You say you are stuck; where?I'm trying to practice writing proofs and i've been stuck on this one for a while.
For all sets A, B, and C, if [MATH]A \cap B \subseteq C, Then (A - C) \cap B =\varnothing [/MATH]
HINT: oppose that \(\exists t\in (A\setminus C)\cap B\).I'm trying to practice writing proofs and i've been stuck on this one for a while.
For all sets A, B, and C, if [MATH]A \cap B \subseteq C, Then (A - C) \cap B =\varnothing [/MATH]
Keep going; you're almost there! Now you have an x that is in A and B, but not in C, ...How is this so far:
Suppose not, that is suppose that sets [MATH]A,B, and [/MATH] [MATH]C[/MATH] exist such that [MATH] A \cap B \subseteq C [/MATH] and[MATH] (A-C) \cap B \neq \varnothing [/MATH]
Our supposition implies [MATH]\exists x \in (A-C) \cap B[/MATH]By definition of intersection, [MATH]x \in (A-C) \cap B \to x \in(A-C) \land x \in B[/MATH]By definition of difference, [MATH]x \in (A-C) \to x\in A \land x \notin C[/MATH]
That doesn't make any sense to me. The parentheses are meaningless, and a statement about sets can't directly imply anything about a particular element.We were given [MATH]A \cap B \subseteq C [/MATH][MATH]A \cap (B \subseteq C) \to x \in A \land x \in B \land x \in C [/MATH]In Summary, [MATH]x \in A \land x\in B \land x \in C \to x \in A \land x\in B \land x \notin C [/MATH]Therefore, [MATH] x \in C \land x \notin C [/MATH] is a contradiction.
correct?