Re: ~ Sets of functions 1

From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 07/20/04 

Date: Tue, 20 Jul 2004 16:15:42 +0000 (UTC)

In article <x_WKc.1067868$Ar.423145@twister01.bloor.is.net.cable.rogers.com>,
Adam <addam@rogers.com> wrote:
>
>"Arturo Magidin" <magidin@math.berkeley.edu> wrote in message
>news:cdhdri$n4e$1@agate.berkeley.edu...
>> f and g are supposed to be sets of pairs of elements satisfying the
>> relevant conditions for them to be "functions" that is:
>>
>> f is a subset of X x A, such that
>> (i) for every x in X there exists a in A such that (x,a) is in f;
>> and
>> (ii) for every x in X, if there exist a,b in A such that (x,a) and
>> (x,b) are both in f, then a=b.
>>
>> g is a subset of X x B such that
>> (i) for every x in X there exists b in B such that (x,b) is in g;
>> and
>> (ii) for every x in X, if there exist b,c in B such that both (x,b)
>> and (x,c) are in g, then b=c.
>>
>>
>> So the triple (f,X,A) tells you three things: (i) the domain for the
>> function; (ii) the codomain for the function; and (iii) the set of
>> pairs that is the function.
>
>Oh! That is what is meant by f. Right, okay. In the definition that I use it
>states that there exists F <= X x A, where for each x in X there is one and
>only one a in A such that (x,a) in F. But I never use that set of pairs in
>my proofs because I usually only need the definition of the function since
>that "gives" the pairs in F.

Which is quite sensible, just as you probably never use the fact that
the ordered pair (x,y) is "really" supposed to be the set
{ {x}, {x,y} }; because the way we define a function as a set of
ordered pairs is just a way to define it so that it has the properties
we want, and then we "forget" how it was defined and merely use the
properties we wanted.

In this respect, it is very much like "object oriented programming" in
computer science: we don't care about the exact way in which
"function" or "ordered pair" is implemented; all we care about are
their properties.

But, every once in a while, what the things actually ->are<- can come
into play (for good or ill), as it did in your attempts. It is "clear"
that if A is a subset of B, then F(X,A) is a subset of F(X,B); but as
sets, this is not the case when A is not equal to B (because the
'function' carries information about its domain and codomain). So
instead, if you want to be really technical, you have to say that
there is a "natural" map from F(X,A) to F(X,B) which is one-to-one,
and identifies any function f:X->A with the function f:X->B with the
exact same values at every x in X. Which is, after much discussion and
my usual long-winded-ness, what Julien was saying. (-: 

-- 
======================================================================
"It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu

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