Let $S$ be a non-empty set. Consider the statement $$\forall A\subseteq S \quad \forall B\subseteq S\quad\exists D\subseteq S \ (D\neq \emptyset \ \wedge \ D\cap (A\cup B)=\emptyset) $$ The negation is $$\exists A \subseteq S \quad \exists B\subseteq S \quad \forall D \subseteq S \ (D=\emptyset \ \vee \ D\cap (A \cup B)\neq \emptyset). $$ I have to determine which statement is true.
I had a long think and thought that the second is true since I think the first is false.
Let $S = \{1,2,3\}$.
Then if I pick $A= \{1,2\}$ and $B=\{2,3\}$. If $D$ is empty then the statement is false. If $D$ is not empty, then $D$ must be a set which has common members with $A$ or $B$ and so the intersection is nonempty.
Is this correct or is there a better way to think about it?

  • 3
    $\begingroup$ Yes, this is correct. $\endgroup$ – Mees de Vries Nov 11 '17 at 11:20

You are correct.

To see how this works for any $S$: Pick $A=B=S$. Then $A \subseteq S$, $B \subseteq S$, and $A \cup B =S$. Hence, there cannot be any non-empty $D \subseteq S$ that does not share any elements with $A \cup B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.