Re: What is a proof, exactly?

From: J.E. (troubled6man_at_yahoo.com)
Date: 11/29/04 

Date: 29 Nov 2004 09:34:48 -0800

Han de Bruijn <Han.deBruijn@DTO.TUDelft.NL> wrote in message news:<co4j3f$n0a$1@news.tudelft.nl>...
> J.E. wrote:
>
> > [ ... ] As for your question of "how do you equal
> > objects which are not sets?", the first thing I wonder is "what are
> > you discussing that isn't a set?" A formula? Why not discuss the
> > formalization of the formula into a set? A theory? Same thing with
> > the formalization of the list of theorems. I really haven't come
> > across a need for a wider equality in my work, so I'd be interested in
> > your experience.
>
> In conclusion, could I say that everything in (mainstream) mathematics
> is a set? That only the equality of sets may be considered relevant?

There are things in the field of set theory (like the operation of
ordinal arthimetic) that aren't sets (too big), things in category
theory, things in algebraic topology, and I'm sure other fields too.
But for many things (e.g. functions between sets) the things are sets
(e.g. the set that is the graph of the function is considered "THE
function" itself), so set equality works.

> OK. Back to equivalence relations. If I have an equivalence relation
> which is related to a set, then that equivalence relation induces a
> classification of that set. Right? Such a classification means that
> the original set is partitioned into subsets where every subset is

A binary relation R on a set A is (usually) a subset of {{{a},{a,b}}:
a in A & b in B}, yes. From it you could construct, for any a in A,
the subset of all b in A such that {{a},{a,b}} in R. Which would be
nonempty if {{a}} in R (which would be true for an equivalence
relationship). If you let B={S in P(A): there is some a (a in A and
S={b in A: {{a},{a,b}} in R})}, then I think that would be what I
think you consider the partition, yes.

> non-empty, disjunct from any other subset in the partition and the
> union of all these subsets is the original set. Am I still right?

The sets in B are nonempty for a reflexsive relation. They are
disjoint for a transitive refelxsive symmetric relation. And their
union is A for a refelxisive relation, yes.

> You say that equality is always an equality between sets. Then any of
> the members of a set must also be a set in itself. Now Cantor is saying
> that a set is a collection of _distinct_ elements. Since elements are
> just sets, this can be given a precise meaning: elements are distinct
> iff their intersection is empty.

I think you are confusing disjoint and distinct.

> All of the elements together make up
> the whole set. And why not demand that the elements are "something",
> which means not just empty.

Not empty MEANS that they contain something. The axiom of equality
says that the sets are equal iff the contain the same things, which
means that ALL that matters is WHAT is in them. So a blue set with
the number five in it and a yellow set with the number five in it are
the same assuming the number five is the same as the number five and
that they each contain nothing else. How do we tell if they are the
same? We check what is inside the five in the yellow set and compare
it to what is inside the five in the blue set. If you don't have sets
all the way down, then evetually you get to something that doesn't
have equality defined and it causes equality to become undefined for
everything that contained it, and everything that contained that and
so on transfinitely.
 
> The difference between a "true" equality and an equivalence relation
> becomes quite subtle now. So subtle that I would even dare to ask if
> these two concepts do not denote one and the same thing. Summarizing:
>
> Equality = Equivalence relation
> Member = Class in a partition
>
> Terse mathematics, that would be a good thing ...
>
> Han de Bruijn

I think that's been done before. For instance you could contruct the
non-zero rationals as equavalence classes of pairs of non-zero
integers, with {{p},{p,q}} in the same class as {{r},{r,s}} iff
p*s=r*q. But since these evivalence classes are themselves sets, then
the set of all the classes is the set all "non-zero rationals". And
the members of the class are themselves sets, so the set equality
works, so if you call the set of classes Q*, then a in Q* and b in Q*
are written the "normal way" as is a=b written the normal way and all
mean the usual things. The thing is that set equality itself is an
equivalence relation itself that is "too big" to be a set. So sets
can do relations on specific sets, but sets can't do relations on all
sets in general.

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