Let $A \subset \Bbb{Z}$. $A$ is a called an anti-ideal when
- $x, y \in A \implies x - y \notin A$.
- $z \in \Bbb{Z}, z \neq \pm 1, a \in A \implies za \notin A$.
Matroids are something in graph theory that becomes important when studying greedy algorithms for instance. But there are many applications and the theory of matroids has books written about it. I know because I own one called "Matroid Theory" by D.J.A. Welsh, but I don't read it, lol.
Conjecture. The collection $\mathcal{A}$ of anti-ideals of $\Bbb{Z}$ form a matroid.
Proof. $\varnothing \in \mathcal{A}$ satisfies 1. & 2. vacuously. And clearly $\mathcal{A}$ is closed under taking subsets. Thus all we need to prove is that if $A, B \in \mathcal{A}$ are anti-ideals such that $|A| \gt |B|$, then there exists $x \in A \setminus B$ such that $B \cup \{x\} \in \mathcal{A}$. This is true since if no such $x$ exists then we can say that either $A - B = \{ a - b : a \in A, b \in B\} \subset B$ or $z A \subset B$ for all $z \in \Bbb{Z}$ each of which implies $|A| \leq |B|$ since $f(x) = x - b$ and $g(x) = zx : A \to B$ are always injective. $\square$
Can you please verify this argument for me?
Thanks! :D
Corollary. All maximal anti-ideals are in bijection.
By the way, the set of of all odd primes is an anti-ideal.
There is another way for no such $x$ to exist: if $A \subset \{zb: z \in \Bbb Z_{\text{non_unit }}, b \in B\} \cup B$. This tells you nothing about the relative sizes of $A$ and $B$; in particular, if $B = \{-1, 1\}$ then $\{zb: z \in \Bbb Z_{\text{non_unit }}, b \in B\} = \Bbb Z \setminus B$.