FOM: RE: 1-1 correspondence

Dennis E. Hamilton dennis.hamilton at
Fri Aug 16 16:23:52 EDT 2002

[cleaned up and refined for the FOM list:]

Nice metaphor.

The ordinals are often characterized in a way similar to what you have
described.  The ordinals can be developed in a manner comparable to f(p1) =
empty, f(p2) = {p1}, f(p3) = {p1, p2}, with the idea that each
(set-theoretic) ordinal is the set of all of its predecessors, with 0
represented by the empty set.  That's the gist of it.  It is not precisely
the same as what you are doing with pebbles, but I think you can see where
the harmony is.

And there is a condition on finitude that is similar to what you say.
Basically it has to do with there being a well-ordering (a premise about the
pebbles) and having it that each non-empty subset of a finite set has a
minimal element.  Another one is that every subset of a finite set is a
finite set.

I'd say you've captured most of that in your pebbles interpretation.

In my digging around for a grasp on what Cantor was up to, and why, I found
useful discussion of the ordinals, the denumerable, and the transfinite
cardinals in

Suppes, Patrick Colonel.  Axiomatic Set Theory.  D. Van Nostrand (New York:
1960).  Unabridged and corrected republication with new preface and section
8.4, Dover Publications (New York: 1972).  ISBN 0-486-61630-4 pbk.
Section 4.2 deals with finite sets, and section 5.2 deals with the finite
(set-theoretic) ordinals and the axiom of infinity.

For axiomatic set theories to provide for transfinite sets, the existence of
infinite sets has to be postulated, with the axiom of infinity being
essentially a declaration that the set of all of the ordinals (or some
equally denumerable set) exists.   And then off we go ...

I notice that elsewhere in Suppes it is establised (section 5.1 Theorem 12)
that, without the axiom of infinity, there can be no set of all the
ordinals.   It comes from how (set-theoretic) ordinals are developed there,
and the proof is by contradiction.  WIth the axiom of infinity, Suppes gets
to the (infinite) set of all natural numbers in a way which appears to
prevent that set from itself being a natural number, by virtue of a
well-ordering condition (Section 5.2 Definition 7 and then Definition 8).
This appears to constrain the set of (set-theoretically-represented) natural
numbers to only contains the *finite* (set-theoretic) ordinals and, in
particular, the set of all ordinals isn't a member of the set.

I don't know if this has held up over time, but I am keen to get a better
grasp of it.

-- orcmid

Dennis E. Hamilton
mailto:dennis.hamilton at
tel. +1-206-932-6970
cell +1-206-779-9430
     The Miser Project:

-----Original Message-----
From: Dean Buckner
Sent: Thursday, August 15, 2002 11:13
Subject: FOM: 1-1 correspondence

I have a bucket which I fill with pebbles.  I construct a function as

    f(p1) = {p1}
    f(p2) = {p1,p2}
    f(p2) = {p1,p2,p3}
    and so on

[ ... ]

It can be shown, I think, that no such set can be equinumerous with any
proper subset.  If so, it's a neat definition of finitude, because it
captures our intuition about finite sets - if we count out the pebbles we
come to the end at some point - without having to appeal to the existence of
any such process.  All that is required is that there be a mapping of the
right kind.

But is this correct?  Would be grateful for views.

[ ... ]

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