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Does this make sense? |
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| Aug5-07, 07:14 AM | #1 |
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Does this make sense?
A,B are sets
A + B=AuB + AnB Does it make sense to add sets? I know union and intersections are possible. |
| Aug5-07, 09:01 AM | #2 |
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| Aug5-07, 03:38 PM | #3 |
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No. I am talking about sets in measure theory.
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| Aug5-07, 04:02 PM | #4 |
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Does this make sense? |
| Aug5-07, 04:03 PM | #5 |
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| Aug5-07, 04:43 PM | #6 |
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honestrosewater: I take that to mean that you want an element that is in both A and B to show up twice in the sum of A and B?
I take it that he wants to say: A +B = A union B-A intersection B. |
| Aug5-07, 06:14 PM | #7 |
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| Aug5-07, 07:44 PM | #8 |
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If we take the sets {1,2,3} + {2,3,4} = {1,2,3,4}= A U B, which for n=1 to 4 is the whole set. Thus [tex]A\cup B+A\cap B =A\cup B [/tex] (I don't think measure theory has any effect on that.)
However if we thought of these as collections, then we would have: {1,2,3}+{2,3,4} = {1,2,2,3,3,4} (From Wikipedia: When two or more collections are combined into a single collection, the number of objects in the single collection is the sum of the number of objects in the original collections. ) This is easier to follow if we were thinking of collections of furniture like lamps, rugs, etc. So I believe that you are correct about the symmetric difference of sets. |
| Aug5-07, 09:47 PM | #9 |
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The original definition, "A + B=AuB + AnB" appears to be circular since the symbol that it is defining is used in the definition, so who knows. Normally, when you add two things, the result includes, in a loose sense, all of what you started with. For sets, this would seem to simply be union, but I assume the OP had something more than union in mind. You at least don't usually lose, or subtract, things when you add, so I assume the OP was thinking that the sum of two sets should include everything that was in those sets in some way that union doesn't, i.e., by including any duplicates. |
| Aug5-07, 10:08 PM | #10 |
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Since you say measure theoretic, perhaps you mean measure(a) + measure(b) = measure(a union b) + measure (a intersect b)? (for finitely additive measures, of course!)
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| Aug6-07, 12:59 AM | #11 |
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Yes, you are right. Measure is a mathematical concept, so we can use the plus or minus sign. So that in general: [tex] A\cup B = A+B-A\cap B [/tex]
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| Aug6-07, 05:52 AM | #12 |
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But that equation doesn't say anything about measure! Do you mean having first defined A+ B as [tex] A\cup B + A\cap B [/tex]. Of course, as has been pointed out, that is just equal to [tex] A\cup B[/itex]
It WOULD make sense if you would do what people have been asking you to do and write the "measure": [tex] measure(A\cup B) = measure(A)+ measure(B)-measure(A\cap B) [/tex] |
| Aug6-07, 09:04 AM | #13 |
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| Aug6-07, 04:25 PM | #14 |
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In what sense is that better? It is clearly wrong.
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| Aug6-07, 05:05 PM | #15 |
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| Aug6-07, 06:38 PM | #16 |
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I think I had read your post the wrong way round (i.e. so that it agreed with the wrong assertion that m(A+B)=m(A)+m(B)+M(AnB). Sorry.)
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| Aug6-07, 10:04 PM | #17 |
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Yeah, I was agreeing with m(A)+m(B) = m(A u B) + m(A n B)...just a tad better than the other way around, m(A)+m(B) - m(A n B) = m(A u B) as that's not quite true if m(A n B) is infinite. I wasn't agreeing with the other formulations.
The idea of "adding" sets though... How could addition be defined so that additive inverses might exist? |
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