I am having problems understanding the definition of graph $k$-degeneracy. The problem has been asked multiple times in this board, but I cant find a explanation that clarifies the problem for me.
So, let me walk you through my problem with understanding.
First, we have the definition of $k$-degeneracy:
A graph $G$ is said to be $k$-degenerate if every subgraph of $G$ has a vertex of degree at most $k$. (from Definition of degeneracy)
and similar:
A $k$-degenerate graph is an undirected graph in which every induced subgraph has minimum degree at most $k$. The degeneracy of a graph is the smallest $k$ for which it is $k$-degenerate. (from https://en.wikipedia.org/wiki/Glossary_of_graph_theory_terms#degeneracy)
A big problem I have here is the part with "every subgraph". Lets just quickly check the definition of "subgraph":
A subgraph of a graph $G$ is another graph formed from a subset of the vertices and edges of $G$. The vertex subset must include all endpoints of the edge subset, but may also include additional vertices [...]. (from https://en.wikipedia.org/wiki/Glossary_of_graph_theory_terms#subgraph)
or, similar:
A subgraph of a graph $G=(V,E)$ is a graph $G'=(V',E')$ such that $E'$ is a subset of $E$, $G'$ is a subset of and all the vertices connected by the edges in $E'$ are in the subset $G'$. (from Definition of a subgraph)
So, if $k$-degeneracy is the number $k$ of a graph, from which EVERY (induced) subgraph has at most the degree $k$, what about the (ALWAYS existing) minimal subgraph of ZERO vertices and ZERO edges?
Formally: $G_{sub} := (V_{sub}, E_{sub})$, with $V_{sub} = \emptyset$ and $E_{sub} = \emptyset$
$\Delta(G_{sub})$ of this subgraph is zero. Same goes if the subgraph has only ONE vertex. As it always exists, it is ONE OF the (induced) subgraphs of G. As $\Delta(G_{sub})$ is zero, $\delta(v)$ is zero for any $v \in V_{sub}$, hence the degeneracy is also always zero. For all graphs.
Obviously, I am missing a crucial part of the definition here. Anyone here able to point this out to me?
Thanks in advance!!!