Set proof

**shilz222** · August 10th 2007, 05:09 PM

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

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

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

Thanks

**Plato** · August 11th 2007, 06:41 AM

Originally Posted by shilz222

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

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

**topsquark** · August 11th 2007, 07:16 AM

Originally Posted by shilz222

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

Thoughts/Work: I tried proof by contradiction. Suppose that $B \cap C \neq \emptyset$ . Then $x \in B$ and $x \in C$ .

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

**shilz222** · August 11th 2007, 07:29 PM

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

**topsquark** · August 11th 2007, 08:10 PM

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

**shilz222** · August 11th 2007, 08:29 PM

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

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