Re: Transitive Closure

Paul <paul_at_test.com> wrote in message news:<Hu2qc.4347$wI4.496108_at_wards.force9.net>...
> x wrote:
> >>The commutative property does not make sense with unary operators like
> >>TClose. Unary operators are always commutative because there is only
> >>one operand's order.
> >
> >
> > Commutativity does make sense with unary operators.
> > M--f-->M--g-->M
> > fg = gf or fg != gf
>
> I think the confusion here is that in something like group theory we say
> a binary operator "*" is commutative if a*b = b*a. Or writing it as a
> function we say:
>
> *(a,b) = *(b,a)
>
> What we're talking about here which I think is what Alfredo is
> misunderstanding is commutativity of the "composition of operators"
> operator. So say we have unary operators f and g we can define the
> operator "f*g" to be:
>
> (f*g)(x) = f(g(x)) for all x.
>
> Commutativity of this "*" operation just means that:
>
> f*g = g*f
>
> ie. f(g(x)) = g(f(x)) for all x.
>
> For this to make sense f and g must take arguments of the same type, and
> return values of that same type as well.

I'm confused myself.

So we are talking about different algebra, that has a universe with elements being unary relational algebra operators only -- selection, projection, Cartesian Power (as product is, unfortunately, binary operator), transitive closure, and, perhaps, negation. In this tiny 4 element universe we define one binary operation: composition of unary relational operators. So we can speak of this operation commutativity, although, it's kind of wierd to say that operation commutes on these 2 particular elements of the universe, and doesn't commute on those. In traditional algebra commutativity is all or nothing proposition: it's either commutative on the whole universe, or isn't (if there is at least a pair of elements such that ab!=ba).

Can people familiar with Category Theory comment? (Diagram by "x" provoked this thought). Received on Mon May 17 2004 - 20:58:58 CEST