Why does $\|(-1/\lambda \cdot m)-x_0\|\ge r$ hold when $\|m-x_0\|\ge r $ holds?

Question

$M$ is a closed subspace of the normed linear space $X$ and $x_0 \in X\setminus M$. Thus there is an $r>0$ for which:
$\|m-x_0\|\ge r $ $\forall m \in M$.
My question is, why does the following hold for $m \in M$ and for a scalar $\lambda \in \mathbb{R} $:
$\|(-1/\lambda \cdot m)-x_0\|\ge r$
How can we be sure, that the norm is still $\ge r$, when we multiply $m$ by $-1/\lambda$?

Answer 1

$M$ is a linear subspace so it is closed under scalar multiplication, i.e. if $m \in M$ then also $-\frac1{\lambda}\cdot m \in M$.

Therefore $\left\|-\frac1{\lambda}\cdot m - x_0\right\| \ge r$ for all $m \in M$.