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The answer is Yes. Suppose $M$ is minimal, but not strongly minimal. Then there is $S\subseteq M, S \neq \emptyset$ and $K\subseteq E$ such that $\bigcup K \supseteq \bigcup S$; $\text{card}(K) < …

We say that $C\subseteq E$ is a covering if $\bigcup C = V$, and we set $$\text{mult}(C) = \{v\in V:|\{e\in C: v\in e\}| >1 \}.$$ Given a covering $C\subseteq E$, is there a covering $C_0\subseteq E$ … $\text{mult}(C_0)\subseteq \text{mult}(C)$, and for every covering $D\subseteq E$ with $\text{mult}(D) \subseteq \text{mult}(C_0)$ we have $\text{mult}(D)=\text{mult}(C_0)$. …

If $G=(V,E)$ is a simple, undirected graph, $C\subseteq V$ is said to be a vertex cover if for every $e\in E$ we have $C\cap e \neq \emptyset.$ If $G=(V,E) $ is infinite, is there necessarily a vertex …

Let $X\neq \emptyset$ be a set. We say $C \subseteq {\cal P}(X)$ is a cover of $X$ if $\bigcup C = X$. For covers $C, D$ of $X$ we say that $C$ is more efficient than $D$ if $|C\setminus D| < |D \setm …

Let $X$ be an infinite set, and let ${\cal A}\subseteq{\cal P}(X)$ be a family of non-empty sets. We say $S\subseteq X$ is a cover for ${\cal A}$ if $A\cap S \neq \emptyset$ for all $A\in{\cal A}$. S …

Motivation. The following has a real-life (!) inspiration from a discussion about how to connect lamps and switches in an efficient way. Question. Let $n\in\mathbb{N}$ be a positive integer and let $\ …

Let $(L,\land,\lor)$ be a complete distributive lattice. Given $x\neq y \in L$, is there a finite set ${\cal I}$ of closed intervals in $L$ such that no member of ${\cal I}$ contains both $x$ and $y …

Let $X\neq \emptyset$ be a set. We say ${\cal C} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ is a cover of $X$ if $\bigcup {\cal C} = X$. A subset $S\subseteq X$ is a choice set for ${\cal C}$ if $|S …

A hypergraph $H=(V,E)$ consists of an non-empty set $V$ and a collection $E\subseteq {\cal P}(V)\setminus \{\emptyset\}$ of non-empty subsets of $V$. A transversal of $H$ is a set $T\subseteq V$ such …

Let $G=(V,E)$ be a simple, undirected graph with $\bigcup = E$ (that is, there are no isolated vertices). We say that $C\subseteq E$ is an edge cover of $G$ if $\bigcup C = V$. For any edge cover $C$ …

A boring example is the set of all constant functions which always has the same cardinality as $X$ - and taking away one member of the collection of all constant functions destroys the "covering" property …

Let $H=(V,E)$ be a hypergraph, that is $V$ is a set and $E\subseteq \mathcal{P}(V)$. We say that $C\subseteq E$ is a cover of $H$ if $\bigcup C = V$. A cover $M\subseteq E$ is said to be strongly min …

If $G=(V,E)$ is a simple, undirected graph, then $C\subseteq V$ is an edge cover if $C\cap e \neq \emptyset$ for all $e\in E$. The "best" covers in some sense are subsets $C\subseteq V$ that meet eve …

If $x_0\in X$, we let the covering number of $x_0$ be $\text{cov}_{\cal C}(x_0) = |\{A \in {\cal C}: x_0\in A\}|$. …

This is a follow-up question to an older question. 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 $| …