Let $G=(V,E)$ be a graph with minimum degree $\delta(G)=n\lt\aleph\_0$. Does $G$ necessarily have a spanning subgraph $G'=(V,E')$ which also has minimum degree $\delta(G')=n$ and is minimal with that property?


This question is easily answered in the affirmative if $G$ is locally finite or if $n\le1$. It already seems difficult for $n=2$, but I am not very clever and may be missing something obvious.


The question also seems to make sense for hypergraphs:


Let $m,n\in\mathbb N$. Let $E$ be a family of sets, each of cardinality at most $m$. If $E$ is an $n$-cover of a set $V$ (each element of $V$ is in at least $n$ elements of $E$), does $E$ contain a minimal $n$-cover of $V$?


I would expect such simple questions to have been asked and answered 100 years ago.


**Where are these questions considered in the literature?**


  
**P.S.** The following proof for the simple case of a (non-hyper) graph with $\delta=1$ is probably a dead end, as it does not seem to generalize in any obvious way. I'm putting it here anyway because it's quite simple.


  
**Theorem.** A graph with no isolated points has a minimal spanning subgraph with no isolated points.


  
**Proof.** Let $G$ be a graph with no isolated points. Let $H$ be a maximal spanning subgraph of $G$ not containing $K\_3$ or $P\_4$ as a subgraph, induced or otherwise. Then $H$ is a star-forest, possibly with some isolated points. For each isolated vertex $v$ of $H$, choose an edge of $G$ which is incident with $v$ and add it to $H$. This results in a spanning subgraph of $G$ in which each component is a nontrivial tree of radius at most $2$.The proof is completed by observing that any nontrivial tree of radius at most $2$ has a minimal spanning subgraph with no isolated points.



Graphs with minimum degree $\delta(G)\lt\aleph_0$
[Fedor Petrov](https://mathoverflow.net/users/4312/fedor-petrov) pointed out in a comment that the hypergraph question for $n=1$ was settled nicely by [Taras Banakh](https://mathoverflow.net/users/61536/taras-banakh) in [his answer](https://mathoverflow.net/questions/308265/minimal-covers-in-hypergraphs-with-finite-edges/308417#308417) to [Dominic van der Zypen](https://mathoverflow.net/users/8628/dominic-van-der-zypen)'s question [Minimal covers in hypergraphs with finite edges](https://mathoverflow.net/questions/308265/minimal-covers-in-hypergraphs-with-finite-edges), and he mentioned some ideas for generalizing Banakh's argument to $n$-covers. This is my attempt to answer the general question in the affirmative using the ideas of Banakh and Petrov.


In order for the induction to work, it seems necessary to work with multihypergraphs (hypermultigraphs?) instead of simple hypergraphs. Therefore I will consider a hypergraph as a triple $H=(V,E,I)$ consisting of a vertex set $V$, an edge set $E$, and a vertex-edge incidence relation $I\subseteq V\times E$. For any vertex $v\in V$, let $E\_v=\{e\in E:(v,e)\in I\}$, the set of all edges incident with $v$. For any edge $e\in E$, let $V\_e=\{v\in V:(v,e)\in I\}$, the set of all vertices incident with $e$.


  
**Banakh–Petrov Theorem.** Consider a hypergraph $(V,E,I)$. Let $m\in\omega$ and $\varphi:V\to\omega$ satisfy the conditions:
$$\forall e\in E\ |\{v\in V\_e:\varphi(v)\gt0\}|\le m;$$
$$\forall v\in V\ |E\_v|\ge\varphi(v).$$
Then there is a set $E'\subseteq E$ such that:
$$\forall v\in V\ |E'\cap E\_v|\ge\varphi(v);$$
$$\forall e\in E'\ \exists v\in V\_e\ |E'\cap E\_v|=\varphi(v).$$


  
**Proof.** We use induction on $m$. The case $m=0$ is trivial, so we assume that $m\gt0$ and that the theorem holds with $m$ replaced by $m-1$. We may assume without loss of generality that $\varphi(v)\gt0$ for all $v\in V$.


By Zorn's lemma there is a set $D\subseteq E$ which is maximal with the property that
$\forall v\in V\ |D\cap E\_v|\le\varphi(v)$.


We will apply the induction hypothesis to the hypergraph $(V,F,J)$ where $F=E\setminus D$ and $J=I\cap(V\times F)$, and the function $\psi:V\to\omega$ defined by setting $\psi(v)=\varphi(v)-|D\cap E\_v|$.


First, if $e\in F$, then by the maximality of $D$ there is a vertex $v\in V\_e$ such that $|D\cap E\_v|=\varphi(v)$. Hence $\psi(v)=0\lt\varphi(v)$, so that $|\{v\in V\_e:\psi(v)\gt0\}|\le|\{v\in V\_e:\varphi(v)\gt0\}|-1\le m-1$.


Second, if $v\in V$, then $|D\cap E\_v|+|F\cap E\_v|=|E\_v|\ge\varphi(v)$, so that $|F\cap E\_v|\ge\varphi(v)-|D\cap E\_v|=\psi(v)$.


Therefore, by the inductive hypothesis, there is a set $F'\subseteq F$ such that
$$\forall v\in V\ |F'\cap E\_v|\ge\psi(v)=\varphi(v)-|D\cap E\_v|;$$
$$\forall e\in F'\ \exists v\in V\_e\ |F'\cap E\_v|=\psi(v)=\varphi(v)-|D\cap E\_v|.$$


Let $C=D\cup F'$; then we have
$$\forall v\in V\ |C\cap E\_v|\ge\varphi(v);$$
$$\forall e\in F'\ \exists v\in V\_e\ |C\cap E\_v|=\varphi(v).$$


Let us call a subset $S\subseteq D$ *bad* if $|(C\setminus S)\cap E\_v|\lt\varphi(v)$ for some $v\in V$, *good* otherwise. Since $D\cap E\_v$ is finite for each $v\in V$, every bad subset of $D$ contains a finite bad set. Therefore, by Zorn's lemma, there is a maximal good set $S\subset D$. The set $E'=C\setminus S$ has the desired properties.


