Question
Show for every $k \in \mathbb N$ that every graph of minimum degree $2k$ has a $(k+1)$-edge-connected subgraph.
Existing post (Edited)
Let $k∈N$ and $G$ be a graph with $δ(G) = 2k$. Show that $G$ has a $(k + 1)$-edge-connected subgraph.
There is already a post for this question, however the solution proves a stronger result rather than proving the statement directly. In this post, the solution proves the statement for any graph with sum of degrees at least $2nk$ which in turn shows that the statement always hold too for graphs with minimum degree at least $2k$.
I was wondering if there is a way to prove the statement more directly?
My attempt
For those of you familiar with the Diestel textbook, I have tried to mimic the proof of Theorem 1.4.3 (Mader 1972). I have linked a photo at the bottom of this post for reference. I have generally outlined my attempt below.
- Put $\gamma = \varepsilon(G)$ and consider the subgraphs $G' \subseteq G$ such that $$ \delta(G) \geq 2k \text{ and } \Vert G' \Vert > \gamma (|G'|-2k) \tag{*}.$$ Such $G'$ must exist as $G$ itself satisfies (*), and so consider $H$ the one of smallest order.
- Suppose $H$ is not $(k+1)$-edge-connected. Then there must exist an edge set $E$ with $|E| \leq k$ such that removing $E$ from $H$ disconnects $H$. Let $X$ be the set of vertices corresponding to the endvertices of $E$.
We know $X$ must have order at most $2k$. This occurs when $E$ has order $k$ and every edge in $E$ introduces two new vertices.
- Let $\{U_1, U_2\}$ be a proper separation of $H$ such that $U_1 \cap U_2 = X$, and put $H_1 = H[U_1]$ and $H_2 = H[U_2]$. I wanted to then show that $\delta(H_1), \delta(H_2) \geq 2k$, and this is where I had problems.
For any vertex $v \in U_1 \setminus U_2$, clearly all its $d(v) \geq \delta(H) \geq 2k$ neighbours lie in $H_1$. Similarly for $v \in U_2 \setminus U_1$.
The problem arises for vertices $v \in X$. A vertex $v \in X$ could have its neighbours lie in $X$ or in $U_1 \setminus U_2$ or $U_2 \setminus U_1$, and so we cannot say that vertices $v \in X \cap U_1$ have all its $d(v)$ neighbours lie in $H_1$. This makes it difficult to claim that vertices in $H_1$ have minimum degree of $2k$ and similarly for $H_2$.
A realisation that made me temporarily hopeful (and may still be useful) is that every vertex $v \in X$ must have its neighbours lie entirely in ($X$ and $U_1 \setminus U_2$) or in (X and $U_2 \setminus U_1$) but not both. This is because if a vertex $v \in X$ had neighbours in both $U_1 \setminus U_2$ and $U_2 \setminus U_1$, then the removal of edge set $E$ would NOT disconnect $H$. I was wondering based off of this fact if there is some other way to choose $U_1$ and $U_2$ to make this argument work.
- Suppose for a moment that the above argument did work. (i.e. we could show that $\delta(H_1), \delta(H_2) \geq 2k$). Then the remaining argument is as follows.
By the minimality of $H$, $H_1$ and $H_2$ cannot also satisfy (*). Thus we must have, $$ \Vert H_i \Vert \leq \gamma(|H_i|-2k), $$ for $i=1,2$.
Then we would have, $$ \Vert H \Vert \leq \Vert H_1 \Vert + \Vert H_2 \Vert \leq \gamma(|H_1|+|H_2|-4k) \leq \gamma(|H|-2k),$$ which contradicts the fact that $H$ satisfies (*). Note the last inequality above holds because $|H|=|H_1|+|H_2|-|H_1 \cap H_2|$ and $|H_1 \cap H_2|$ is at most $2k$.
End of my attempt
Please let me know if you have any ideas that are completely different to my attempt or any ideas on how to finish off what I have started. Also feel free to ask any questions for clarification!
Screenshot of Proof of Theorem 1.4.3:
