Return to Answer

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 $x_0\ne x_1$.

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.

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 $A_n\cap\{x_i:i\in\mathbb N\}=\{x_0,\dots,x_n\}$.) Note that, since $\mathcal C$ has no weakly mininal subcover, whenever $x\in X$ and $A\in\operatorname{cov}_\mathcal C(x)$, there is some $y\in X\setminus A$ with $\operatorname{cov}_\mathcal C(y)\subseteq\operatorname{cov}_\mathcal C(x)$.

Choose $x_0\in X$ and choose $A_0\in\operatorname{cov}_\mathcal C(x_0)$. Choose $x_1\in X\setminus A_0$ with $\operatorname{cov}_\mathcal C(x_1)\subseteq\operatorname{cov}_\mathcal C(x_0)$ and choose $A_1\in\operatorname{cov}_\mathcal C(x_1)$. Choose $x_2\in X\setminus A_1$ with $\operatorname{cov}_\mathcal C(x_2)\subseteq\operatorname{cov}_\mathcal C(x_1)$ and choose $A_2\in\operatorname{cov}_\mathcal C(x_2)$.

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 $x_0\ne x_1$.

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 $x_0\ne x_1$.

The answer is yes. I will prove the contrapositive. Suppose $\mathcal C$ is a cover of $X$ with no weakly minimal subcover; I claim that $\mathcal C$ is nonlinear. Note that, since $\mathcal C$ has no weakly mininal subcover, whenever $x\in X$ and $A\in\operatorname{cov}(x)$, there is some $y\in X\setminus A$ with $\operatorname{cov}(y)\subseteq\operatorname{cov}(x)$.

Choose $x_0\in X$ and choose $A_0\in\operatorname{cov}(x_0)$. Choose $x_1\in X\setminus A_0$ with $\operatorname{cov}(x_1)\subseteq\operatorname{cov}(x_0)$ and choose $A_1\in\operatorname{cov}(x_1)$. Choose $x_2\in X\setminus A_1$ with $\operatorname{cov}(x_2)\subseteq\operatorname{cov}(x_1)$ and choose $A_2\in\operatorname{cov}(x_2)$.

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 $x_0\ne x_1$.

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 $A_n\cap\{x_i:i\in\mathbb N\}=\{x_0,\dots,x_n\}$.) Note that, since $\mathcal C$ has no weakly mininal subcover, whenever $x\in X$ and $A\in\operatorname{cov}_\mathcal C(x)$, there is some $y\in X\setminus A$ with $\operatorname{cov}_\mathcal C(y)\subseteq\operatorname{cov}_\mathcal C(x)$.

Choose $x_0\in X$ and choose $A_0\in\operatorname{cov}_\mathcal C(x_0)$. Choose $x_1\in X\setminus A_0$ with $\operatorname{cov}_\mathcal C(x_1)\subseteq\operatorname{cov}_\mathcal C(x_0)$ and choose $A_1\in\operatorname{cov}_\mathcal C(x_1)$. Choose $x_2\in X\setminus A_1$ with $\operatorname{cov}_\mathcal C(x_2)\subseteq\operatorname{cov}_\mathcal C(x_1)$ and choose $A_2\in\operatorname{cov}_\mathcal C(x_2)$.

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 $x_0\ne x_1$.