Set theory operators compared to Algebra operators #543 Correcting Math
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From: Jesse F. Hughes on 28 Mar 2010 15:36 A <anonymous.rubbertube(a)yahoo.com> writes: > On Mar 28, 4:16 am, Archimedes Plutonium > <plutonium.archime...(a)gmail.com> wrote: [...] >> Now I find it rather intriguing that Intersection as a set operator is >> viewed as multiplication. >> Now how that comes to be is rather puzzling to most people because it >> seems to be a >> smaller end result from either set A or set B, and as we know in >> mathematics that the >> multiplication is larger than addition, in the usual cases. So why >> does Set theory have >> intersection as multiplication. >> > > > > 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). (This is from my own half-memories and I didn't bother to check it, so take this at your own risk.) -- Jesse F. Hughes "Please. I was a philosophy major. Nobody can 'know' anything. And I DO know." -- George Greene embarrasses philosophy majors everywhere. |