Identical affine spaces
Task number: 4464
Solution
\(\Leftarrow\): For any vector \(\boldsymbol{w}\in U\), it holds that \(\boldsymbol{u}+U = (\boldsymbol{u}+\boldsymbol{w})+U\), because for any vector \(\boldsymbol{z}\) from a vector space in which \(U\) is a subspace and \(u+U\) is an affine subspace, \(\boldsymbol{z}-\boldsymbol{u}\in U \Leftrightarrow \boldsymbol{z}-\boldsymbol{u}- \boldsymbol{w}\in U \Leftrightarrow \boldsymbol{z}-(\boldsymbol{u}+\boldsymbol{w})\in U\).
If we substitute \(\boldsymbol{w}=\boldsymbol{v}-\boldsymbol{u}\), we get \(\boldsymbol{u}+U = (\boldsymbol{u}+(\boldsymbol{v}-\boldsymbol{u}))+U=\boldsymbol{v}+U\) and from the assumption \(U=V\) we have immediately \(\boldsymbol{v}+ U=\boldsymbol{v}+V\).
\(\Rightarrow\): It holds that \(\boldsymbol{v}=\boldsymbol{v}+\mathbf{e} \in \boldsymbol{v}+V\) and by assumption \(\boldsymbol{v}+V=\boldsymbol{u}+U\) is \(\boldsymbol{v}\in \boldsymbol{u}+U\) and hence \(\boldsymbol{v}- \boldsymbol{u}\in U\).
Using the equivalence from the previous proof of the reverse implication and the same substitution, we get \(\boldsymbol{v}+U =\boldsymbol{u}+U =\boldsymbol{v}+V\). Now for each vector \(\boldsymbol{z}\), \(\boldsymbol{z} \in U \Leftrightarrow \boldsymbol{z}+\boldsymbol{v}\in \boldsymbol{v}+U \Leftrightarrow \boldsymbol{z}+\boldsymbol{v}\in \boldsymbol{v}+V \Leftrightarrow \boldsymbol{z} \in V\), thus \(U=V\).
