Prove subspace $V=C-x_{0}=\left\{x-x_{0} | x \in C \subseteq R^n \right\}$ associated with the affine set $C$ does not depend on the choice of $x_0$

Question

Subspace $V=C-x_{0}=\left\{x-x_{0} | x \in C \subseteq R^n \right\}$ associated with the affine set $C$ does not depend on the choice of $x_0 \in C$, i.e. $V$ are the same regardless of $x_0$

My try:
For $V_0=C-x_{0}=\left\{x-x_{0} | x \in C\right\}$ and $V_1=C-x_{1}=\left\{x-x_{1} | x \in C\right\}$
suppose $v_0 \in V_0$, I try to show that $v_0 + x_1 \in C$,but I don't know how to show it

Any hint?

Answer 1

First, note that $x_1-x_0\in V$.

Indeed, since $x_1\in C$, we have $x_1=v+x_0$, for some $v\in V$. Therefore, $x_1-x_0=v\in V$.

Now, observe that, for $v_0\in V$

$$v_0+x_1=v_0+(x_1-x_0)+x_0=v_1+x_0,$$ where $v_1=v_0+(x_1-x_0)\in V$.

Thus, $v_0+x_1\in C$.