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 n-tuples).
>>> Only if n=1.
>>
>> Only if D is finite.
>
> My point was that a relation over a single set is a set of 1-tuples; (01)
I am not following you. In standard, classical logic and set theory,
an n-place relation over a set is a set of n-tuples of members of S.
So a 1-place relation -- i.e., a property -- is a set of 1-tuples
over S (or, more typically, just a subset of S). A 2-place relation
is a set of 2-tuples (i.e., ordered pairs) over S, i.e., a subset of
SxS; a 3-place relation is a set of 3-tuples, i.e., a subset of
SxSxS; and so on. (02)
> ...
>> 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. (03)
You appear to be using "relation" to mean "property". Is that right? (04)
> (But see above.)
>
>> To cover the general case of D any size, insert "at least" above
>> after "There are".
>
> At least, unless n=1. (05)
Huh? That's not even true if "relation" is understood to mean
"property". The above claim holds for any finite n -- assuming
"relation" is understood as it is standardly used in logic and set
theory. (06)
-chris (07)
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