# Thread: Set proof

1. ## Set proof

Problem: Prove that $\displaystyle (A \cup C) - B = (A-B) \cup C \Leftrightarrow B \cap C = \emptyset$.

Thoughts/Work: I tried proof by contradiction. Suppose that $\displaystyle B \cap C \neq \emptyset$. Then $\displaystyle x \in B$ and $\displaystyle x \in C$. So $\displaystyle x \in (A \cup C) - B \Rightarrow x \in A$ or $\displaystyle x \in C$ and $\displaystyle x \not \in B$. Then $\displaystyle x \in A$ and $\displaystyle x \not \in B$ or $\displaystyle x \in C$. But this contradicts the fact that $\displaystyle B \cap C \neq \emptyset$. So $\displaystyle B \cap C = \emptyset$ for equality between the two sets. Suppose that $\displaystyle (A \cup C) - B \neq (A - B) \cup C$. Then $\displaystyle x \in A$ or $\displaystyle x \in C$ and $\displaystyle x \not \in B \Rightarrow x \not \in A, x \in B$ or $\displaystyle x \not \in C$. So $\displaystyle x \in B$ and $\displaystyle x \not \in C$ or $\displaystyle x \in C$ and $\displaystyle x \not \in B$. This is a contradiction and so $\displaystyle (A \cup C) - B = (A - B) \cup C$. QED

I am not sure if I made this too complicated. Does this look ok?

Thanks

2. Originally Posted by shilz222
Problem: Prove that $\displaystyle (A \cup C) - B = (A-B) \cup C \Leftrightarrow B \cap C = \emptyset$.
I am not sure if I made this too complicated. Does this look ok?
Frankly, I do not follow what you have done. I am not sure that you quite understand the process we some times call “pick a point proofs”. Below I have given two types of proof.

First: Suppose that $\displaystyle \left( {A \cup C} \right) - B = \left( {A - B} \right) \cup C$, what if $\displaystyle \exists z \in \left( {B \cap C} \right)$ then because $\displaystyle z \in B$ we can conclude that $\displaystyle z \notin \left( {A \cup C} \right) - B$. But by the equality that means that $\displaystyle z \notin \left( {A - B} \right) \cup C$ and that contradicts the fact that $\displaystyle z \in C$; thus $\displaystyle \left( {B \cap C} \right) = \emptyset$.

Second: Suppose that $\displaystyle \left( {B \cap C} \right) = \emptyset$.
This means that $\displaystyle C \subseteq B'\quad \Rightarrow \quad C \cup B' = B'$.
$\displaystyle \left( {A - B} \right) \cup C = \left( {A \cap B'} \right) \cup C$
$\displaystyle = \left( {A \cup C} \right) \cap \left( {B' \cup C} \right)$
$\displaystyle = \left( {A \cup C} \right) \cap \left( {B'} \right)$
$\displaystyle = \left( {A \cup C} \right) - B$.
This means that $\displaystyle \left( {A \cup C} \right) - B = \left( {A - B} \right) \cup C$

3. Originally Posted by shilz222
Problem: Prove that $\displaystyle (A \cup C) - B = (A-B) \cup C \Leftrightarrow B \cap C = \emptyset$.

Thoughts/Work: I tried proof by contradiction. Suppose that $\displaystyle B \cap C \neq \emptyset$. Then $\displaystyle x \in B$ and $\displaystyle x \in C$.
If I am not mistaken (and I easily may be) what you are saying in the last sentence here is that x must belong either to B or C and this is not true. Since the union of B and C is not empty then there does exist at least one x belonging to both B and C, but an arbitrary x does not have to fulfill this property.

-Dan

4. Yes TopsQuark that is what I thought. I don't see why my method is incorrect. I arrive at contradictions in both directions.

5. Originally Posted by shilz222
Yes TopsQuark that is what I thought. I don't see why my method is incorrect. I arrive at contradictions in both directions.
My thought is that you will have to do two proofs:
1) x belongs to the intersection of B and C
2) x does not belong to the intersection of B and C

You have already completed the first step, now you will need to do the second.

Or you can be satisfied with Plato's more elegant proof that doesn't share this difficulty.

-Dan

6. Well I assumed that $\displaystyle B \cap C \neq \emptyset$ for the '$\displaystyle \Rightarrow$' direction and arrived at a contradiction. So in that direction, I have shown that $\displaystyle B \cap C = \emptyset$. So I don't think (2) is necessary. In the '$\displaystyle \Leftarrow$' direction I arrived at a contradiction by assuming that $\displaystyle (A \cup C)-B \neq (A-B) \cup C$ and showed that there must be equality. But I agree Plato's method is more clear.