# Thread: Need help on a Set Theory proof

1. ## Need help on a Set Theory proof

This one must be easy, but I missed the class and we don't have a textbook and I'm just not sure how to approach it. Here goes:

Prove that for all sets A: $A \cap \emptyset = \emptyset$

We've been proving these by converting them into a sort of logical form (i.e. A intersects B turns into (x in the domain of A ^ x in the domain of B). Any suggestions?

2. Perhaps prove by contradiction? Here's something to get you started:

Suppose, for the sake of contradiction, that $A \cap \emptyset \neq \emptyset$. Then there exists $x \in A$ such that $x \in \emptyset$. However, x cannot be an element of the empty set, because the empty set has no elements. Contradiction.

3. Originally Posted by mxrider530
This one must be easy, but I missed the class and we don't have a textbook and I'm just not sure how to approach it. Here goes:

Prove that for all sets A: $A \cap \emptyset = \emptyset$

We've been proving these by converting them into a sort of logical form (i.e. A intersects B turns into (x in the domain of A ^ x in the domain of B). Any suggestions?
$A \cap \emptyset = \{x \in A : x \in \emptyset\}$

Assume $A \cap \emptyset \neq \emptyset$. Then, there must be some element $x_0 \in (A\cap \emptyset) \Rightarrow x_0 \in \emptyset \Rightarrow \emptyset \neq \emptyset$ and therefore that is a contradiction.