# Thread: set question

1. ## set question

A,B,C are sets with A union B = B intersection C. The question asks what do these operations imply about A and C. I let x be an element satisfying the equation and found that any element in A must be in B and any element in B must be in C. Does this make sense and is there anything else I can say about A and C?

2. Originally Posted by PvtBillPilgrim
A,B,C are sets with A union B = B intersection C. The question asks what do these operations imply about A and C. I let x be an element satisfying the equation and found that any element in A must be in B and any element in B must be in C. Does this make sense and is there anything else I can say about A and C?
Did you try drawing a picture?
I get that $A\subseteq B\subseteq C$.
But we need to be careful, pictures can decieve us.

See if you can prove the above statement.

3. Hello,
Originally Posted by PvtBillPilgrim
A,B,C are sets with A union B = B intersection C. The question asks what do these operations imply about A and C. I let x be an element satisfying the equation and found that any element in A must be in B and any element in B must be in C. Does this make sense and is there anything else I can say about A and C?
You have proved that any element of A will be in C. This means that $A \subseteq C$

And this is all you can say about A and C.

4. Nice question.

Let $x \in A$.

Then $x \in A \cup B$ because $A \subseteq A \cup B$ (not sure whether you'd be expected to prove the latter).

So as $A \cup B = B \cap C$, we have $x \in B \cap C$.

But because $B \cap C \subseteq C$ (again something you may need to prove separately), it follows that $x \in C$.

Thus $x \in A \rightarrow x \in C$ and thus $A \subseteq C$.

5. Thanks for the replies.

Can you say A is a subset of B and B is a subset of C (therefore A is a subset of C)? Or can you say nothing about B?

6. Yes, if set A is a subset of set B and set B is the subset of set C, then set A is indeed the subset of set C.

Well, to make things more interesting than usual, it is not wrong to say that set A is an element of the set B.

7. Yes it is.

$A \subseteq B$ does NOT mean that $A \in B$. Sorry, but that is completely wrong.