I want to prove the following claim:
A connected graph, $G=(V,E)$ with $|V|\geq 4$ and without a $K_4$-minor has a cut-set of cardinality at most $2$.
For the proof, I divided the problem to multiple cases:
- $G$ is a tree
- $G$ is a circuit
- G includes a circuit
I could easily show the first case, but stuck at the second case. How can we prove the part when $G$ is itself a circuit, with at least $4$ notes, and without a $K_4$ minor?
Proof: Let $G$ be a $3$-connected graph with no $K_4$ as subgraph. Assume two distinct vertices $u,v \in V(G)$. $G$ is $3$-connected so there exist three disjoint paths $P$,$Q$ and $R$ from $u$ to $v$, and vertices $p \in V(P)−\{u, v\}$ and $q \in V(Q) − \{u, v\}$. By connectivity there exists a path $S$ from $p$ to $q$ in $G-\{u,v\}$. Now let S be the shortest such path. If $z$ in the interior of $S$ belongs to $V(P) \cup V(Q) \cup V(R)$, then $S$ is not a shortest path, and a subpath of S with ends $p$ and $z$ or $q$ and $z$ contradicts the choice of $S$. therefore $S$ is internally disjoint from $P$, $Q$, and $R$. Now $P \cup Q \cup R \cup S$ is a $K_4$-subdivision in $G$, and it's contradiction!
Therefore $G$ has a cut-set of cardinality at most $2$