On Apr 14, 2008, at 12:05 , TSa wrote:


Xavier Noria wrote:
  {0, 1} X {{}} = {(0, {}), (1, {})}
which, you see, is different from {0, 1}. They have different elements. The fact that there's a clear mapping that sort of identifies them has nothing to do with set equality.

But X is cooperating with , in Perl 6:

 (0,1) X (()) === ((0,()),(1,())) === (0,1)

That is, X strips the outer list and comma concatenates the
inner empty list away.

With your definition X is not even associative because @a X @b X @c does
not produce a list of triples but either
   (@a X @b) X @c  === ( ((@a[0],@b[0]), @c[0]), ...)
    @a X (@b X @c) === ( ( @a[0], (@b[0],@c[0])), ...).
That is two lists of differently  structured pairs.

Oh yes thank you. There was a subthread about the Cartesian product in Set Theory itself. I didn't mean that applied to the Perl operator.

-- fxn

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