Prove that any parallelopided is a convex set. 
Definition
The parallelopided generated by a point $O\in\Bbb R^n$ and $k$ linearly independent vectors $\vec v_1,\dots,\vec v_k$ is the set $\mathcal P_O(\vec v_1,\dots,\vec v_k)$ defined through the condition
$$
\mathcal P_O(\vec v_1,\dots,\vec v_k):=\big\{P\in\Bbb R^n|\,\,\,\text{there exist}\,\alpha^1,\dots,\alpha^k\in[0,1]\,\text{such that}\,\,P=O+\alpha^i\vec v_i\big\}
$$

So clearly with respect this definition any parallelotope is homeomorphic to the unitary cube $[0,1]^k$ so that it is connected but unfortunately it is not possible to claim that it is convex since homeomorphisms do not preserve the convexity as here showed. So I ask if any parallelopided is a convex set. Could someone help me, please?
 A: So first of all we remember that the product of convex sets is convex too (see here for details) and moreover we remember that any interval of the real line $\Bbb R$ is convex: so any rectangle is convex being product of convex sets and thus the unit cube $[0,1]^k$ is convex. Now any linear map and any translation between t.s.v. preserve the convexity since if $f:V\rightarrow W$ is a such map then
$$
f\big(x+t(y-x)\big)=f(x)+t\big(f(y)-f(x)\big)
$$
for any $x,y\in V$. So we conclude that any $k$-parallelopiped is convex because it is homeomorphic to the unit cube $[0,1]^k$ via the composition of a translation and a linear map.
Anyway it is possible (if it interest) to prove the statement via other argumentations I show to follow.
So if $x,y\in\mathcal P_O(\vec v_1,\dots\vec v_k)$ then there must exist $\xi^i,\eta^i\in[0,1]$ for $i=1,\dots,k$ such that
$$
x=O+\xi^i\vec v_i\,\,\,\text{and}\,\,\,y=O+\eta^i\vec v_i
$$
and thus we have to prove that
$$
x+t\cdot(x-y)=\big(O+\xi^i\vec v_i\big)+t\cdot(\eta^i-\xi^i)\vec v_i\in\mathcal P_O(\vec v_1,\dots,\vec v_k)
$$
for any $t\in[0,1]$. So observing that
$$
\big(O+\xi^i\vec v_i\big)+t\cdot(\eta^i-\xi^i)\vec v_i=O+\big(\xi^i+t\cdot(\eta^i-\xi^i)\big)\vec v_i=O+\big((1-t)\cdot\xi^i+t\cdot\eta^i\big)\vec v_i
$$
for any $t\in[0,1]$ we observe that
$$
\begin{cases}0\le\xi^i,\eta^i\le1\\0\le t\le1\end{cases}\Rightarrow\begin{cases}0\le\xi^i,\eta^i\le1\\0\le t\le 1\\0\le1-t\le1\end{cases}\Rightarrow\\
\begin{cases}(1-t)\cdot\xi^i\ge0\\t\cdot\eta^i\ge0\\t\cdot\eta^i\le t\le1\\(1-t)\cdot\xi^i\le(1-t)\le1\end{cases}\Rightarrow0\le(1-t)\cdot\xi^i+t\cdot\eta^i\le(1-t)+t=1
$$
for any $i=1,\dots, k$ and this proves the statement.
