# Thread: Real Analysis - Sets and Proofs

1. ## Real Analysis - Sets and Proofs

I have 2 problems that are annoying the heck out of me...

#1. If A is contained in B prove that (C\B) is contained in (C\A) Either prove the converse is true or give a counterexample. Don't even know where to start. It is only the first week of the course, so I just need pretty basic help.

#2. Under what conditions does A\(A\B)=B. Again, how do I start this?

2. I point you towards a website I've been contributing to for some few weeks now ... here's the page containing some proofs that might help.

Subset Equivalences - Proofwiki.org

explore and enjoy.

3. Oi, miss Green !
Originally Posted by kathrynmath
I have 2 problems that are annoying the heck out of me...

#1. If A is contained in B prove that (C\B) is contained in (C\A) Either prove the converse is true or give a counterexample. Don't even know where to start. It is only the first week of the course, so I just need pretty basic help.
To prove a set P is contained in another one, Q, prove :
if x is in P, then x is in Q.

If you want to know whether it's true or false, draw a diagram. Just to give you the correct direction !

Note, $\displaystyle A \backslash B=A \cap \overline{B}$

4. Originally Posted by Moo
Oi, miss Green !

To prove a set P is contained in another one, Q, prove :
if x is in P, then x is in Q.
Does this involve assuming P=Q?

5. Originally Posted by Moo
Oi, miss Green !

To prove a set P is contained in another one, Q, prove :
if x is in P, then x is in Q.

If you want to know whether it's true or false, draw a diagram. Just to give you the correct direction !

Note, $\displaystyle A \backslash B=A \cap \overline{B}$
So, I figured out #1 now.
For #2, is it something with A equaling B?

6. Not quite.

$\displaystyle A - (A - B) \Longrightarrow A \cap B$ (sorry, I use $\displaystyle -$ for set difference)

The above may need some thought, it's not trivial.

But that means $\displaystyle A \cap B = A$ which is one of the classic subset equivalences:

$\displaystyle A \cap B = A \iff A \subseteq B$

That's the direction you want to go in, but proving them needs thought.

7. Hmmm, this question is still frustrating me.

8. Is B a null set?
Because say A={1, 2, 3, 4, 5} and B={1, 2,3}.
We get:
A\({4,5}
={1,2, 3}

If B is a null set, and A is the same, we get:
A\{1, 2, 3, 4, 5}.
={}
Therefore, we have B.

Also, what about A=B? A = {1, 2 , 3, 4, 5} and B={1,2, 3, 4, 5}
A\({}
{1,2,3,4,5}
Therefore, we get B.

So, these 2 cases appear to work. these are my thoughts so far.

9. Sorry, I thought I just told you: A has to be a subset of B.

Note that if $\displaystyle A=B$ then it is also true that $\displaystyle A \subseteq B$ so yes, the fact that $\displaystyle A=B$ does make your equation work, it's just not the full answer.

10. So, B has to be contained in A? For example A={1,2,3,4,5}, B={1,2,3}
A\(A\B)
A\{4,5}
{1,2,3,4,5}\{4,5}
{1,2,3}
B

That works....

11. Did you actually check out the link I gave you a few postings back in the thread?

12. Originally Posted by kathrynmath
I have 2 problems that are annoying the heck out of me...

#1. If A is contained in B prove that (C\B) is contained in (C\A) Either prove the converse is true or give a counterexample. Don't even know where to start. It is only the first week of the course, so I just need pretty basic help.

#2. Under what conditions does A\(A\B)=B. Again, how do I start this?
I just looked at #1.
Clearly, we have the following identities

1. if $\displaystyle A\subset B$, then $\displaystyle B^{c}\subset A^{c}$, where $\displaystyle ~^{c}$ stands for the compliment of the sets.
2. if $\displaystyle A\subset B$, then $\displaystyle A\cap C\subset B\cap C$
3. $\displaystyle A\backslash B=A\cap B^{c}$

Hence, from the properties above, we get $\displaystyle A\subset B$ implies $\displaystyle B^{c}\subset A^{c}$. Intersection both sides with $\displaystyle C$, we get $\displaystyle C\cap B^{c}\subset C\cap A^{c}$, which is equivalent to $\displaystyle C\backslash B\subset C\backslash A$.

I hope you may easily find #2 by using 3. for twice and thinking on the resultant situation.

Note. The easiest way to work on such problems may be using Venn diagrams.