How Can You Prove Set Equality for (A-B) ∩ C and (A ∩ C) - B?

[chica911]

1. Let A , B and C be sets. Prove that (A-B) ∩ C = (A ∩ C)- B = (A ∩ C) – (B ∩ C).


3. Proof:
1st part: Let A, B and C be sets where (A-B) ∩ C. Let X be a particular, but arbitrary element of C. Since C and (A-B) ∩, X € (A-B) and X € C. Therefore, X € A but X ∉ B. Since X is an element of A and C, A ∩ C and since X ∉ B, (A ∩ C)-B. Therefore (A-B) ∩ C=(A ∩ C)-B.
2nd part: Let A, B and C be sets where (A ∩ C)- B.
3rd part: Suppose (A-B) ∩ C=(A ∩ C)-B and (A ∩ C)- B = (A ∩ C) – (B ∩ C) By the transitive property (A-B) ∩ C=(A ∩ C) – (B ∩ C) and therefore (A-B) ∩ C = (A ∩ C)- B = (A ∩ C) – (B ∩ C).

The 2nd part is what I need help with. I am not sure how to prove (A ∩ C)- B = (A ∩ C) – (B ∩ C).
please help

 

[SteveL27]

The 2nd part is what I need help with. I am not sure how to prove (A ∩ C)- B = (A ∩ C) – (B ∩ C).
please help

The way to attack these is first, to break it into two parts, and second, for each part just read it left to right in English. Let me show you what I mean.

You have to prove two things:

(=>) If x is in (A ∩ C)- B then x is in (A ∩ C) – (B ∩ C) and

(<=) If x is in (A ∩ C) – (B ∩ C) then x is in (A ∩ C) - B.

Ok let's do the => direction. Suppose x is in (A ∩ C) - B. What does that mean in English? x is in A, and x is in C, but x is not in B.

Now on the right side, we must show that (A ∩ C) – (B ∩ C). We are given that x is in A, and we are given that x is in C; so x is in (A ∩ C). But we're given that x is not in B, so x is not in (B ∩ C). And that's what we had to show: that x is in (A ∩ C) – (B ∩ C). So the => is done.

Now you do the <= direction.

 

[Deveno]

1. Let A , B and C be sets. Prove that (A-B) ∩ C = (A ∩ C)- B = (A ∩ C) – (B ∩ C).


3. Proof:
1st part: Let A, B and C be sets where (A-B) ∩ C. Let X be a particular, but arbitrary element of C. Since C and (A-B) ∩, X € (A-B) and X € C. Therefore, X € A but X ∉ B. Since X is an element of A and C, A ∩ C and since X ∉ B, (A ∩ C)-B. Therefore (A-B) ∩ C=(A ∩ C)-B.

if you pick an arbitrary element of C, you cannot claim it must lie in A-B as well, just because A-B and C intersect. horse first, THEN cart. choose x arbitrary in (A-B)∩C (and keep in mind there is no guarantee this set is not empty, because it might be).

what you have done above could be turned into an argument that (A-B)∩C is contained in (A∩C) - B, but the reverse containment needs to be shown as well.

2nd part: Let A, B and C be sets where (A ∩ C)- B.
3rd part: Suppose (A-B) ∩ C=(A ∩ C)-B and (A ∩ C)- B = (A ∩ C) – (B ∩ C) By the transitive property (A-B) ∩ C=(A ∩ C) – (B ∩ C) and therefore (A-B) ∩ C = (A ∩ C)- B = (A ∩ C) – (B ∩ C).

The 2nd part is what I need help with. I am not sure how to prove (A ∩ C)- B = (A ∩ C) – (B ∩ C).
please help

you are supposing what you wish to prove is true. that's a no-no.

what you want to do, is show (A∩C) - B is contained in (A∩C) - (B∩C) and vice versa.

so you suppose x is (A∩C) - B. this then, means x is in both A and C (since x is in A∩C), but not in B. well, if x is not in B, it cannot lie in any subset of B. so...

the other direction will be harder, because you are taking a "bigger bite" out of A∩C when you remove all the elements of B, then just removing all the elements of B∩C. so you need to show that A∩C doesn't contain any elements of B, EXCEPT those elements of B that already intersect C.