# Thread: Vin diagrams? is there any other way..

1. ## Vin diagrams? is there any other way..

Let a & b be sets:

A u (B-A)U(C-A)=(B u C)-A

How do I solve?

The other ones I have are:
(A n B) E A
A E (A U B)
A-B E A
A n (B - A) 0

2. ## be careful

your first one is not even true.
A=(0,1)
B=(1,2)
C=(2,3)

Your LHS yields $(0,1)\cup (1,2) \cup (2,3)$
but the RHS is $(1,2) \cup (2,3)$

you should learn a few TeX codes so set theory makes sense.

\cup is union
\cap is intersection
\subset is subset
\supset goes the other way
\in is element of
\emptyset is the nullset

I have no idea what your E is.

Maybe type your question again correctly so we can help?

3. $$\cup$$ gives $\cup$ is union

$$\cap$$ gives $\cap$ is intersection

$$\subseteq$$ gives $\subseteq$ is subset

$$\in$$ gives $\in$ is element

$$\emptyset$$ gives $\emptyset$ is emptyset

4. Originally Posted by Gamma
I have no idea what your E is.
Unfortunately I have learned to speak some ASCII... E is apparently often equivalent to $\in;$ although, rarely, it is sometimes seen outside of its usual habitat as being equivalent to $\subset$ or $\subseteq,$ which I suspect is what's meant in this case.

Sorry for the mixed metaphor, there.

5. Originally Posted by orendacl
Let a & b be sets:

A u (B-A)U(C-A)=(B u C)-A

How do I solve?

The other ones I have are:
1)(A n B) E A
2)A E (A U B)
3)A-B E A
4)A n (B - A) 0

Okay so I will show you how to do these without vinn diagrams. I will take Alephzero's suggestion and take E as
$\subset$
1)Let $x\in A \cap B$ then x is by definition in both A and B, therefore x is in A, so $x\in A \Rightarrow A \cap B \subset A$ as desired

2) Let $x\in A$, then x is certainly in A or B since it is in A. Thus $x\in A \cup B \Rightarrow A \subset (A\cup B)$ as desired

3) Let $x\in A-B$. Then x is in A, but not in B. But this means $x\in A$ so $A-B \subset A$ as desired.

4) I think this is supposed to say $A \cap (B-A)=\emptyset$

So you show both containments.

Clearly $A \cap (B-A)\supset \emptyset$.

So let $x\in A \cap (B-A)$. Then $x\in A$ and $x\in B-A$. But this means x is in A, and x is in B, but not in A. So clearly no x can both be in A and not be in A, so this is precisely the empty set, so we have the desired $A \cap (B-A)=\emptyset$

QED.

It might be worth at least noting what happens if you cannot chose an x in the set, like for instance if you have the case where like $A-B=\emptyset$ or something, then you just have the containment being trivially true.

As for the first one, you are gonna have to get the correct statement before I can show you how to prove it because as it stands it is not true as my counterexample demonstrates.