Covering relation in the lattice of closed subsets

Question

Covering relation in the lattice of closed subsets

Hello!

I am currently reading Martin Aigner's Combinatorial Theory and I have to say this book is definitely above my level, but (very) slowly things are coming together. Anyway I got to one statement that doesn't seem obvious to me. There is a proof given, but I don't see how it is sufficient. The statement is the following:

$\overline{A\cup p}\ \cdot\!\! >A$ for any $A\in L(S)$, $p\notin A$.

Here we are talking about a matroid on the set $S$. $\overline{A}$ is the closure of $A$ and $L(S)$ denotes the lattice of closed subsets of $S$. We use the symbol $\cdot\!\! >$ for the covering relation.

The proof given is the following:

If $q\notin A$, $q\in \overline{A\cup p}$, then by the exchange property of the closure operator $p\in \overline{A\cup q}$, i. e. $\overline{A\cup p} = \overline{A\cup q}$.

That's it. Now I have this question (among many): Isn't it possible for example to consider another element $r$, such that $r\notin A$, $r\in \overline{A\cup p}$ and $r\notin \overline{A\cup q}$? Now $\overline{A\cup p} \ne \overline{A\cup q}$.

I fear that I am missing something really obvious so I decided to ask.

Answer 1

It might help if you try to write your own proof. The key idea is the same as in the text. I have written this one out in (too much) detail so you can see what I mean.

Suppose $B$ is a closed set such that $A\subseteq B \subseteq \overline{A\cup p}$. If there is $q\in B$ such that $q\not\in A$, then $q\in\overline{A\cup q}\subseteq B \subseteq \overline{A\cup p}$. By the exchange property, $p\in \overline{A\cup q}\subseteq B$. Since $A\subseteq B$ and $p\in B$, we get that $\overline{A\cup p}\subseteq B$, hence $\overline{A\cup p}=B$. This shows that if $B$ is between $A$ and $\overline{A\cup p}$ and $B\neq A$, then $B=\overline{A\cup p}$, i.e. that $\overline{A\cup p}$ covers $A$ in the lattice of closed sets.