FOM: Cantor's Diagonal Argument
Dean.Buckner at btopenworld.com
Sat Jun 29 03:06:52 EDT 2002
To summarise an earlier point.
(*1) Every A is a B, not every B an A
(*2) For every B there is an A
(*3) There is no collection of A's such that they are all the A's
I argued that De Morgan's argument (that if (1) is true, (2) is false),
contains a fallacy. If (3) is false, then De Morgan is right. But if (3)
is true - if there are an infinite number of A's - he is not. (1) - (3) can
all be true.
This depends on whether we can make sense of (3), the medieval axiom (there
are not so many that there are not more), but I'll come to that.
Let's interpret this in terms of Friedman's Cantorean argument. Let there
be some list L and let A be "item on the list", and B "natural number set",
(1a) Every item in the list is a natural number set (we assume)
(1b) Not every natural number set is an item in the list.
This is what the Cantorean argument, as supplied by Harvey, actually proves.
Let's now assume the medieval axiom:
(3) There is no collection of items in the list such that they are all the
items in the list
Note, this is not saying anything like "A is an infinite set". William Tait
complained I was "trying to understand Cantor's theorem while avoiding the
`set-theoretic' language in which it formulated and proved". There's a
reason for this. I am reading "A" as a common noun, not as a proper name
for an "infinite set", for reasons I will explain in a moment. But, unless
we fall into De Morgan's fallacy, we are perfectly entitled to assume
(2) For every natural number set there is an item in the list.
by the logic of my earlier postings. (Which is exactly the same logic that
shows that for every natural number (B) there is a corresponding even number
(A) ). Yet I'm not sure Friedman (or Tait or many others) would agree with
In reply, it could be argued that, in a set-theoretic formulation, we would
use not a noun like "A", but a proper name or designating expression such as
"the A's". If "the A's" now designates all the things that are A, and if
there B that are not one of these A's, I think it follows that we cannot
match up every B with an A. We would want some other undesignated A
available to match, but there aren't any, as we just designated all of them.
This is a perfectly good reply, but there is a difficulty. In order to
state the medieval axiom, in order to say that there are infinitely many
A's, it seems necessary to posit that there is no collective term like "the
A's" that does signify every A.
This follows from the very definition of collective designation. If the
sentence "All the A's are B" is true, there exists some set A* of A's, of
which the sentence says that all them, and _only_ those things, are B. But
in using a quantified noun phrase, as in "every A is B", we are not
asserting of any A that it is B. No A have even to exist, for it is
equivalent to saying "there is no A that is non-B". It is essential to the
collective proposition, in other words, that it state *of* certain things
that they are such and such.
But if (3) above is true, there can be no collective term C, that denotes a
collection of A's. There must always be members different from every member
of such a collection. Whatever A's the term C denotes, there must be other
A's. But then C cannot denote every A that there is.
Note that we can "reach" every A by the general proposition "every A is B".
That is because general propositions are not in any sense "about" the things
they are true of. But when we formulate the Cantorean arguments using such
propositions, they do not seem to work. We need a collective proposition
for the argument to work, but such a proposition does not "get" to all the
A's that we want.
Any suitable reply needs to show at least one of the following
(a) that my restatement of Friedman's argument is flawed. But I can't see
how, since we can replace the terms as follows
item on list = even number
natural number set = number
It is perfectly true, to paraphrase Friedman's statement, that "every
listing of even numbers omits some numbers". But still, for every number
there is a corresponding even number.
(b) That it is possible to restate the argument in a way that is not flawed,
without using what I have called "collective designation" or anything like
(c) That it is possible to define collective designation, or something like
it, in a way that does not rule out the designation of "infinite sets".
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