I asked [this question](https://math.stackexchange.com/q/2811279/660) on Mathematics Stackexchange, but got no answer.


Let $K$ be a field and $n$ a positive integer. To a finite dimensional $K$-vector space $V$, equipped with a family $V\_1,\dots,V\_n$ of subspaces, we attach the map $d:\mathcal P(\{1,\dots,n\})\to\mathbb N$ defined by 
$$
d(S)=\dim\left(\bigcap\_{s\in S}V\_s\right).
$$ 
[Here $\mathcal P(\{1,\dots,n\})$ denotes the set of subsets of $\{1,\dots,n\}$.] Then we have 


$(1)\quad S\subset T\implies d(S)\ge d(T)$, 


$(2)\quad d(S\cap T)\ge d(S)+d(T)-d(S\cup T)$


for all $S,T\in\mathcal P(\{1,\dots,n\})$. [Here $S\subset T$ means that each element of $S$ belongs to $T$.]


Conversely, given $d:\mathcal P(\{1,\dots,n\})\to\mathbb N$ satisfying $(1)$ and $(2)$, is there a family $(V,V\_1,\dots,V\_n)$ as above inducing $d$ in the way just described?


The answer is No in general, a counterexample being given as follows: $K=\mathbb F\_2, n=4$, 
$$
d(\varnothing)=2=\dim V,\quad d(\{i\})=1=\dim V\_i,
$$ 
and $d(S)=0$ if $S$ has at least two elements. [We use the fact that there are only three lines through the origin in $\mathbb F\_2^2$.] There is an obvious analog for any finite field.


In view of the above observations, it seems natural to ask:


Assume that our field $K$ is *infinite*, and that $d:\mathcal P(\{1,\dots,n\})\to\mathbb N$ satisfies $(1)$ and $(2)$. Is there a family $(V,V\_1,\dots,V\_n)$ as above such that 
$$
d(S)=\dim\left(\bigcap\_{s\in S}V\_s\right)
$$ 
for all $S\ ?$



Prescribing the dimension of intersections of sub-vector spaces
I find this easier to think about if we dualize everything: replace $d(S)$ with $\dim V - d(S)$ and each subspace with its annihilator in $V'$. Recall that the annihilator of a sum is the intersection of the annihilators. Thus your question is the following:


Assume $d:\mathcal{P}(\{1,\dots,n\}) \to \mathbb{N}$ is increasing and submodular (this is just the cool name for the condition $d(S \cup T) + d(S\cap T) \leq d(S) + d(T)$) and satisfies $d(\emptyset)=0$ and $d(\{1,\dots,n\}) \leq \dim V$. Then does there exists a collection of subspaces $V\_i$ of $V$ such that $\dim \sum\_{i \in S} V\_i = d(S)$ for each $S \subset \{1,\dots,n\}$?


If we further assume that $d(\{x\}) \leq 1$ for each $x$ then this is exactly one of the equivalent definitions of a [matroid](https://en.wikipedia.org/wiki/Matroid#Rank_functions), and your question is whether every matroid is [linear](https://en.wikipedia.org/wiki/Matroid_representation). Apparently the answer is [no](https://en.wikipedia.org/wiki/V%C3%A1mos_matroid).


