Show that a matroid $M$ is connected iff, for every pair of distinct elements of $E(M),$ there is a hyperplane avoiding both.

Question

Here is the question I am thinking about:

Show that a matroid $M$ is connected iff, for every pair of distinct elements of $E(M),$ there is a hyperplane avoiding both.

It is in James Oxley book in chapter 4, section 1. Could anyone give me some hints to the idea of solving it please?

Answer 1

First recall that the complement of a hyperplane is a cocircuit, so this means that for every pair of elements, there exists a cocircuit that contains both. This implies that $M^*$ is connected, using Proposition 4.1.3 in Oxley. If you show that $M$ is connected iff $M^*$ is connected, you are done. To show the later, start with a pair of elements $\{a,b\}\subseteq C$ for a circuit $C$ and then $C\setminus \{a\}$ is independent, hence there is a basis $B$ such that $C\setminus \{a\}\subseteq B$, hence $B^* = X\setminus B$ contains $a$ but it doesn't contain $b$, so consider the fundamental circuit of $B^*$ w.r.t $b$, this will have $a$ and $b$ and so connectivity is transferred by duality.