must "necessarily" be a subset, then anyone can come back with
f:D->Cu{z} where z not a member of C. And it is a theorem:

f:D->C -> f:D->u{z}.

Proof (is this really necessary to give such proof?!):

Suppose f:D->C. So f is a function and domain of f is D and range of f
is a subset of C. And C is a subset of Cu{z}, so f is a function and
domain of f is D and, by transitivity of subsets, range of f is a
subset of Cu(z). So f:D->Cu{z} and so f is a subset of DxCu{z}.

So there are at least these four ways for you:

(1) Let the codomain be the range. (I don't think you want that.)

(2) Let there be more than one codomain for any function. (You don't
want that.)

(3) Let a function be a triple <D C f>.

(3) it is.

(But that conflicts with
saying "Let f(x) = x^2, since really you'd have to say "Let <D C f>(x)
= x^2.)

We don't have to say anything of the sort. We simply give the function
(i.e., triple) a convenient name, e.g., "f". Then we define "f(x) =
x^2" to mean (x,x^2) in S.

(4) Let a function be just what it is: a certain kind of relation; and
let a function-triple be a triple <D C f> such that f:D->C. (Then
every function-triple (or 'function-system' or whatever words you
like) has a unique codomain, viz. the third coordinate of the triple;
but still every function itself has many codomains.

--
David Marcus
.

Relevant Pages

  • Re: Review of Mueckenheims book.
    ... The second factor of the Cartesian product of which the set of ordered ... Cartesian product you're using for this definition is domXrange, ... so that you claim DxC is the Cartesian product of ... Let the codomain be the range. ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... The second factor of the Cartesian product of which the set of ordered ... so that you claim DxC is the Cartesian product of ... domain of f is D and, by transitivity of subsets, range of f is a ... Let the codomain be the range. ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... My point was that in set theory, we can speak of a codomain of f (any ... and s is a superset of the range of f. ... Every function is a subset of a Cartesian product. ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... a given function only has one codomain. ... A function as a subset of a cartesian product has all of these. ... For every set, there is a unique range, but not a unique codomain, ... Do you consider relations as sets of ordered pairs equally amorphous as ...
    (sci.math)