Set theory operators compared to Algebra operators #543 CorrectingMath
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From: Rotwang on 28 Mar 2010 16:00 Jesse F. Hughes wrote: > A <anonymous.rubbertube(a)yahoo.com> writes: >> >> [...] >> >> What does this even mean, for set theory to "have intersection as >> multiplication"? >> > > AP doesn't really know what it means, I'm sure, but one way to > understand his half-remembered thought is this: Let S be a set. Then > the powerset of S, P(S), is a ring, where union is addition, > intersection is multiplication, the empty set the 0 element and S the > unit (i.e., 1). Part of the definition of a ring is having an additive inverse for each element, which P(S) doesn't. It does form a "semiring" though. @A: another way in which intersection can be thought of as a product (though I don't know of anybody using the term "multiplication" in this context) is in the sense of category theory: the partial ordering on P(S) makes it into a category, in which the intersection of any two sets is a product (i.e. a greatest lower bound). |