Intersection of subspaces by construction Let be $S \subset V$, subspace of a vector space $V$ over the field $\mathbb{K}$, $dim(V)=n$, and, $dim(S)=k<n$. For every $r\in \mathbb{N},$  $1≤r≤n-k$. Prove that $S$ is the intersection of $m$ subspaces of $V$ of dimension $n-r$. 
Aplication: If $n=2016$, $k=1000$, $r=5$ What is the minimum value of $m$ satisfying the above paragraph?
Good evening, do not really understand what I say is the problem and I have no ideas to attack him, but I do not know if it relates to the Archimedean property, any help would be greatly appreciated. Thank you!
 A: My answer here will only show that such a $m$ exists and give an upper bound for $m$.
Fix an $r \in \{1,...,n-k\}$. 
Choose a basis for $S$, say $s_1,...,s_k$, and extend to a basis of $V$, say by adding vectors $v_1,..., v_{n-k}$. 
Observe that the space
$$ \operatorname{span} \left( s_1 , ... , s_k, v_{\sigma(1)} , ... , v_{\sigma(n - k - r)} \right)$$ 
has dimension $n - r $ for any distinct elements $\sigma(1) , ... , \sigma(n - k - r) \in \{1,...,n-k\} $.
Now intersecting every space of the form above yields,
$$I = \bigcap_{\sigma \in S_{n-k}} \operatorname{span} \left( s_1 , ... , s_k, v_{\sigma(1)} , ... , v_{\sigma(n - k - r)} \right)$$
I claim that $I = S$. 
Certainly $S \subseteq I$, since $S \subseteq \operatorname{span} \left( s_1 , ... , s_k, v_{\sigma(1)} , ... , v_{\sigma(n - k - r)} \right)$.
For the inclusion $I \subseteq S$, note that $v_1 \notin \operatorname{span} \left( s_1 , ... , s_k, v_{2} , ... , v_{\sigma(n - k - r + 1)} \right)$ and so $v_1 \notin I$. Similarly $v_i \notin I$ for each $i$ and hence $I \subseteq \operatorname{span} \left( s_1 , ... , s_k \right) = S $
Thus $S$ is the intersection  of subspaces of $V$ of dimension $n−r$, as desired.
This procedure produces a value of $m =$$ n-k \choose n - k -r $ since that is the number of spaces of the form $\operatorname{span} \left( s_1 , ... , s_k, v_{\sigma(1)} , ... , v_{\sigma(n - k - r)} \right)$. However this incorporates a lot of redundancy and is not minimal.
