Houjun Liu

affine subset

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 \(v \in V\):

\begin{equation} v + U = \{v+u : u \in U\} \end{equation}

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:

  1. \(v-w \in U\)
  2. \(v+U = w+U\)
  3. \((v+U) \cap (w+U) \neq \emptyset\)

\(1 \implies 2\)

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 \(u_2-u_1 \in U \implies v-w \in U\).