How to prove these equalities:

A∆(B∆C) = (A∆B)∆C

A∩(B∆C) = (A∩B)∆(A∩C)

A∩B = (A∆B)∆(A∪B)

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- November 12th 2010, 11:45 PMGarasA∆(b∆c) = (a∆b)∆c
How to prove these equalities:

A∆(B∆C) = (A∆B)∆C

A∩(B∆C) = (A∩B)∆(A∩C)

A∩B = (A∆B)∆(A∪B) - November 13th 2010, 01:35 AMtonio
- November 13th 2010, 01:58 AMGaras
I'm not sure that i understand what you are saying.

- November 13th 2010, 02:30 AMFernandoRevilla
- November 13th 2010, 02:51 AMGaras
I haven't studied that kind of function

- November 13th 2010, 04:55 AMtonio
- November 13th 2010, 05:30 AMGaras
I know some basics of theory and definition of symmetric difference, but i think that i don't know some transformations which are needed in order to prove those equations, because when i'm trying to solve this it becomes to complicated and i dont know how to move father from that point.

- November 13th 2010, 06:38 AMPlato
**Traditionally this is one of the most dreaded proofs on all of basic set theory.**

There is one way around the messy proof by cases, many cases.

That is the use of the characteristic function.

But you say you have not seen that function.

So in that case, you have to do it by cases.

It is very easy to get lost in all the cases. - November 14th 2010, 05:50 AMtom@ballooncalculus
Just in case a picture helps reduce disorientation and dread...

Imagine two sets B and C (as shaded circles) floating towards each other to form a combined venn diagram. But as the two float over each other, any shading in one that meets shading in the other fizzles momentarily then disappears. Then we get shading in either set but not both - as per symmetric difference.

http://www.ballooncalculus.org/draw/venn/one.png

Now if we imagine the same with set A and our picture of B∆C...

http://www.ballooncalculus.org/draw/venn/onea.png

... we notice the result is symmetrical, and might just as well have been produced from either of the other two combinations, e.g...

http://www.ballooncalculus.org/draw/venn/oneb.png

Now suppose the momentary fizzling succeeds in validating the shading, so that it drops out otherwise... then the shading in the result shows the intersection...

http://www.ballooncalculus.org/draw/venn/onec.png

This isn't so symmetrical, but we notice we can reach the same picture by getting the symmetric difference of two simpler intersections, as per your second equality...

http://www.ballooncalculus.org/draw/venn/oned.png

And for the third one...

http://www.ballooncalculus.org/draw/venn/onee.png - November 14th 2010, 10:14 AMGaras
It's is easy with diagrams but i don't think that that is a preferred method. This was one of the home work problems and i don't think that many people will solve that. Professor maybe thought that this is a easy problem without trying to solve it.

Can someone try to solve this using characteristic function? - November 14th 2010, 10:38 AMDrexel28