Prove or provide a counterexample:

For all sets A, B, C, if A is subset of B and B is a subset of C^c (complement of C), then A intersection C= { }.

done

I tried using a proof by contradiction, but I am not sure if that is actually the correct method to do it.

did you see the solution I offered to your first post?

To prove or provide a counterexample for this statement, we first need to understand its logical structure. The statement asserts that for any three sets A, B, and C, if A is a subset of B and B is a subset of the complement of C, then the intersection of A and C is an empty set.

To prove this statement, we need to show that the given conditions always lead to an empty intersection. Conversely, to provide a counterexample, we need to find specific sets A, B, and C that satisfy the given conditions, but their intersection is not empty.

Let's consider the statement and explore both cases:

Case 1: Proving the Statement
To prove the statement, we assume that A is a subset of B, and B is a subset of C^c. We want to show that A intersection C is an empty set.

To verify this, we start by assuming that there is an element x that belongs to A intersection C. This means that x belongs to both A and C. Since A is a subset of B, x must also belong to B. However, since B is a subset of C^c, x cannot belong to C. This contradicts x belonging to both A and C. Therefore, our initial assumption was incorrect, and there cannot exist an element in A intersection C. Hence, A intersection C is an empty set, which proves the statement.

Case 2: Providing a Counterexample
To provide a counterexample, we need to find specific sets A, B, and C satisfying the given conditions, where A intersection C is not an empty set.

Let's consider an example:
A = {1}, B = {1, 2}, C = {2}

In this case, A is a subset of B since all elements in A are also in B. B is a subset of the complement of C since all elements in B (1 and 2) do not belong to C. However, A intersection C is not an empty set because 1 belongs to A and C simultaneously.

Therefore, A intersection C = {1}, which violates the given statement. Thus, this example serves as a counterexample.

Conclusion:
We have both proven the statement and provided a counterexample, showing that the statement is true for all sets A, B, and C.