Prove the statement P:
For all sets $A$, $B$, and $C$, if $A-B \subseteq A – C$ then $ A \cap C = \varnothing $
My attempt to answer:
This statement is true, and here is a proof:
Proof: Suppose A, B, and C are sets such that $A-B \subseteq A – C$. We want to prove that $A \cap C = \varnothing $ by contradiction.
Suppose $A \cap C \neq \varnothing $. That is there is $x \in A \cap C$. Also, since $A-B \subseteq A – C$, suppose there is a $x \in A – B$, then $x \in A – C$. That is, $x \notin C$. Therefore, that implies that $x \notin A \cap C$ which contradicts with given that $x \in A \cap C$. Therefore, concluding by contradiction: $A \cap C = \varnothing $. End of proof.
Is this proof correct? Any issues?.
My proof is wrong, got it based on your comments:
Let $A = \{1, 2\}, B = C = \{2\}$, then $A-B \subseteq A – C$, but, $A \cap C \neq \varnothing = \{2\}$. Can someone, tell me what I did wrong in my proof above? Is it because, I assumed $x \in A – B$?
part b) Write the converse and proof/disproof that.
The converse is: For all sets $A$, $B$, and $C$, if $ A \cap C = \varnothing $ then $A-B \subseteq A – C$
Best Answer
A counterexample: Let $A=B=C=\{17\}$.
The converse is true: If $A\cap C=\emptyset$, then $A\setminus B\subseteq A\setminus C$. For if $A\cap C=\emptyset$ then $A\setminus C=A$.
Remark: The added example that tries to show the converse does not hold is not correct. For you let $A=\{1\}$ and $C=\{1\}$, which makes $A\cap C$ non-empty.