an affine subset of V is a subset of V that is the sum of a vector and one of its subspace; that is, an affine subset of V is a subset of V of the form v+U for v \in V and subspace U \subset V. for v \in V and U \subset V, an affine subset v+U is said to be parallel to U. that is, an affine subset for U \subset V and v \in V:

additional information two affine subsets parallel to U are either equal or disjoint Suppose U is a subspace of V; and v,w \in V, then, if one of the following is true all of them are true: v-w \in U v+U = w+U (v+U) \cap (w+U) \neq \emptyset 1 \implies 2 Given v-w \in U…. For an element in v+U, we have that v+u = (w-w)+v+u = w+((v-w)+u) \in w + U. This is because U is closed so adding v-w \in U and u will remain being in U. w-w=0 just by everything being in V. We now have v+u \in w+U\ \forall u \in U; we now can reverse the argument to argue in a similar fashion that w+u \in v+U\ \forall u \in U. So, we have that v+U \subset w+U and w+U \subset v+U. So v+U = w+U, as desired. 2 \implies 3 By definition of v+U=w+U as long as v+U and w+U is not empty sets, which they can’t be because U is a vector space so guaranteed nonempty. 3\implies 1 Given (v+U) \cap (w+U) \neq \emptyset, we have that there exists some u_1, u_2 \in U such that v+u_1 = w+u_2. Because everything here is in V, we can add their respective inverses (“move them around”) such that: v-w = u_2-u_1. Therefore u_2-u_1 \in U \implies v-w \in U.