## Identical affine spaces

Show that two affine spaces coincide: $$\boldsymbol{u}+ U=\boldsymbol{v}+V$$, if and only if $$\boldsymbol{u}-\boldsymbol{v}\in U$$ and $$U=V$$.
$$\Leftarrow$$: 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 $$U=V$$.