set theory intersection is a multiplication (due to semigroup)#544 Correcting Math
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From: Jan Burse on 28 Mar 2010 17:18 Archimedes Plutonium wrote: > Now, returning to the product of two sets. As is well known, the > frequently used notation for the intersection of two sets A nd B is > plain AB. Regardless of the notations, it's a semigroup operation. Well, surely at least a semigroup. But for example string concatenation is a also a semigroup. But you don't want intersection and as well multiplication to behave like a string concatenation, observe: "a" + "b" = "ab" <> "ba" = "b" + "a". So I would add that the semigroup is commutative (or abelsch). What is a little bit problematic to identify set intersection with multiplication, is the fact that set intersection, although most often conceived as acting on an infinite domain, has still order 2, i.e.: a^2 = a * a = a The notion of semiring is needed when we have two operations, for example multiplication and addition. And the addition comes from an abelsch semigroup and the multiplication comes from a semigroup. And when these operations are connected via distributivity: a * (b + c) = a * b + a * c (a + b) * c = a * c + b * c So the claims that multiplication is a semigroup, or is part of a semiring, is not wrong. But interpreting multiplication *directly* as set intersection, even when acting only on some subdomains, does not work, since order 2 holds for all elements of the domain. Bye |