Collections of Sets, Problem 3

Solution part (b)

Let A, B, and C be sets.

Suppose that \(x \in (A - C)\). Then \(x \in A\) and \(x \not \in C\). There are two cases:

Case 1: \(x \not \in B\). Then \(x \in A\) and \(x \not \in B\). So \(x \in (A - B)\).

Case 2: \(x \in B\). Then \(x \in B\) and \(x \not \in C\). So \(x \in (B - C)\).

So \(x \in (A - B)\) or \(x \in (B - C)\). Therefore \(x \in (A - B) \cup (B - C)\).