# Homework Help: Abstract Algebra set theory

1. Feb 23, 2012

### iHeartof12

Let A, B and C be sets.
Prove that if A$\subseteq$B$\cup$C and A$\cap$B=∅, then A$\subseteq$C.

My attempted solution:
Assume A$\subseteq$B$\cup$C and A$\cap$B=∅.
Then $\vee$x (x$\in$A$\rightarrow$x$\in$B$\cup$x$\in$c).

I'm not sure where to start and how to prove this. Any help would be greatly appreciated. Thank you.

2. Feb 23, 2012

### Staff: Mentor

....disregard earlier post...

3. Feb 23, 2012

### jbunniii

Suppose $x \in A$. The goal is to show that this implies $x \in C$.

Since $x \in A$ and $A \subset B \cup C$, it follows that $x \in B$ or $x \in C$. Can you exclude one of these possibilities?

4. Feb 23, 2012

### iHeartof12

Since A$\bigcap$B=∅, x$\in$A or x$\in$B.
Thus x$\in$A, x$\notin$B and x$\in$C.
Therefor A$\subseteq$C.

Is that a good way to show how to exclude the possibility of x$\in$B?

5. Feb 23, 2012

### jbunniii

You have the right idea, but the wording is a little unclear. The following is not true: "Since A$\bigcap$B=∅, x$\in$A or x$\in$B."

Suppose $x \in A$. The goal is to show that this implies $x \in C$.

Since $x \in A$ and $A \subset B \cup C$, it follows that $x \in B$ or $x \in C$. However, $x$ cannot be in $B$, because if it were, then we would have $x \in A \cap B = \emptyset$, which is impossible. Therefore...

6. Feb 23, 2012

### iHeartof12

Ok I think I get it tell me if I worded this correctly:

Suppose $x \in A$ and $A \cap B = \emptyset$
Since $x \in A$ and $A \subset B \cup C$, this means that $x \in B$ or $x \in C$. Consequently, $x$ cannot be in $B$, because if it were, then we would have $x \in A \cap B = \emptyset$, which is impossible. Therefore $x \in C$. Thus $x \in A$ implies $x \in C$ so it follows that $A \subseteq C$

7. Feb 23, 2012

### jbunniii

Looks good to me.

8. Feb 23, 2012

### iHeartof12

ok thank you for your help