Let $M$ be a matroid with circuit $C$, and $x,y \in C$. Prove that there is a cocircuit $D$ so that $x,y \in D$. Contradiction?

Question

Let $M$ be a matroid with circuit $C$, and $x,y \in C$. Prove that there is a cocircuit $D$ so that $x,y \in D$.

My problem: We learned in class that $Y$ is dependent in $M^*$ iff for every basis $B \in M$, $Y \cap B=\emptyset$.

Yet we can define $X=C \setminus \{y\}$, which is an independent set. Which means there is a basis $B_0$ so that $X \subseteq B_0$. That means that if such $D$ exists, and is dependent as a cocircuit, we'll get $D \cap B_0 \neq \emptyset$, which is a contradiction.

Am I getting anything wrong here?

Thanks in advance for any assistance!

Answer 1

You may want to double-check your notes. A cocircuit $C^*$ of $M$ is a set that is dependent in the dual $M^*$, and one property of cocircuits is that they intersect every basis of $M$. You probably meant $Y \cap B \neq \emptyset$.

Answer 2

We know that $E-C$ is a hyperplane in $M^*$. Choose a basis, $B^*$, in $M^*$ of $E-C$. Then $r(B^*)=r(M^*)-1$. Since $x\not\in E-C$, we get $r(B^*\cup x)=r(M^*)$, and likewise, $r(B^*\cup y)=r(M^*)$, so each of $B^*\cup x$ and $B^*\cup y$ are independent. Then $r(B^*\cup \{x,y\})=r(M^*)$, so $B^*\cup \{x,y\}$ is dependent in $M^*$, and thus contains a circuit, $D$ in $M^*$. We can't have $D\subset B^*\cup x$ or $D\subset B^*\cup y$, so $\{x,y\}\subseteq D$.

Answer 3

since there is a circuits that contains x and y, so M has a connected component that contains x and y. Now since if a matroid be connected its dual is also connected, so there is a cocircuit that contains x and y.