counterexample. S is not always a base

main.tex

@@ -89,7 +89,7 @@ We choose an edge $e$ with probability proportional to its capacity $c(e)$.
 Note that $B$ is taken from the support of the optimal base packing, then $\sum_{B:e\in B} \frac{\Pr[B]}{c(e)}$ are the same for all elements and every $B$ contains $r(E)-r(F^*)$ edges in $E\setminus F^*$.
 
 In graphic matroids it follows easily that $S$ is a spanning tree.
-However, $S$ may not be independent in general matroids. (can we find an example or prove that $S$ is a base in all matroids?)
+However, $S$ may not be independent in general matroids.\footnote{Here is a systematical way to construct counterexamples. Let $B$ be a base of $M|F^*$ and let $X\subset E-F^*$ be an independent set such that $B\sqcup X$ is a base of $M$. Let $B'$ be the counterexample with largest intersection with $B$. We can assume the circuit $C\subset B'\sqcup X$ contains $B'\setminus B$ since otherwise we can do multiple symmetric exchange with $B$ and $B'$. Then we can divide $C$ into 3 parts, $R=B'\setminus B, S=C\cap B, T=C\cap X$. One can set $R=\set{(0,0,1)^T,S=\set{(1,1,1)^T}}$ and $T=\set{(1,0,0)^T,(0,1,0)^T}$. Then it is easy to add more dimensions and get the desired $F^*$.}
 $S$ is independent does not imply that $M$ is a direct sum of $M|F^*$ and $M\setminus F^*$ since the rank of $M\setminus F^*$ can be larger than $r(E)-r(F^*)$. 
 Characterization of matroids where $S$ is a base is another interesting problem.
 \note{From now on we assume $S$ is a base. This should holds in all $(k,2k-1)$-sparsity matroids.}