Proving Set Equality: A-B, A∩B, B-A

[simmonj7]

Homework Statement

Prove the following identities involving complementation of sets.
1. A ∩ [(A ∩ B)^C] = A - B
2. A ∩ [(A ∩ B^C)^C] = A ∩ B
3. (A ∪ B) ∩ A^C = B - A

Homework Equations

A^C = A complement. If A is contained in some understood universal set U, then the complement of A is the set A^C = {x ∈ U: x ∉ A}.
A - B = { x ∈ A and x ∉ B}
B - A = { x∈ B and x ∉ A}
A ∪ B = {x ∈ A or x ∈ B}
A ∩ B = { x ∈ A and x ∈ B}

The Attempt at a Solution

Since I am dealing with set equality and trying to show when one set equals another, I have been trying to show that x ∈ the first set if and only if x ∈ the second set in each of these cases.

I am not actually going to put this in legit proof form here but I will just summarize my steps for you and know that I plan on doing it better later once I figure it out entirely.

For part 1, I started off by choosing an arbitrary x in A ∩ [(A ∩ B)^C]. Then, if x is in this set, x ∈ A and x ∉ A ∩ B. This means that x ∈ A and x ∉ A or x ∉ B. Well I am trying to show that this is equal to A - B, where x ∈ A and x ∉ B so I have the parts I need, however, I don't know what to do with the part where it says x ∉ A.

For part 2, I started off by choosing an arbitrary x in A ∩ [(A ∩ B<sup>C/SUP])C/SUP]]. Then, if x is in this set, x∈ A and x ∉ A ∩ BC/SUP]. This means that x ∈ A and x ∉ A or x ∉ B^C. Once again, this means that x∈A and x∉A or x∈ B. Well I am trying to show that this set is equal to A ∩ B, where x ∈ A and x ∈ B. But, once again, I have that extra part in the middle where x ∉ A that I don't know what to do about.

For part 3, I chose an arbitrary x in (A ∪ B) ∩ AC. Then, if x is in this set, x ∈ A or x ∈ B and x ∈ AC. Well, this means x ∈ A or x ∈ B and x ∉ A. This time, I have that other term in the front saying x ∈ A when I really need to get to B - A, where x∈ B and x ∉ A.

Am I completely missing something here?
Thanks.</sup>

 

[Dick]

That sort of proof gets tedious after while. Do you know deMorgan's laws applied to sets? http://planetmath.org/encyclopedia/DeMorgansLaws.html and the distributive laws of union and intersection?

 

[simmonj7]

All we really know about de Morgan's laws is that, prior to this problem, we had to prove de Morgan's second law.

 

[Dick]

All we really know about de Morgan's laws is that, prior to this problem, we had to prove de Morgan's second law.

Well, ok then. For the first part if x is not in A, then x is not in A intersect anything, nor is it in A minus anything. So x is not in either side. So x is in one side iff x is in the other side. Because it's not in either. That's how the negative side of these things works. These things are a lot easier just using set algebra than juggling words. Hope you get to that phase soon.

 

[simmonj7]

We are actually completely done with discussing sets so I don't know it we are going to get to there. :/

 

[Dick]

Too bad, maybe next course? Here's the first one. An(AnB)^C=An(A^C U B^C)=(An(A^C))U(An(B^C)=An(B^C), since An(A^C) is empty. And An(B^C)=A-B. Done.

 

[simmonj7]

That actually makes some sense to me but I wonder if I will even be away to get away with set algebra since we haven't covered it.
But thank you so much for that.

 

[Dick]

That actually makes some sense to me but I wonder if I will even be away to get away with set algebra since we haven't covered it.
But thank you so much for that.

You can do it just by logic too. Take the first one, you got up to "This means that x ∈ A and x ∉ A or x ∉ B." If x is in A, then x can't NOT be in A too. So the second part of the 'or' must be true. So you get x ∈ A and x ∉ B. Which is A-B.

 

[simmonj7]

Thank you.
It's all solved now.
:)