Inequality relating connectivity, edge connectivity and minimum degree (Whitney's theorem)

Question

The following is a theorem given in Bella Bollobas's Modern graph theory (Springer 2002) Page-73

If $G$ is nontrivial (that is, has at least two vertices), then the parameters $\delta(G)$, $\lambda(G)$ and $\kappa(G)$ satisfy the following inequality: $\kappa(G) \le \lambda(G) \le \delta(G)$, where $\delta(G)$ is the minimum degree of a vertex, $\lambda(G)$ is the edge connectivity, and $\kappa(G)$ is the vertex connectivity.

Proof:

Indeed, if we delete all the edges incident with a vertex, the graph becomes disconnected, so the second inequality holds. To see the other inequality, note first that if $G$ is complete then $\kappa(G) = \lambda(G) = |G| - 1$, and if $\lambda(G) \le 1$ then $\lambda(G) = \kappa(G)$. Suppose now that $G$ is not complete, $\lambda(G) = k \ge 2$, and $\{x_1 y_1, x_2 y_2, \ldots , x_k y_k\}$ is a set of edges disconnecting $G$. If $G - \{x_1, x_2, \ldots , x_k\}$ is disconnected then $\kappa(G) \le k$. Otherwise, each vertex $x_i$ has degree at most $k$ (and so exactly $k$). Deleting the neighbours of $x_1$, we disconnect $G$. Hence $\kappa(G)\le \lambda(G)$.

I did not understand the following part of the proof: "Otherwise, each vertex $x_i$ has degree at most $k$ (and so exactly $k$)" . Why is this true? I would really appreciate an elaborate answer to this.

Answer 1

I don’t see how to justify the statement in question, but I do see how to justify a weaker statement that is still sufficient to prove the theorem.

Suppose that $G-\{x_1,x_2,\ldots,x_k\}$ is connected and and $\deg v_i>k$ for each $i\in\{1,\ldots,k\}$. Then each $v_i$ has at least $k+1$ neighbors in $G$, and at most $k-1$ of them can be in $\{x_1,x_2,\ldots,x_k\}$, so there must be at least two vertices $u_i$ and $v_i$ in $G-\{x_1,x_2,\ldots,x_k\}$ that are neighbors of $x_i$ in $G$. At least one of $u_i$ and $v_i$ must be different from $y_i$; say $u_i\ne y_i$. I claim that this implies that $G-\{x_1y_1,\ldots,x_ky_k\}$ is connected, which is a contradiction.

Let $v$ and $w$ be any two vertices of $G$. If $v,w\notin\{x_1,\ldots,x_k\}$, then there is a path from $v$ to $w$ in $G-\{x_1,\ldots,x_k\}$ and hence in $G-\{x_1y_1,\ldots,x_ky_k\}$. If $v=x_i$ and $w\notin\{x_1,\ldots,x_k\}$, there is a path $P$ from $w$ to $u_i$ in $G-\{x_1y_1,\ldots,x_ky_k\}$, and we can add the edge $u_ix_i$ to $P$ to get a path from $w$ to $x_i$ in $G-\{x_1y_1,\ldots,x_ky_k\}$. Finally, if $v=x_i$ and $w=x_j$ with $i\ne j$, there is a path $P$ from $u_i$ to $u_j$ in $G-\{x_1y_1,\ldots,x_ky_k\}$, and we can extend it with the edges $x_iu_i$ and $u_jx_j$ to get a path from $x_i$ to $x_j$ in $G-\{x_1y_1,\ldots,x_ky_k\}$. This proves the claim.

This contradiction shows that there must be at least one $i\in\{1,\ldots,k\}$ such that $\deg x_i\le k$, and since we already know that $k=\lambda(G)\le\delta(G)\le\deg x_i$, we must have $\deg x_i=k$. Deleting the neighbors of $x_i$ disconnects $G$, so $\kappa(G)\le k=\lambda(G)$.