# Math Help - Sets

1. Originally Posted by shilz222
I was asked to prove: $(A-B) \cap C = \emptyset \Leftrightarrow A \cap C \subseteq B$

So there are two directions.

In one direction ( $\Leftarrow$) we are given: $A \cap C \subseteq B$ and $(A-B) \cap C \neq \emptyset$ for proof by contradiction.

In the other direction ( $\Rightarrow$) we are given: $(A-B) \cap C = \emptyset$ and then do what you said in your earlier post.
I apologize if I am misinterpreting something. All of my comments are directed toward your $\implies$ proof by contradition. I have not addressed any comments to the proof in the other direction.

-Dan

2. Originally Posted by shilz222
I was asked to prove: $(A-B) \cap C = \emptyset \Leftrightarrow A \cap C \subseteq B$ So there are two directions.
YES, YES, YES! That is correct.

Originally Posted by shilz222
In one direction ( $\Leftarrow$) we are given: $A \cap C \subseteq B$ and $(A-B) \cap C \neq \emptyset$
NO, NO, NO This not correct.

One direction: Given $\left( {A \cap C} \right) \subseteq B$ prove that $\left( {A - B} \right) \cap C = \emptyset$.

The other direction: Given $\left( {A - B} \right) \cap C = \emptyset$ prove that $\left( {A \cap C} \right) \subseteq B$.

3. Why is it not correct?

We assume that $(A-B) \cap C \neq \emptyset$ for proof by contradiction.

One direction: Given that $(A \cap C) \subseteq B$ and $(A-B) \cap C \neq \emptyset$ use proof by contradiction to show that $(A-B) \cap C = \emptyset$ (my thinking).

So now is this correct?

4. Originally Posted by shilz222
Why is it not correct?

We assume that $(A-B) \cap C \neq \emptyset$ for proof by contradiction.

One direction: Given that $(A \cap C) \subseteq B$ and $(A-B) \cap C \neq \emptyset$ use proof by contradiction to show that $(A-B) \cap C = \emptyset$ (my thinking).

Okay, so to be clear. You agree that your proof of the $\implies$ direction is incorrect? And now you are working on the $\leftarrow$ proof. And you have decided to try to prove it by contradiction. Is this all correct?

-Dan

5. I asked help for the $\Longrightarrow$ direction.

Originally Posted by Plato
This is a very important concept: $X \cap Y = \emptyset \text{ if an only if }X \subseteq Y'$.
That is: two sets are disjoint iff each is a subset of the complement of the other.

$\left( {A - B} \right) \cap C = \left( {A \cap B'} \right) \cap C = \left( {A \cap C} \right) \cap B'$, thus if $\left( {A - B} \right) \cap C = \emptyset$ then $\left( {A \cap C} \right) \subseteq B$.

Now you do the other way.

Trying to teach oneself this material can be very dangerous if one wants to do higher mathematics.
I agree with this method that Plato has done.

For the $\Longleftarrow$ direction I already proved it by contradiction.

Suppose that $(A-B) \cap C \neq \emptyset$. If $x \in A \cap C$ then $x \in A$ and $x \in C \Rightarrow x \in B$. Then $\exists x \in (A-B) \cap C$ such that $x \in A$ and $x \in C$ but $x \not \in B$. But this is a contradiction and so $(A-B) \cap C = \emptyset$.
So is this not correct?

6. Basically I used 2 different techniques to prove the directions.

7. Here it is.
Given $\left( {A \cap C} \right) \subseteq B$ that means $\left( {A \cap C} \right) \cap B' = \emptyset$.
But $\left( {A \cap C} \right) \cap B' = \left( {A \cap B'} \right) \cap C = \left( {A - B} \right) \cap C = \emptyset$

8. I already know that is correct. But why isnt the proof by contradiction correct?

9. I guess what I am asking is that is it because my proof is awkward that its not correct (i.e. using direct and proof by contradiction)?

10. Originally Posted by shilz222
I guess what I am asking is that is it because my proof is awkward that its not correct (i.e. using direct and proof by contradiction)?
Simply because the direct proof is shorter. From what I have seen (I may be wrong) proof by contradiction is usually only used if a direct proof is either lacking or obscenely difficult.

-Dan

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