Technically the Cartesian cross operator doesn't have an identity value. There
is no set X such that
A x X = A. Now any singleton set gives a result that is naturally isomorphic
to the original set, I.e, there is a obvious bijection between the two sets,
but they are not equal sets.
--
Mark Biggar
[EMAIL PROTECTED]
[EMAIL PROTECTED]
[EMAIL PROTECTED]
-------------- Original message ----------------------
From: Darren Duncan <[EMAIL PROTECTED]>
> Adriano, I think perhaps what Tsa is trying to get at is the identity value
> for the X operator, and I believe I know what it is.
>
> In the relational model of data, both the version of the model where tuples
> have unordered named attributes/elements (which I prefer), and the version
> where tuples have ordered attributes/elements (which Perl 6 seems to be
> using in its X operator), the relational cross product operator is
> analogous to numeric multiplication or logical 'and' in its properties
> (except that the ordered version isn't commutative).
>
> With respect to relational join being like multiplication, the special
> values 0 and 1 are represented by the nilary (zero attribute) relation with
> either 0 or 1 tuples respectively, which in common Perl array-of-hash (or
> array-of-array) notation for rowsets is
>
> []
>
> and
>
> [ {} ] or [ [] ]
>
> respectively; I'll call them R0 and R1 for now. Joining any relation R
> with R0 gives R0 (or alternately a relation with the same attributes as R
> but zero tuples; its zero tuples either way), and joining any relation R
> with R1 gives R.
>
> So R1 is the identity value for cross product, meaning the identity value
> for X, in Perl 6 would be a one-element array|seq whose element is the
> empty array|seq. That is,
>
> [X] ()
>
> equals this:
>
> ( () )
>
> Larry et al, on a related note, the list of identity values for reduce in
> S03 should be updated to account for this; [X] wasn't on the list last I
> looked.
>
> -- Darren Duncan
