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Does this make sense?

by pivoxa15
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pivoxa15
#1
Aug5-07, 07:14 AM
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A,B are sets

A + B=AuB + AnB

Does it make sense to add sets? I know union and intersections are possible.
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CRGreathouse
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Aug5-07, 09:01 AM
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Quote Quote by pivoxa15 View Post
A,B are sets

A + B=AuB + AnB

Does it make sense to add sets? I know union and intersections are possible.
What do you mean that to do? Additive number theory has addition of sets like [itex]X+Y=\{x+y:x\in X,y\in Y \}[/itex] (so that {1, 2, 3} + {10, 40} = {11, 12, 13, 41, 42, 43}). Is that what you want?
pivoxa15
#3
Aug5-07, 03:38 PM
P: 2,268
No. I am talking about sets in measure theory.

mathman
#4
Aug5-07, 04:02 PM
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Does this make sense?

Quote Quote by pivoxa15 View Post
A,B are sets

A + B=AuB + AnB

Does it make sense to add sets? I know union and intersections are possible.
You ccan define "+" to mean anything you want. What is the point of your definition?
honestrosewater
#5
Aug5-07, 04:03 PM
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Quote Quote by pivoxa15 View Post
A,B are sets

A + B=AuB + AnB

Does it make sense to add sets? I know union and intersections are possible.
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? The sum then could not be a set since there are no dupllcates in sets. What kind of object do you want the sum to be, a bag, a.k.a. multiset?
robert Ihnot
#6
Aug5-07, 04:43 PM
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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.
honestrosewater
#7
Aug5-07, 06:14 PM
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Quote Quote by robert Ihnot View Post
I take it that he wants to say: A +B = A union B-A intersection B.
Oh. So symmetric difference (more) then?
robert Ihnot
#8
Aug5-07, 07:44 PM
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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.
honestrosewater
#9
Aug5-07, 09:47 PM
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Quote Quote by robert Ihnot View Post
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.)
Is this what you meant previously? You seem to have just defined this addition to be union. The symmetric difference is "the set of elements belonging to one but not both of two given sets", i.e., "A union B-A intersection B", which I assume you meant as "(A union B) - (A intersection B)", with "-" denoting set difference (A - B = {x | x in A and x not in B}).

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.
CRGreathouse
#10
Aug5-07, 10:08 PM
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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!)
robert Ihnot
#11
Aug6-07, 12:59 AM
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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]
HallsofIvy
#12
Aug6-07, 05:52 AM
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Thanks
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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]
phoenixthoth
#13
Aug6-07, 09:04 AM
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Quote Quote by CRGreathouse View Post
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!)
this form might be better because it works even if measure(a intersect b) is infinite.
matt grime
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Aug6-07, 04:25 PM
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In what sense is that better? It is clearly wrong.
phoenixthoth
#15
Aug6-07, 05:05 PM
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Quote Quote by matt grime View Post
In what sense is that better? It is clearly wrong.
It is? Counterexample, please.
matt grime
#16
Aug6-07, 06:38 PM
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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.)
phoenixthoth
#17
Aug6-07, 10:04 PM
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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?
matt grime
#18
Aug7-07, 12:37 AM
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You need to define the boolean operations properly. You need to use the symmetric difference. Every element is self inverse.


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