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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.
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