Proof regarding the basis of a quotient space (of a vector space and a subspace) I'm not sure how to approach this question. We let $U$ be a subspace of a vector space $V$ over a field $F$ and also let $(v_{1},\dotsc, v_{n})$ be a basis for $V$ . We must show that there are $1 \le k_{1} \lt \dotsc \lt k_{m} \le n$ such that $(v_{k_{1}} + U, \dotsc, v_{k_{m}} + U)$ is a basis for the quotient space $V/U$. I'm quite confused about what this question is asking so if someone could explain what we are trying to show and give an outline on how to show it that would help greatly. Thanks, Lauren!
 A: 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. 
A: The set $\{v_1+U,v_2+U,\dots,v_n+U\}$ is a spanning set for $V/U$.
The space $V/U$ is a red herring: the following result holds in any vector space, so in particular in $V/U$.
Theorem. Any spanning set for a vector space contains a basis.
(Note: this holds also for infinite dimensional vector spaces, but I'll show it only for finite dimension.)
Proof. Suppose there is a spanning set that contains no basis for the vector space $W$. So, let $\{w_1,w_2,\dots,w_m\}$ be such a spanning set with the minimal possible number of elements. Thus any spanning set with $m-1$ elements will contain a basis.
The given set is not linearly independent, otherwise it would be a basis. So one of the vectors is a linear combination of the other elements; it's not restrictive to assume that it is $w_m$, so
$$
w_m=\beta_1w_1+\dots+\beta_{m-1}w_{m-1}
$$
Let $w\in W$. By assumption we can write $w=\alpha_1w_1+\dots+\alpha_mw_m$, because the given set spans $W$. But then
\begin{align}
w&=
\alpha_1w_1+\dots+\alpha_{m-1}w_{m-1}+\alpha_m(\beta_1w_1+\dots+\beta_{m-1}w_{m-1})\\
&=(\alpha_1+\alpha_m\beta_1)w_1+\dots+(\alpha_{m-1}+\alpha_m\beta_{m-1})w_{m-1}
\end{align}
so also $\{w_1,\dots,w_{m-1}\}$ is a spanning set. By the minimality of $m$, this set contains a basis. Contradiction. QED
Alternative proof.
A finite spanning set $S=\{v_1,\dots,v_n\}$ of the vector space $V$ has at least a linearly independent subset, namely the empty set. Take a linearly independent subset with maximal number of elements; without loss of generality, we can assume it is $B=\{v_1,\dots,v_k\}$. I claim that this is a basis. If all vectors among $v_{k+1},\dots,v_n$ are in the span of $B$, then $B$ is a spanning set, hence a basis.
Otherwise, we can assume, without loss of generality, that $v_{k+1}$ is not in the span of $B$. But then $B'=\{v_1,\dots,v_k,v_{k+1}\}$ is linearly independent. Contradiction to maximality of $B$.
The same technique (with the help of Zorn's lemma) proves the statement for infinite dimensional vector spaces. The main point is that the union of a chain (under set inclusion) of linearly independent subsets is still linearly independent. A maximal linearly independent subset then exists by Zorn's lemma; it is a basis with the same argument as before.
