# Proof subset?

#### leilei

Proof subset????

Given three sets A, B, and C, set X = (A-B) U (B-C) U (C-A) and
Y = (A∩B∩C) complement C. Prove that X is subset of Y. Is Y necessarily a subset of X? If yes, prove it. If no, why???
---When I draw the two venn diagrams X and Y, they are the same, but I don't know how to prove it...

Can someone help me out here...
Thanks in advance!

Related Linear and Abstract Algebra News on Phys.org

#### CompuChip

Science Advisor
Homework Helper
The usual proof for such a statement is: Let x be an element from X and try to prove that it is also an element of Y. So if x is in X, you know that it is in A but not in B, and/or it is in B but not in C, and/or it is in C but not in A. You want to show that it must be in (A∩B∩C)C. If it is in (A∩B∩C) then it would be in A and B and C, so it being in the complement means it is at least not in A or not in B or not in C. So you could suppose it is both in A and B and show that it cannot be in C.

So let me write this out in the right order:
Let $x \in X$. Suppose that $x \in A, x \in B$. Then definitely, $x \not\in A - B, x \not\in C - A$, because if it is in A it cannot be in any set from which we remove all elements of A (and similarly for B). But it must be in one or more of (A-B), (B-C) and (C-A), so it must be in (B - C). That is, x is in B (which we knew) but not in C. So if x is not in C, it cannot be in the intersection of C with whatever set you make up. In particular, it is not in $A \cap B \cap C$. Therefore, it must be in the complement of that set, which is called Y.

Now try to do the same reasoning for $Y \subset X$. You have already shown by your Venn diagram that if x lies in Y, it must lie in X. So try to prove it in the same way as I just did.

#### leilei

Thanks alot !!!!!

#### HallsofIvy

Science Advisor
Homework Helper
There is, however, a technical problem with "Let $x \in X$"- the proof collapses is X is empty. Far better to start "IF $x \in X$". That way, if X is empty, the hypothesis is false and the theorem is trivially true.

#### CompuChip

Science Advisor
Homework Helper
You are right, obviously the empty set is a subset of any set S (vacuously, all its elements are also in S).

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving