Diestel Proposition 1.2.2: $\delta(H)>\epsilon(H)\ge\epsilon(G)$ This is straight from the fifth edition. I expanded a little to help my understanding. I'm having a hard time understanding the last lines. I've separated them from the body of the proof. How do we know none of our $G_i$ are trivial? I understand why $H$ is then non-empty, but I am not seeing the logic of the previous statement.
($\epsilon(G)=(\frac{1}{2})d(G)$, where $d(G)$ is the average degree of $G$)
Prop 1.2.2: Every $G$ with at least one edge has a sub-graph, $H, \ni  \delta(H)>\epsilon(H)\ge\epsilon(G)$
Construct a sequence $G=G_0\supset G_1\supset \dots$ of induced subgraphs of $G$, $\ni$ if $G_i$ has a vertex $d(v_i)\le\epsilon(G_i)$, let $G_{i+1}:=G_i-v_i$. If not, terminate the sequence and set $H:=G_i$. By choice of $v_i$, we have $\epsilon(G_{i+1})\ge\epsilon(G_i)$, $\forall i$ and then $\epsilon(H)\ge\epsilon(G)$. Now to show $\delta(H)>\epsilon(H)$, observe that the average degree of the complete graph on one vertex is 0 (there are no edges) and then $\epsilon(K^1)<\epsilon(G)$ ($G$ has an edge)
$\implies$ no $G_i$ is trivial, then $H\ne\varnothing$. And since by our construction, H has no suitable vertex for deletion we have, $\delta(H)>\epsilon(H)$
Thanks
 A: Since $G$ has an edge, we have that $\epsilon(G_0) = \epsilon(G) > 0$. In other words, the edge density in $G_0$ is greater than 0.
Since we are removing 1 vertex but at most $\epsilon(G_i)$ edges in constructing $G_{i+1}$ from $G_i$, we get that each iteration from $G_i$ to $G_{i+1}$ either increases the edge density, or leaves it the same, so every $G_i$ must have non-zero edge density. The edge density of the trivial graph is $0$, so no $G_i$ can be trivial.
A: I am reading "Graph Theory 5th Edition" by Reinhard Diestel.
$G=:G_0\supset G_1\supset\dots\supset H.$
$1\leq|E(G_0)|$ by our assumption.
Assume that $1\leq |E(G_i)|$.
If $G_i$ has a vertex $v_i$ of degree $d(v_i)\leq\epsilon(G_i)$, then $d(v_i)\leq\epsilon(G_i)=\frac{|E(G_i)|}{|V(G_i)|}<|E(G_i)|.$
So, in this case, $1\leq |E(G_i)|-d(v_i)=|E(G_{i+1})|.$
Since $|E(G)|=|E(G_0)|$ is finite, $G_i$ eventually has no vertex $v_i$ of degree $d(v_i)\leq\epsilon(G_i)$ for some $i$.
When $G_i$ has no vertex $v_i$ of degree $d(v_i)\leq\epsilon(G_i),$ then $H=G_i$.
And $1\leq |E(G_i)|=|E(H)|.$
So,
$1\leq |E(G)|=|E(G_0)|.$
$1\leq |E(G_1)|.$
$\dots$
$1\leq |E(H)|.$
