Re: Review of Mueckenheims book.

In article <1173382978.119642.245410@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"MoeBlee" <jazzmobe@xxxxxxxxxxx> wrote:

On Mar 8, 1:03 am, Virgil <vir...@xxxxxxxxxxx> wrote:
In article <1173332811.266694.176...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

"MoeBlee" <jazzm...@xxxxxxxxxxx> wrote:
On Mar 7, 9:20 pm, Virgil <vir...@xxxxxxxxxxx> wrote:

Whether that definition, or definitions in genneral, or whatever are
more technical formality does not bear upon the fact that the standard
definition in set theory textbooks of 'is a function' is what I said
it was.

Is it so in "Naive Set Theory" by Halmos?

Halmos only gives the three-place vesion, so there is no chance even
for him to dispute that the definition I gave for a one-place version.

The fact that Halmos does not give any "one-place" definition at all
disputes your claim that it is the standard definition.

Nope, actually Halmos DOES USE the one-place definition without
stating it first. Actually, it's not even a definition, it's a
CONSEQUENCE of the three-place definition.

On page 30, he writes, "If X and Y are sets, a function from (or on) X
to (or into) Y is a relation f such that dom f = X and such that for
each x in X there is a unique y in Y with (x,y)ef. The uniqueness
condition can be formulated explicitly as follows: if (x,y)ef and
(x,z)ef, then y = z."

Later, e.g., on page 48 he writes, "Since it easy to see that u itself
belongs to C, it remains only to prove that u is a function. We are to
prove, in other words, that for each natural number n there exists at
most one element x of X such that (n,x)eu (Explicitly: if both (n,x)
and (n,y) belong to u, then x=y)."

So, to prove "u is a function" we notes that we are to prove that for
every member n of the domain (the set of natural numbers in this case)
we have "if both (n,x) and (n,y) belong to u, then x=y)".

And THAT is just what I said. f is a function iff (f is a relation and
if (n,x)ef and (n,y)ef, then x=y.

And that gives the grounds for proving 'f is a function' with no
mention of a codomain whatsover.

And let's go back to his original definition to DERIVE the one-place
version, not even as a definition, but as a CONSEQUENCE:

A function from X to Y is a relation f such that dom(f) = X and such
that if <x y>ef and (<x z>ef, then y = z.

Now, suppose f is a function from X to Y. So f is a relation & ((<x
y>ef & <x z>ef) -> y=z).

Now suppose f is a relation & ((<x y>ef & <x z>ef) -> y=z). So, f is a
function from dom(f) to range(f). So there exists and X and Y such
that f is a function from X to Y. So f is a function (a fortiori, if f
is a function form X to Y, then f is a function; and note that Halmos
DOES use the termninology 'f is a function' alone as cited on page 48
where he speaks of u being a function and even gives the requirements
for proving that "u is a function").

So f is a function <-> (f is a relation & ((<x y>ef & <x z>ef) ->
y=z)).

And that is a CONSEQUENCE of Halmos's three-place version and is USED
by Halmos HIMSELF such as on page 48.

And it is the standard definition of 'is a function' in set theory.

And that some books in mathematics in general give the three-place
version is not only consistent with the one-place version, but the one-
place version is a CONSEQUENCE of the three place version, as I just
showed.

And, though I don't have Halmos in front of me as I type, I would bet
that in the text he does make use of the principle that a function is
a relation such that for every x in the domain there is exactlhy one y
such that <x y> is an element of the function.

That a function has that single-valued property is certainly accepted by
Halmos, but that differing codomains, so long as they are large enough
to contain the range, define the same function appears not to be
acceptable to Halmos.

WHERE does Halmos say that? And I didn't even say that different
codomains DEFINE the same function. I said that, GIVEN a certain
defintion of 'codomain', a function has more than one codomain. And I
said that *I* am NOT insisting on that definition of 'codomain' but
rather that I welcome YOU to state whatever definition you wish to
state.

So, AGAIN, what is your definition of 'codomain' now?

Halmos is no
counterexample to anything I've said.

That is not not at all clear.

It sure as shootin' is.

As I do not have a copy of Halmos at hand, and thus cannot review the
context from which your quotes were taken, it remains unclear to me.

Unless one is trying to decide whether two functions are identical or
whether two functions can be composed, the issue of which codomain a
function has is largely moot. The Halmos example you cited, not being
either of those, is not sufficient to settle the issue of whether
different codomains produce different functions.
.

Relevant Pages

  • Re: Review of Mueckenheims book.
    ... The fact that Halmos does not give any "one-place" definition at all ... disputes your claim that it is the standard definition. ... And it is the standard definition of 'is a function' in set theory. ... defintion of 'codomain', a function has more than one codomain. ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... mathematics disallows making definitions? ... was even made EXPLICIT to be in regards to set theory. ... implies both a domain and a codomain. ... of course anyone is "allowed" not to use the standard set ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... given the definition of codomain as 'a superset of the ... which is an area of mathematics that is not subsumed by ... Not in those versions of set theory with which I am most familiar. ... I'm just using it in the overwhelmingly standard sense within set ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... is your chance just to think and see that Halmos says a "relation" NOT ... relation and he doesn't say it's a triple, so my formalization may not ... is quite in keeping with ordinary set theory, ... domain, codomain, and set of ordered pairs such that etc. ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... Is it so in "Naive Set Theory" by Halmos? ... for him to dispute that the definition I gave for a one-place version. ... disputes your claim that it is the standard definition. ...
    (sci.math)