# Algebra with set notation and set properties

Suppose that $S$ and $T$ are sets with $S \cap T = \emptyset$

Let $C \subseteq S \cup T$ and let $A = C \cap S$ and $B = C \cap T$. Show that $A \subseteq S$, and $B \subseteq T$.

I said, let $S=\{x |P\}$ and $T = \{x|Q\}$. Then

$x \in A$ iff $x \in \{y|(P \lor Q) \land P\}$

iff $x \in \{y | (P \land P) \lor (P \land Q)\}$

iff $x \in \{y | P\}$ Since $S\cap T = \emptyset$

Therefore $A \subseteq S$. And obviously a similar argument for $B \subseteq T$. However I know this argument is completely wrong because it was marked wrong on my homework. What is wrong with it though?

• There is nothing to prove... $C \cap S \subseteq S$ is trivial, and the same for $C \cap T \subseteq T$. Commented Nov 24, 2014 at 12:25
• Intuitively, I understand that. However, this was assigned as a homework problem and I believe there is a way to prove it somehow. I just can't figure out how if the way I did above is incorrect and I was told it was by my teacher who is very good.
– Sam
Commented Nov 24, 2014 at 12:28
• Your argument is wrong since the beginning. You said $x \in A$ iff $x \in \{y|(P \lor Q) \land P\}$ but this is not true. The definition is $A=C\cap S$. Where did you use the set $C$? Commented Nov 24, 2014 at 12:38
• $x \in C$ iff $x \in \{y | P \lor Q\}$ is that not correct?
– Sam
Commented Nov 24, 2014 at 12:41
• No, it is not correct, since the second set is not related to $C$. Commented Nov 24, 2014 at 12:48

## 1 Answer

If $S=\{x|P\}$ and $T=\{x|Q\}$ then there is no reason to find why $x\in A$ should be equivalent with $x\in\{y|(P\vee Q)\wedge Q\}$. The first statement involves set $C$ and the second does not.