Sets

**topsquark** · August 13th 2007, 09:06 AM

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

**Plato** · August 13th 2007, 09:11 AM

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 $\left( {A \cap C} \right) \subseteq B$ .

**shilz222** · August 13th 2007, 09:15 AM

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).

Your other direction looks good.

So now is this correct?

**topsquark** · August 13th 2007, 09:28 AM

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).

Your other direction looks good.

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

**shilz222** · August 13th 2007, 09:32 AM

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?

**shilz222** · August 13th 2007, 10:47 AM

Basically I used 2 different techniques to prove the directions.

**Plato** · August 13th 2007, 11:06 AM

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

**shilz222** · August 13th 2007, 11:19 AM

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

**shilz222** · August 13th 2007, 05:39 PM

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)?

**topsquark** · August 13th 2007, 06:18 PM

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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