Affine space/Affine subspace/Characterization/Fact/Proof

If ${\displaystyle {}F}$ is empty, then all three conditions are true. So we assume that ${\displaystyle {}F}$ is not empty. ${\displaystyle {}(1)\Rightarrow (2)}$. Let ${\displaystyle {}F=P+U}$ with ${\displaystyle {}P\in F}$ and a linear subspace ${\displaystyle {}U\subseteq V}$. Then, ${\displaystyle {}P_{i}=P+u_{i}}$ with some ${\displaystyle {}u_{i}\in U}$. Due to the definition of a barycentric combination, it follows that

${\displaystyle {}\sum _{i=1}^{n}a_{i}P_{i}=P+\sum _{i=1}^{n}a_{i}{\overrightarrow {PP_{i}}}=P+\sum _{i=1}^{n}a_{i}u_{i}\,}$

is an element of ${\displaystyle {}F}$.

${\displaystyle {}(2)\Rightarrow (3)}$. This is a weakening of the condition.

${\displaystyle {}(3)\Rightarrow (1)}$. We choose a point ${\displaystyle {}P\in F}$, and consider

${\displaystyle {}U:={\left\{{\overrightarrow {PQ}}\mid Q\in F\right\}}\subseteq V\,.}$

We have ${\displaystyle {}0\in U}$. For ${\displaystyle {}Q,Q'\in F}$, due to the condition, also ${\displaystyle {}-P+2Q}$ and ${\displaystyle {}-P+2Q'}$ belong to ${\displaystyle {}F}$. Therefore, also

${\displaystyle {}{\frac {1}{2}}{\left(-P+2Q\right)}+{\frac {1}{2}}{\left(-P+2Q'\right)}=-P+Q+Q'\,}$

belongs to ${\displaystyle {}F}$, where this equality rests on exercise. This point equals

${\displaystyle {}P-{\overrightarrow {PP}}+{\overrightarrow {PQ}}+{\overrightarrow {PQ'}}=P+{\overrightarrow {PQ}}+{\overrightarrow {PQ'}}\,,}$

so that ${\displaystyle {}{\overrightarrow {PQ}}+{\overrightarrow {PQ'}}}$ belongs to ${\displaystyle {}U}$. Hence, ${\displaystyle {}U}$ is closed under the vector addition. Let ${\displaystyle {}Q\in F}$ and ${\displaystyle {}s\in K}$. Then, due to the condition, also

${\displaystyle {}(1-s)P+sQ=P+s{\overrightarrow {PQ}}\,}$

belongs to ${\displaystyle {}F}$, and, therefore, ${\displaystyle {}s{\overrightarrow {PQ}}}$ belongs to ${\displaystyle {}U}$. Thus, ${\displaystyle {}F=P+U}$ with a linear subspace ${\displaystyle {}U}$.