(Weakly) minimal subcovers of linear covers

Question

Motivation. The starting point of this question is the trivial observation that if we cover $\mathbb{N}$ with $$\big\{\{0,\ldots n\}: n\in \mathbb{N}\big\},$$ then this cover doesn't have a minimal subcover (from which no more sets can be taken away). So in this question we focus on covers in which the members of the cover have minimum overlap.

Formal setting. Let $X\neq \emptyset$ be a set. We say that ${\cal C}\subseteq {\cal P}(X)$ is a cover if $\bigcup {\cal C} = X$, and we call ${\cal C}$ linear if $|A \cap B| \leq 1$ whenever $A\neq B \in {\cal C}$. If $x_0\in X$, we let the covering number of $x_0$ be $\text{cov}_{\cal C}(x_0) = |\{A \in {\cal C}: x_0\in A\}|$.

We say ${\cal C}$ is minimal if for all $A \in {\cal C}$ the set ${\cal C}\setminus \{A\}$ is no longer a cover. Note that ${\cal C}$ is minimal if for all $A\in {\cal C}$ there is $a\in A$ with $\text{cov}_{\cal C}(a) = 1$. We call ${\cal C}$ weakly minimal if there is $x_0\in X$ such that $\text{cov}_{\cal C}(x_0) = 1$.

Question. If $X$ is an infinite set and ${\cal C}$ is a linear cover of $X$, is there necessarily a weakly minimal cover ${\cal C}'\subseteq {\cal C}$?

Answer 1

The answer is yes. I will prove the contrapositive. Suppose $\mathcal C$ is a cover of a nonempty set $X$ with no weakly minimal subcover; I claim that $\mathcal C$ is nonlinear. (In fact there is an infinite sequence of distinct points $x_n\in X$ and corresponding sets $A_n\in\mathcal C$ such that $A_n\cap\{x_i:i\in\mathbb N\}=\{x_0,\dots,x_n\}$.)

For $x\in X$ let $\mathcal C_x=\{A\in\mathcal C:x\in A\}$. Note that, since $\mathcal C$ has no weakly mininal subcover, whenever $x\in X$ and $A\in\mathcal C_x$ there is some $y\in X\setminus A$ with $\mathcal C_y\subseteq\mathcal C_x$.

Choose $x_0\in X$ and choose $A_0\in\mathcal C_{x_0}$. Choose $x_1\in X\setminus A_0$ with $\mathcal C_{x_1}\subseteq\mathcal C_{x_0}$ and choose $A_1\in\mathcal C_{x_1}$. Choose $x_2\in X\setminus A_1$ with $\mathcal C_{x_2}\subseteq\mathcal C_{x_1}$ and choose $A_2\in\mathcal C_{x_2}$.

Now $A_1,A_2\in\mathcal C$, and $A_1\ne A_2$ since $x_2\in A_2\setminus A_1$, and $|A_1\cap A_2|\ge2$ since $x_0,x_1\in A_1\cap A_2$ and $x_0\ne x_1$.

P.S. The lemma in my partial answer to the subsequent question Does every linear cover contain a minimal cover? is a stronger form of the proposition proved here.