Separable items and matroids

Let $E$ be a finite set and $I$ a downwards-closed set of subsets of $E$ (let's call them "independent sets").

Say that two elements $x,y\in E$ are separable if there exists a partition of $E$ into two independent sets $X, Y\in I$, such that $x\in X$ and $y\in Y$.

If $(E,I)$ is a matroid, and there exists any partition of $E$ into two independent sets, then any two items are separable. This follows from the symmetric basis exchange property.

Is the opposite also true? Suppose all pairs of elements in $E$ are separable. Is $(E, I)$ necessarily a matroid?

Note that, without the requirement that $I$ is downwards-closed, the answer is "no". For example, suppose $E = \{1,2,3\}$ and $I$ contains all subsets of adjacent integers (that is, $I = 2^E\setminus \{ \{1,3\}\}$). Then $(E,I)$ is not a matroid since it is not downwards-closed. But all pairs of elements of $E$ are separable: $1,2$ and $1,3$ are separable by the partition $\{1\},\{2,3\}$, and $2,3$ are separable by the partition $\{1,2\},\{3\}$.

I found a counter-example. Suppose $E = \{1,2,3,4\}$, and $I$ contains all subsets of $E$ with a sum of at most $7$. It is clearly downwards-closed, but it is not a matroid, since there are maximal independent sets with different cardinalities: $\{1,2,4\}$ and $\{3,4\}$. But all pairs of items are separable:

• 1 and 2 are separated by the partition $\{1,3\},\{2,4\}$;
• 3 is separated from the rest by the partition $\{3\},\{1,2,4\}$;
• 4 is separated from the rest by the partition $\{4\},\{1,2,3\}$.