The number of axioms is subject to taste and debate (for me there is just one: A vector space is an abelian group on which a field acts). You should not want to distinguish by noting that there are different criteria. Actually, there is a reason why a subspace is called a subspace: It is also a vector space and it happens to be (as a set) a subset of a given space and the addition of vectors and multiplicataion by scalars are "the same", or "inherited" from that other space. So this way there is no real difference, and one should better introduce and define the notion of subspace per "vectorspace that is contained (the way I describe above) in a vector space" instead of "subset with operations that have some magical other properties". Rather the fact that "nonempty and closed under multiplication and addition" are (necessary and) sufficient conditions for a subset to be a subspace should be seen as a simple theorem, or a criterion to see when a subset of a vector space is in fact a subspace. It gives you a simple recipe to check whether a subset of a vector space is a supspace.
Well,think about what it means to have a basis for a vector space V over a field F. A basis B is a set of vectors in V for which for every vector v in V, there exists S= ${v_1,v_2,.....v_n}\subseteq B$ and ${a_1,a_2,...,a_n}\in F$ such that $\sum_{i=1}^n a_iv_i = v$ and $\sum_{i=1}^n a_iv_i = 0$ iff for every $i$, $a_i=0$. So let's check. Let U be a subspace of V and consider the quotient space V\U. Consider $S'= S + U\subseteq V/U $ and let's see if this is a basis for V/U. Let's see if it spans V/U. For every $u\in V/U$, u = v + w where v is an arbitrary vector in V where $u-v=w\in U$. Since B is a basis for V, there exists ${v_1,v_2,.....v_i}\in B$ and ${a_1,a_2,...,a_i}\in F$ where $1\leq i \leq n$ such that u = $\sum_{j=1}^i a_jv_j$. Also, since B is a basis for V and U is a subspace of V, there exists ${v_1,v_2,.....v_k}\in B$ and ${b_1,b_2,...,b_k}\in F$ such that for every $w\in U$, w = $\sum_{l=1}^k b_lv_l$ where $1\leq k \leq n$. But this means u = v + w = $\sum_{j=1}^i a_jv_j$ + $\sum_{l=1}^k b_lv_l$ = $\sum_{m=1}^{j+l}(a_j + b_l)v_m$ where $1\leq j+l \leq n$ . But this means S' spans V/U. Since for every m where $1\leq m\leq n$, $v_m\in B$, then $\sum_{i=1}^m a_iv_i = 0$ iff for every $i\leq m$, $a_i=0$. But that means S' is a linearly independent set of vectors in V and that means S' is a basis for V/U. Q.E.D.
The notation of my proof in the indices may be a little sloppy. I'll go over it later, but the basic logic is correct.
Best Answer
The vectors that form a basis for the column space are certainly in the column space. You can choose one of these.