Set Theory Proof

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  • #1
cmajor47
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Homework Statement


Prove that for all sets A, B, and C, (A-C) [tex]\cap[/tex] (B-C) [tex]\cap[/tex] (A-B) = ∅


Homework Equations





The Attempt at a Solution


Proof: Suppose A, B, and C are sets
Let x [tex]\in[/tex] (A-C) [tex]\cap[/tex] (B-C) [tex]\cap[/tex] (A-B)
Since x [tex]\in[/tex] (A-C), by definition of difference, x [tex]\in[/tex] A and x [tex]\notin[/tex] C
Since x [tex]\in[/tex] (B-C), x [tex]\in[/tex] B and x [tex]\notin[/tex] C
Since x [tex]\in[/tex] (A-B), x [tex]\in[/tex] A and x [tex]\notin[/tex] B
Then by definition of intersection, if x [tex]\in[/tex] A then x [tex]\notin[/tex] C and x [tex]\notin[/tex] B
Also, if x [tex]\in[/tex] B then x [tex]\notin[/tex] C
Therefore there is no intersection of sets A, B, and C
Therefore, the intersection of (A-C) [tex]\cap[/tex] (B-C) [tex]\cap[/tex] (A-B) = ∅

Is this proof correct, I feel like I am missing something?
 

Answers and Replies

  • #2
Dick
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Homework Statement


Prove that for all sets A, B, and C, (A-C) [tex]\cap[/tex] (B-C) [tex]\cap[/tex] (A-B) = ∅


Homework Equations





The Attempt at a Solution


Proof: Suppose A, B, and C are sets
Let x [tex]\in[/tex] (A-C) [tex]\cap[/tex] (B-C) [tex]\cap[/tex] (A-B)
Since x [tex]\in[/tex] (A-C), by definition of difference, x [tex]\in[/tex] A and x [tex]\notin[/tex] C
Since x [tex]\in[/tex] (B-C), x [tex]\in[/tex] B and x [tex]\notin[/tex] C
Since x [tex]\in[/tex] (A-B), x [tex]\in[/tex] A and x [tex]\notin[/tex] B

You are doing fine up to here. Do you see anything contradictory in those conditions you have on x?
 
  • #3
cmajor47
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It states that x [tex]\in[/tex] B and x [tex]\notin[/tex] B, which isn't possible.
Do I just say that since this is a contradiction, the intersection is the null set?
 
  • #4
Dick
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Sure. There is no x that can satisfy those two conditions.
 

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