[FOM] Slater's position
Hartley Slater
slaterbh at cyllene.uwa.edu.au
Tue Sep 30 23:49:35 EDT 2003
Randall Holmes (FOM Digest Vol 9 Issue 32) is still wanting the set
of those sets with n members to be the number n, i.e. {x|(ny)(y isin
x)}=n. He writes
>If I read your notation this way, then your sentence (nx)(Px) expands
>to {x | Px} \in n. So the numeral n is a set (or class, or
>superclass) whose elements are exactly the extensions with n elements.
Certainly if (x)(Px iff x isin {y|Py}), then (nx)Px is equivalent to
(nx)(x isin {y|Py}), and so maybe to: {y|Py} isin {x|(ny)(y isin x)},
i.e. the set of Ps is in the set of those sets with n members. But
that is not to say that (y|Py) isin n.
>In other words, your numerical quantifier n is exactly the same thing
>as my Frege natural number n,
It would be if one could say what Holmes wants to say, namely
{x|(ny)(y isin x)}=n, but how can one say that? Holmes needs to
provide a general reformulation of 'x has n elements' which does not
involve a quantifiable place 'n', and there is no such. There is no
problem with '(1x)Fx', '(2x)Fx', '(3x)Fx', etc. since these can be
turned into standard predicate logic expressions which do not involve
'1', '2' ,'3'. But not so with '(nx)Fx', since given the recursive
definition which includes that (nx)Fx iff
(Ey)(Fy.([n-1]x)(Fx.-(x=y))), one cannot get rid of the 'n'. Hence
in formulating Holmes' 'set ... whose elements are exactly the
extensions with n elements' one must allow the 'n' place to remain,
and his n={x|(ny)(y isin x)} cannot be a definition of the left hand
side.
>I would answer that _standard_ syntax does not allow quantification
>over quantifiers.
>But if quantifiers are to be taken as first class objects, I insist on
>interpreting them as classes of extensions.
Here is the nub of the matter, as I see it. It is not just that
Holmes is here appealing to what is 'standard' (and elsewhere to what
is 'generally accepted') when he is admitting to wanting to say a lot
of non-standard things himself, like '2 is a member of 3' etc. For
the 'n' in the quantifier 'nx' is a numeral, and it is that place
which needs to be quantified over to get the generality required, in
'iota-n(nx)Fx', etc. So Holmes simply does not want to quantify over
such numerical places, and allow that the second-order properties
which those numerals denote exist. 'To be is to be the value of a
variable', and Holmes does not want to recognise that numbers, as
opposed to sets of things with numbers of elements, exist. But such
'extensionalisation' quite generally is pointless, since one does not
remove reference to the property F if one replaces 'x has the
property F' with 'x is in the set of things with the property F'.
And isn't '(nx)Fx' standard syntax? 'The set of Fs has n members' is
standard English, and quantification over the numerical places in
such expressions comes out of Frege, see David Bostock's 'Logic and
Arithmetic' (O.U.P. Oxford 1974, Vol 1, Ch 3), and the paper by Mayo
in JSL 2002 previously referred to, 'Frege's Unofficial Arithmetic'.
Frege 'officially' preferred 'Nx:Fx = n' to '(nx)Fx', but with
'Nx:Fx' as 'iota-m(mx)Fx' these are interderivable as I showed before.
--
Barry Hartley Slater
Honorary Senior Research Fellow
Philosophy, M207 School of Humanities
University of Western Australia
35 Stirling Highway
Crawley WA 6009, Australia
Ph: (08) 9380 1246 (W), 9386 4812 (H)
Fax: (08) 9380 1057
Url: http://www.arts.uwa.edu.au/PhilosWWW/Staff/slater.html
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