Set Proof

1. Feb 15, 2014

scorpius1782

1. The problem statement, all variables and given/known data
The whole problem:
If $(B \cap C) \subseteq A$ then $[(B \cup C)-A] \subseteq [(B \cup C)-(B \cap C)]$

2. Relevant equations
$A-B=A\cap \bar{B}$
$N \subseteq M=\bar{M} \subseteq \bar{N}$

3. The attempt at a solution
Initially I'm having troubles with $[(B \cup C)-(B \cap C)]$. I know it should be all the elements that are not shared between B and C but I don't know how to write this out in set notation. I'm hoping that once this is clear I can begin to work the problem backwards so to speak.

For reference the way I've thought of things so far:
I don't really know set problems at all yet so I've begun by thinking of elements inside the sets. So, to start here $(B \cap C) \subseteq A$ means there is an object 'x' in all sets A, B, C.
$[(B \cup C)-A]$ Means there is an object 'y' that is in B or C but not in A. And finally: $[(B \cup C)-(B \cap C)]$ 'y' is in a set that is in B or C.

Last edited: Feb 15, 2014
2. Feb 16, 2014

Dick

No, $(B \cap C) \subseteq A$ means EVERY element x of A is also in B and C. If y is an element of $[(B \cup C)-A]$ then it's in B and C but not in A. And if y is an element of $[(B \cup C)-(B \cap C)]$ then it's in B and C but not in both.

3. Feb 17, 2014

scorpius1782

I don't understand this. Did you mean to say B or C but not in both?

4. Feb 17, 2014

Dick

Yes, that's what I meant. Sorry.

5. Feb 17, 2014

scorpius1782

Thats okay, I'm still working on this. I'm just having a real hard time thinking about it.

I find everything after 'then' to be confusing. It seems like we're saying mutually exclusive things about 'y'. It can't be in B and C but not in A. And in B or C but not in both. Unless reusing 'y' for the final statement was incorrect.

Edit: Reading my son a bedtime story so won't be back for a bit.

6. Feb 17, 2014

Dick

I might not be helping, I've found more typos in my previous quote. Let me try again.

No, $(B \cap C) \subseteq A$ means EVERY element x that's in both B and C is also in A. If y is an element of $[(B \cup C)-A]$ then it's in B or C but not in A. And if z is an element of $[(B \cup C)-(B \cap C)]$ then it's in B or C but not in both.

Last edited: Feb 17, 2014
7. Feb 17, 2014

scorpius1782

Okay, then y may, or may not, equal z from those statements alone. However, y must equal z if $y \subseteq z$. Then (x,y) must be in either B or C but not A. What does that have to do with (B∩C)⊆A? It seems to me that at a minimum A contains all the shared elements but it doesn't preclude the possibility that it is much larger and contains all the values they don't share as well. Unless that first statement is an attempt to define A somehow.

8. Feb 17, 2014

scorpius1782

Trying to start the proof but I'm not sure I'm allowed to do things this way:
Using $A-B=A\cap \bar{B}$
then $(B \cap C) \subseteq A$ is the same as $(B \cap C) \cup \bar{A}$
and $[(B \cup C)-(B \cap C)]$ becomes $[B \cup C \cap \bar{B \cap C}]$=$[B \cup C \cap \bar{B} \cup \bar{C}]$

So then: $(B \cap C) \cup \bar{A} \subseteq [(B \cup C) \cap (\bar{B} \cup \bar{C})]$

Then from $(B \cap C) \cup \bar{A}$ y' is in either B or C but not A
and from $[(B \cup C) \cap (\bar{B} \cup \bar{C})]$ x' is in B or C but not Both.
From this I should be able to assert that y' must be in x'
But how does this relate?: $(B \cap C) \subseteq A$turns into $A \subseteq (\bar{B} \cup \bar{C})$ from the other equation above.
Which says that w' is element in A that is not shared by B and C.

Last edited: Feb 17, 2014
9. Feb 18, 2014

Dick

That's getting off to a bad start right there. $A-B=A\cap \bar{B}$ defines a set. $(B \cap C) \subseteq A$ isn't a set. It's a statement that one set is a subset of the other.

Try this. Can you show that if $E \subseteq F$ then for any set D, $D-F \subseteq D-E$?

10. Feb 18, 2014

scorpius1782

I think so:
There exists an element x in D but not F. And since E is in F that means x is also not in E. Therefore the statement is true.
I'm not sure how to write mathematically a proof for this though.

11. Feb 18, 2014

scorpius1782

I made a mistake before. I should have written that [(B∪C)−A]=$(B∪C)\cap\bar{A}$

12. Feb 18, 2014

Dick

You understand it I think, but you aren't quite saying it right. If x is any element of D-F then x is in D but not in F. Since E is a subset of F then x is also not in E. So x is also an element of D-E. That proves D-F is a subset of D-E. Do you see how this applies to your problem?

13. Feb 18, 2014

scorpius1782

I understand that because of the first condition we know that A contains all the shared elements from B and C. Then when we have [(B∪C)−A] we're only left with elements that are in either B or C but not both which is really the same set as [(B∪C)−(B∩C)] is all elements that they do not share. Because of [(B∪C)−A] this element may or may not be smaller depending on whether A contains any elements other than the shared elements between B and C. But clearly [(B∪C)−A]⊆[(B∪C)−(B∩C)] is true.

14. Feb 18, 2014

Dick

Except for a few turns of phrase, like "this element may or may not be smaller" which are a little awkward, I think that's a perfectly fine proof.

15. Feb 18, 2014

scorpius1782

Thank you, this helped me a lot. The class ramped up in difficulty for me rather quickly when they introduced sets. I'm a lot better at reading and understanding them with your help! Sorry for being a little dense, I'm an unbearably slow learner and this isn't the sort of thing I'm good at to begin with. Thanks, again.