Prove that A B and B C, then A C, when A B denotes \"A is a proper subset of B\". Solution Suppose A B and B C. We will first show that A C. Let a be an arbitrary element of A. Since A B, we have that a B. Yet B C, and so a C. It follows that A C. Thus, either A C or A = C. Assume that A = C. Since A is a proper subset of B, there is an element b of B that is not contained in A, but because B C, we have that b C. However, A = C, and so b A, which is a contradiction. Therefore A C..