Sets

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• Aug 12th 2007, 07:09 PM
shilz222
Sets
Prove that $(A-B) \cap C = \emptyset \Leftrightarrow A \cap C \subseteq B$.

Proof: 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 $(A-B) \cap C = \emptyset$. How would I go about proving that $A \cap C \subseteq B$?

I've tried writing $(A-B) \cap C = \emptyset$ as $(A \cap B') \cap C = \emptyset$ and then $C \subseteq (A \cap B')$ or $C \cup (A \cap B') = A \cap B'$. But from here I get stuck.

Any help is appreciated.

Sorry for sounding like an idiot but I am learning the material by myself for fun and have no one to guide me.

Thanks
• Aug 13th 2007, 03:51 AM
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.
• Aug 13th 2007, 07:18 AM
shilz222
thanks

but is my contradiction method all right? why would it be 'dangerous'?
• Aug 13th 2007, 07:52 AM
Plato
Quote:

Originally Posted by shilz222
but is my contradiction method all right? why would it be 'dangerous'?

Actually NO! Those steps make no sense at all!

Quote:

Originally Posted by shilz222
Proof: Suppose that $(A-B) \cap C \neq \emptyset$. If $x \in A \cap C$ then $x \in A$ and $x \in C \Rightarrow x \in B$.

That is a totally invalid conclusion! Nothing follows from $x \in A \cap C$ that has anything to do with $x \in B$. If I were grading that proof, then I would have stopped at that point. I would assign 0 points, because that mistake shows real confusion about this kind a proof. There is no reason to use a “pick-a-point” proof on a straightforward question about set manipulation.

Now you asked why is it dangerous? Well that is just one example. Unless you learn how to do these various kinds of proofs well, you can forget doing higher mathematics.
• Aug 13th 2007, 07:57 AM
topsquark
Quote:

Originally Posted by shilz222
Suppose that $(A-B) \cap C \neq \emptyset$. If $x \in A \cap C$ then $x \in A$ and $x \in C \Rightarrow x \in B$.

Let A = {1, 2, 3, 4, 5}, B = {2, 3, 4} and C = {4, 5, 6}.

$(A - B) \cap C = \{ 1, 5 \} \cap \{ 4, 5, 6 \} = \{ 1, 4, 5, 6 \} \neq \emptyset$

Let $x \in A \cap C = \{ 1, 2, 3, 4, 5, 6 \}$. Specifically let $x = 5$. We have that $x \in A \cap C$ and we have $x \in A$ and $x \in C$, but $5 \not \in B$.

-Dan
• Aug 13th 2007, 08:09 AM
topsquark
Plato I study Mathematics on my own all the time. I agree that it is a "risky" proposition because of things like shilz222's example. It is very easy to get a concept wrong, or misapply it, and start drawing the wrong conclusions. Drawing from my own experience even if I get the methods right I find I often don't fully appreciate or understand the fine details of the field.

However that doesn't mean that self-study is a bad thing. When I am confused I ask questions. I check other texts and post questions on fora (like this one) to see that the information that I think is correct is truly correct. I work the problems at the ends of the sections and check my answers. I work out the skipped steps in the middle of the sections. It can take me weeks to get through a chapter. (Months in some cases!) But I do learn the material, and even though I don't necessarily learn the material fully I learn what I am capable of learning and know just what material I have mastered. I stretch myself in this way so that I can learn even more.

shilz222 is doing the right thing in asking if his/her proof is correct. Doubtless (s)he will make mistakes in the future, but isn't that what learning is about?

-Dan
• Aug 13th 2007, 08:10 AM
shilz222
$A \cap C \subseteq B$ right? So doesn't it follow that $x \in A \cap C \Rightarrow x \in B$ from the very definition of a subset?

We are given: $A \cap C \subseteq B$ and $(A-B) \cap C \neq \emptyset$.

Our goal is a contradiction in the $\Leftarrow$ direction. We find that $(A-B) \cap C = \emptyset$.

So... is it still wrong?
• Aug 13th 2007, 08:14 AM
topsquark
Quote:

Originally Posted by shilz222
$A \cap C \subseteq B$ right? So doesn't it follow that $x \in A \cap C \Rightarrow x \in B$ from the very definition of a subset?

We are given: $A \cap C \subseteq B$ and $(A-B) \cap C = \emptyset$.

-Dan
• Aug 13th 2007, 08:36 AM
Plato
Quote:

Originally Posted by topsquark
Plato I study Mathematics on my own all the time.
However that doesn't mean that self-study is a bad thing.

Nor did I ever say that “self-study is a bad thing”.
Dan, at your level of mathematical maturity it is a very good thing.
However, most mathematics departments have a foundations course no later than the first semester of the sophomore year where students learn this material. This material is not required for any mathematics taken by non-mathematics majors. So one could do through multivariable calculus without having had such a course. However, doing any course is geometry, topology, abstract algebra, or even probability would be out of the question without a through grounding in set-theoretical proofs. Now I do think that for most students self-training in this particular area is not a good idea. Being in a classroom where these proofs are done is important. My comments were prompted by a question in a different posting by this same student as to whether real analysis or abstract algebra would be the next logical step.
• Aug 13th 2007, 08:38 AM
topsquark
Quote:

Originally Posted by Plato
Nor did I ever say that “self-study is a bad thing”.
Dan, at your level of mathematical maturity it is a very good thing.
However, most mathematics departments have a foundations course no later than the first semester of the sophomore year where students learn this material. This material is not required for any mathematics taken by non-mathematics majors. So one could do through multivariable calculus without having had such a course. However, doing any course is geometry, topology, abstract algebra, or even probability would be out of the question without a through grounding in set-theoretical proofs. Now I do think that for most students self-training in this particular area is not a good idea. Being in a classroom and where these proofs are done is important. My comments were prompted by a question in a different posting by this same student as to whether real analysis or abstract algebra would be the next logical step.

Fair enough. :)

-Dan
• Aug 13th 2007, 08:43 AM
shilz222
Topsquark I understand what you have done, but you constructed those sets right?

But if we are given $A \cap C \subseteq B$, then we cannot say that $x \in A \cap C \Rightarrow x \in B$? I am not saying that it is always true as you pointed out. But if we are given those 2 things, then our sets have to obey those two things right? You can find a counterexample but then it doesn't agree with the givens.

Thats why we have the words "Suppose that so and so is true" right?
• Aug 13th 2007, 08:49 AM
Plato
Quote:

Originally Posted by shilz222
We are given: $A \cap C \subseteq B$ and $(A-B) \cap C \neq \emptyset$.

No, that was not given! You were asked to prove the following:
$A \cap C \subseteq B \text{ if and only if } (A-B) \cap C \neq \emptyset$
• Aug 13th 2007, 08:54 AM
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.
• Aug 13th 2007, 09:02 AM
topsquark
Quote:

Originally Posted by topsquark
Let A = {1, 2, 3, 4, 5}, B = {2, 3, 4} and C = {4, 5, 6}.

$(A - B) \cap C = \{ 1, 5 \} \cap \{ 4, 5, 6 \} = \{ 1, 4, 5, 6 \} \neq \emptyset$

Let $x \in A \cap C = \{ 1, 2, 3, 4, 5, 6 \}$. Specifically let $x = 5$. We have that $x \in A \cap C$ and we have $x \in A$ and $x \in C$, but $5 \not \in B$.

-Dan

Quote:

Originally Posted by shilz222
Topsquark I understand what you have done, but you constructed those sets right?

But if we are given $A \cap C \subseteq B$, then we cannot say that $x \in A \cap C \Rightarrow x \in B$? I am not saying that it is always true as you pointed out. But if we are given those 2 things, then our sets have to obey those two things right? You can find a counterexample but then it doesn't agree with the givens.

Thats why we have the words "Suppose that so and so is true" right?

(chuckles) Of course I constructed those sets in a way that I would be able to refute your statement. It would have been a lousy counterexample if I hadn't! :p

If it were given that $A \cap C \subseteq B$ then I believe your inference must be correct. But the conditions you stated at the outset of the problem are that
1) $(A - B) \cap C \neq \emptyset$
2) $x \in A \cap C$

Nowhere in there is the requirement that $A \cap C \subseteq B$. (This is a good thing. Otherwise you would have over-specified your proof into uselessness.)

-Dan
• Aug 13th 2007, 09:04 AM
shilz222
$A \cap C \subseteq B$ is given in the $\Leftarrow$ direction. (1) is given in the $\Leftarrow$ direction because I wanted to do a proof by contradiction.

In an "if and only if" statement we have to prove both directions.
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