Consider some subset $\mathcal S$ (containing at least two elements) of some one-dimensional subspace $\mathcal V^1$ of some multi-dimensional vector space $\mathcal V^k \equiv (V, +, \mathbb R)$, $k \ge 2$; whereby $\mathcal S \subseteq \mathcal V^1 \subset \mathcal V^k$.
As far as I understand, for any two (not necessarily distinct) elements $\mathbf a, \mathbf b \in \mathcal S$, holds
- $\mathbf a$ and $\mathbf b$ are equal (in the specific sense of algebraic operations referred to as vector space $\mathcal V^k$) if and only if $\mathbf a$ and $\mathbf b$ denote the same element of $\mathcal S$:
$$ (\mathbf a = \mathbf b) \iff (\mathbf a \equiv \mathbf b), \tag{1} $$
- of any two distinct and therefore unequal elements $\mathbf a, \mathbf b \in \mathcal S$, at most one may be the null-element (of vector space $\mathcal V^k$, as well as of the one-dimensional subspace $\mathcal V^1$):
$$ ((\mathbf a \ne \mathbf b) \text{ and } (\mathbf a + \mathbf a = \mathbf a) \text{ and } (\mathbf a + \mathbf b = \mathbf b)) \implies ((\mathbf a + \mathbf b \ne \mathbf a) \text{ and } (\mathbf b + \mathbf b \ne \mathbf b) \text{ and } (\mathbf a = 0 \, \mathbf b)), \tag{2} $$
- any two distinct and therefore unequal elements $\mathbf a, \mathbf b \in \mathcal S$, neither of which is the null-element, are real multiples of each other:
$$ ((\mathbf a \ne \mathbf b) \text{ and } (\mathbf a + \mathbf b \ne \mathbf a) \text{ and } (\mathbf a + \mathbf b \ne \mathbf b)) \implies (\exists \, s_{ba} \in \mathbb R : (s_{ba} \ne 0) \text{ and } (\mathbf b = s_{ba} \, \mathbf a) \text{ and } (\mathbf a = \left(\frac{1}{s_{ba}}\right) \, \mathbf b)) \tag{3}. $$
My questions:
Is it true and by conditions $(1, 2, 3)$ justified to call any two distinct elements of $\mathcal S$ "colinear" ?
And if so:
How do we best (distinctly and crisply) call a set (like $\mathcal S$), which contains at least two distinct elements, and any two distinct elements of which are colinear ?? ...