Can Set Theory Identity (A-B) ∪ (B-C) = (A ∪ B) - (B ∩ C) Be Proven?

[SpatialVacancy]

Help me construct a proof!

Consider the following set property: For all sets $$A$$, $$B$$, and $$C$$, $$(A-B) \cup (B-C) = (A \cup B) - (B \cap C)$$.

a) Use an element argument to derive this property.

b) Use an algebraic argument to derive this property.


Ok, for part (a), I know that I need to show that:
$$(A-B) \cup (B-C) \subseteq (A \cup B) - (B \cap C)$$, and
$$(A \cup B) - (B \cap C) \subseteq (A-B) \cup (B-C)$$ (right?).

To do this, i need to show that $$\forall x$$, if $$x \ \epsilon \ (A-B) \cup (B-C)$$, and $$x \ \epsilon \ (A \cup B) - (B \cap C)$$. From here I do not know where to go.

For part (b), any help you can give me I would appreciate. I have written several pages of calculations and have yet to come up with anything.

Please help! This assignment is due 2/25 at 1:30 EST.

Thanks

 

[MathStudent]

hint: you need to show that if x is an element of the LHS then x is an element of the RHS, for all x. to start, break it down using the definitions of Union and "-"

for example:

if $$x \ \epsilon \ (A-B) \cup (B-C)$$

then $$x \ \epsilon \ (A-B) \ \ \ \ or \ \ \ \ x \ \epsilon \ (B-C)$$

you will end up with special cases of what x can be and you just need to show that for each case, x is also a member of the RHS. You will also need to do this same process with the RHS showing that every element of the RHS is also an element of the LHS.

for part b, I think by "algebraicly" they just want you to use the distributive properties of union etc... on both sides and simplify the set as much as possible to show that they are really just different ways of writing the same set.

 

[thepatientmental]

for reaching out for help with constructing a proof. It can definitely be a challenging task, so let's break it down step by step.

a) To show that (A-B) \cup (B-C) \subseteq (A \cup B) - (B \cap C), we need to show that for any element x, if x is in the left-hand side set, then x is also in the right-hand side set.

So let's start with an arbitrary element x in (A-B) \cup (B-C). This means that x is either in (A-B) or in (B-C).

Case 1: x \ \epsilon \ (A-B)
This means that x is in A, but not in B. So x is definitely in A \cup B. Now, since x is not in B, it cannot be in the intersection B \cap C. Therefore, x is also in (A \cup B) - (B \cap C).

Case 2: x \ \epsilon \ (B-C)
Similarly, this means that x is in B, but not in C. So again, x is in A \cup B. And since x is not in C, it is not in the intersection B \cap C. Therefore, x is also in (A \cup B) - (B \cap C).

Since we have shown that for any element x in (A-B) \cup (B-C), x is also in (A \cup B) - (B \cap C), we have shown that the left-hand side set is a subset of the right-hand side set.

To show the other direction, we need to show that (A \cup B) - (B \cap C) \subseteq (A-B) \cup (B-C). This means that for any element x, if x is in the right-hand side set, then x is also in the left-hand side set.

So let's start with an arbitrary element x in (A \cup B) - (B \cap C). This means that x is in A or B, but not both, and it is not in the intersection B \cap C.

Case 1: x \ \epsilon \ A
Since x is in A, it is definitely not in B. Therefore, x is also in (A-B).

Case 2: x \ \