Christopher Menzel wrote:
> On Jun 15, 2007, at 6:39 PM, Waclaw Kusnierczyk wrote:
>> Christopher Menzel wrote:
>>>>> There are (as of course John and Pat know) 2^card(D) relations over
>>>>> any set (taking relations here to be sets of ntuples).
>>>> Only if n=1.
>>> Only if D is finite.
>> My point was that a relation over a single set is a set of 1tuples;
>
> I am not following you. In standard, classical logic and set theory,
> an nplace relation over a set is a set of ntuples of members of S.
> So a 1place relation  i.e., a property  is a set of 1tuples
> over S (or, more typically, just a subset of S). A 2place relation
> is a set of 2tuples (i.e., ordered pairs) over S, i.e., a subset of
> SxS; a 3place relation is a set of 3tuples, i.e., a subset of
> SxSxS; and so on. (01)
OK. I remembered a definition of a relation over sets S, S', S'', ...
as a set of tuples from the Cartesian product S x S' x S'' x ...
If there are 1 sets (there is one set), the relation is a set of 1tuples. (02)
But of course, this may be wrong. (03)
>>> I must admit that, like any good platonist, I was thinking of D as
>>> infinite, in which case what I say is true for all n.
>> A relation over an infinite set is still an (infinite) set of 1
>> tuples.
>
> You appear to be using "relation" to mean "property". Is that right? (04)
Yes. (05)
>>> To cover the general case of D any size, insert "at least" above
>>> after "There are".
>> At least, unless n=1.
>
> Huh? That's not even true if "relation" is understood to mean
> "property". (06)
? If n=1, then there are exactly 2^card(D) properties over D? (07)
> The above claim holds for any finite n  assuming
> "relation" is understood as it is standardly used in logic and set
> theory. (08)
We read the same statements in different ways.
I shut up. (09)
vQ (010)
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