# Why is the intersection of two flats a flat?

I'm a beginner at matroids - so please forgive the banality of the following question.

Defining a matroid in terms of closed sets: Why is it that the intersection of two closed sets (flats) is a flat, while the union of two flats is not nesceassarily a flat? (This is relevant when defining the join and meet in the lattice of flats of a given matroid.)

Can anyone recommend a good book for getting started on matroids?

Thanks a lot.

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mathoverflow.net/questions/67736/… –  JavaMan Jul 24 '11 at 19:50

Since no one else has done so, I’ll take a crack at the non-reference part of the question; whether this answer actually helps will probably depend on just what definitions you’re using.

Let $M$ be a matroid on a set $E$. The fact that the intersection of two closed sets is closed follows from the fact that the closed sets in $M$ can be defined in terms of a closure operator, i.e., a function $\sigma$ defined on subsets of $E$ that satisfies the following three conditions:

• $X \subseteq \sigma(X)$ for all $X \subseteq E$ ($\sigma$ is expansive);
• $\sigma(Y) \subseteq \sigma(X)$ whenever $Y \subseteq X \subseteq E$ ($\sigma$ is order-preserving); and
• $\sigma(\sigma(X)) = \sigma(X)$ for all $X \subseteq E$ ($\sigma$ is idempotent).

The closed sets are then those $X \subseteq E$ such that $\sigma(X) = X$. Whenever a notion of closed set is defined from a closure operator in this way, it will have the property that the intersection of two closed sets is always a closed set. For suppose that $A$ and $B$ are closed sets. Then $$A \cap B \subseteq \sigma(A \cap B) \subseteq \sigma(A) = A,$$ and $$A \cap B \subseteq \sigma(A \cap B) \subseteq \sigma(B) = B,$$ so $A \cap B \subseteq \sigma(A \cap B) \subseteq A \cap B$, $A \cap B = \sigma(A \cap B)$, and $A \cap B$ is closed. Essentially the same argument shows that if $\mathscr{C}$ is any family of closed sets, then $\bigcap\mathscr{C}$ is closed. (The argument doesn’t actually use the idempotence of $\sigma$, but without idempotence there may not be much in the way of closed sets; the only one whose existence is guaranteed is $E$.)

In the case of a matroid the closure operator is the span function $\sigma_M$ defined by $\sigma_M(X) =$ $X \cup \{e \in E:Y + e\mbox{ is a circuit for some }Y \subseteq X\}$; depending on your definition of closed set, it may or may not be very obvious that a set $X$ is closed iff $\sigma_M(X) = X$.

The situation with unions is different simply because the axioms defining a closure operator aren’t strong enough to guarantee that the union of two closed sets is closed. This is most easily seen from examples. For example, let $E = \{a,b,c\}$, and say that $X \subseteq E$ is independent iff $|X| \le 2$. (Equivalently, $E$ is the only circuit.) Then $\{a\}$ and $\{b\}$ are closed, but $\{a,b\}$ is not, and this should be fairly clear no matter what definition of closed set you’re using.

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Thanks for an excellent answer! –  Bart Patzer Jul 29 '11 at 11:43