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If $G=(V,E)$ is a simple, undirected graph, then $C\subseteq V$ is said to be a vertex cover if $C\cap e\neq \varnothing$ for all $e\in E$.

Is there an infinite graph $G=(V,E)$ such that for any vertex cover $C$ there is a vertex cover $C'\subseteq C$ with $C'\neq C$?

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No, by Zorn's Lemma!

It suffices to check that the intersection of a chain of vertex covers is a vertex cover. If the intersection $C$ fails to be a vertex cover, then there is some edge $(v,w)$ such that neither $v$ nor $w$ is in $C$. But then both $v$ and $w$ are excluded at some point in the chain, so not every set in the chain is a vertex cover.

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  • $\begingroup$ Thanks for your argument, somehow I was convinced that the intersection argument wouldn't work - and thanks to you, I learned something today! $\endgroup$ – Dominic van der Zypen Mar 3 at 19:25
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    $\begingroup$ This argument is almost identical to an exercise I assigned recently in my commutative algebra class: every nonzero commutative ring has a minimal prime ideal. $\endgroup$ – Alex Kruckman Mar 3 at 19:26
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A set $C\subseteq V$ is a minimal vertex cover of the graph $G=(V,E)$ if and only if the complement $V\setminus C$ is a maximal independent set; the existence of a maximal independent set is a straightforward consequence of Zorn's lemma.

Here is an alternative proof of the fact that every graph $G=(V,E)$ has a minimal vertex cover, using the well-ordering theorem instead of Zorn's lemma.

Let $\lt$ be a well-ordering of $V$. Define a subset $C\subseteq V$ recursively so that $$v\in C\iff\{u\in V:u\lt v,\ uv\in E\}\not\subseteq C.$$ Plainly $C$ is a minimal vertex cover of $G$.

If we want the minimal vertex cover $C$ to be contained in a given vertex cover $C_0$, we choose a well-ordering in which every element of $V\setminus C_0$ precedes every element of $C_0$, and apply the same construction.

A similar construction works for hypergraphs with (nonempty) finite edges; just define $$v\in C\iff\exists e\in E\ [\max e=v,\ (e\setminus\{v\})\cap C=\emptyset].$$

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  • $\begingroup$ Can't the argument also been taken to hypergraphs with arbitrary non-empty edges that are not necessarily finite? I mean the argument about a maximal independent set - which, as you write, converts to an argument about a minimal covering set. $\endgroup$ – Dominic van der Zypen Mar 4 at 8:33
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    $\begingroup$ A hypergraph with infinite edges need not have a minimal vertex cover or a maximal independent set. Counterexamples abound. If $V$ is an infinite set and the edges are the infinite subsets of $V$, then the independent sets are the finite subsets of $V$. $\endgroup$ – bof Mar 4 at 9:04

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