Majorization of Finitary Infinite Matroids

Question

It is well-known that classical matroid theory doesn't work the same way on infinite underlying sets because of the failure of duality (although recent work gives a fascinating workaround). I'm interested in the classical approach to infinite matroids, namely using the finitary hypothesis. Specifically, I'm interested in studying when one matroid majorizes another and how this relates to the notion of independent subsets.

For concreteness, here is one definition using closure operations.

Definition. Let $S$ be a (not necessarily finite) set. Denote the power set of $S$ by $2^S$. A closure operator on $S$ is a function $D:2^S\to 2^S$ satisfying the following axioms:

1. For all $X\subseteq S$ we have $X\subseteq D(X)$.
2. For all $X\subseteq Y\subseteq S$ we have $D(X)\subseteq D(Y)$.
3. For all $X\subseteq S$ we have $D(D(X))=D(X)$.
4. For all $X\subseteq S$ and $x,y\in S$ if $y\in D(X\cup\{x\})-D(X)$ then $x\in D(X\cup\{y\})$.
5. For all $X\subseteq S$ and $x\in D(X)$, there is a finite subset $X_0\subseteq X$ such that $x\in D(X_0)$.

Given a closure operator $D$ on a set $S$, we say that a subset $X\subseteq S$ is spanning (resp. independent, resp. a basis) if $D(X)=S$ (resp. for all $x\in X$ we have $x\notin D(X- \{x\})$, resp. if it is both spanning and independent). If $S$ is a set with a closure operator $D$, then for each subset $T\subseteq S$, we get a closure operator on $T$ by sending $X\subseteq T$ to $T\cap D(X)$; this closure operator I will denote by $D|_T$. (More generally one can define the notion of the pullback of a closure operator.) Now consider the following proposition, the proof of which is easy.

Proposition. Let $S$ be a set with two closure operators $D, E$. Consider the following conditions on $D$ and $E$.

1. For every subset $X\subseteq S$, we have $E(X)\subseteq D(X)$.
2. For each subset $T\subseteq S$, each $E|_T$-spanning subset of $T$ is $D|_T$-spanning.
3. Each $D$-independent subset of $S$ is $E$-independent.

Then (1) $\iff $(2) $\implies$ (3).

The question I have is: are all these conditions equivalent, i.e. does (3) $\implies$ (1)? This seems reasonable, and we get from this a well-defined partial order of majorization on closure operators on a set. I'm not an expert on matroids/closure operators, so I apologize if this is well-known and would request references then.

Answer 1

Here's a counterexample with two elements.

Let $S = \{a,b\}$.

Define $D(\varnothing) = D(\{a\}) = \{a\}$, $D(\{b\}) = D(S) = S$. (The $D$-closed sets are $\{a\}$ and $S$.)

Define $E(\varnothing) = \varnothing$, $E(\{a\}) = E(\{b\}) = E(S) = S$. (The $E$-closed sets are $\varnothing$ and $S$.)

The only $D$-independent sets are $\varnothing$ and $\{b\}$, and both of these are $E$-independent.

But $E(\{a\}) = S\not\subseteq D(\{a\}) = \{a\}$.

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In the comments, you ask whether two finitary matroids with the same underlying set and the same independent sets are equal.

The answer is yes: If $D$ and $E$ are closure operators on $S$ (so that $(S,D)$ and $(S,E)$ are finitary matroids) and for all $X\subseteq S$, $X$ is $D$-independent if and only if $X$ is $E$-independent, then $D = E$.

Write $\mathcal{I}$ for the set of $D$-independent sets. It suffices to show that $D$ can be defined from $\mathcal{I}$.

I claim that $$D(X) = \{s\in S\mid \exists X_0\subseteq X\text{ with }X_0\in \mathcal{I}\text{ and }X_0\cup \{s\}\notin \mathcal{I}\}.$$

Indeed, if $s\in D(X)$, then by (5), there is a finite $X_0\subseteq X$ with $s\in D(X_0)$. We may assume $X_0$ has minimal size. Then $X_0\in \mathcal{I}$ (otherwise, there is some $x\in X_0$ with $x\in D(X_0\setminus \{x\})$, so $s\in D(X_0\setminus \{x\})$, contradicting minimality), but $X_0\cup \{s\}\notin \mathcal{I}$.

Conversely, suppose there is $X_0\subseteq X$ with $X_0\in \mathcal{I}$ and $X_0\cup \{s\}\notin \mathcal{I}$, and assume for contradiction that $s\notin D(X)$. Then $s\notin D(X_0)$, so there is some $x\in X_0$ with $x\in D((X_0\cup \{s\})\setminus \{x\})$. Since $X_0\in \mathcal{I}$, $x\notin D(X_0\setminus \{x\})$. By exchange, $s\in D(X_0)$, contradiction.