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.

5more comments