0

Here's what I have:

The statement is true. Assume A, B, and C are sets, and that $A \setminus (B$ intersect $C) = \emptyset$. We will prove $(A \setminus B = \emptyset$ and $A - C = \emptyset)$ by assuming $(A \setminus B \not= \emptyset$ or $A \setminus C \not= \emptyset$ and deriving a contradiction. We have two cases:

Case 1: $A \setminus B = \emptyset$, therefore $A \setminus C \not= \emptyset$. And I don't know how to derive a contradiction from here.

Case 2: $A \setminus B \not= \emptyset$. And I don't know how to derive a contradiction from here.

I'm not entirely sure where to go next using the set definitions. Am I on the right track? And if so, how can I derive a contradiction?

  • Editorial comments: $\cap$ will give you the intersection symbol. Also, one should avoid using both $A-C$ and $A\setminus C$ in the same problem, unless they are meant to be different things. – halrankard2 Oct 18 '20 at 10:59

1 Answers1

1

Work from first principles to derive a contradiction.

Suppose $A\backslash B \neq \emptyset$. Then there is at least one element $x \in A\backslash B$. That means $x \in A$ and $x \not \in B$. But then $x \not \in B\cap C$ either. Therefore $x \in A\backslash (B \cap C)$. That contradicts the requirement that $A\backslash(B \cap C) = \emptyset $. Therefore the supposition that $A\backslash B \neq \emptyset$ must be false and we conclude $A \backslash B = \emptyset$.

The original suppositon is symmetric in $B$ and $C$ so we can also conclude $A\backslash C = \emptyset$.

WA Don
  • 4,488