Let $G$ be a connected graph containing a cut-vertex $v$ and let $G1$ be a component of $G − v$.
Show that the induced subgraph $G[V(G_1) \{v\}]$ of $G$ is connected.
I'm a bit stuck on where to start in this.
My initial proof was the following:
We know that $G$ is a connected graph and $G_1$ is a component. Let $S=V(G_1) \cup \{v\}$.
Then the induced subgroup $G[S]$ contains all the vertices in $G$, so $G[S]$ is connected.
I think the proof is wrong but I don't know where I went wrong. Any hints or how to do it would be great
Best Answer
Assumptions
Since $v$ is a cut-vertex, then by definition we must have (the number of components of graph $G$) $\kappa(G - v) > 1 = \kappa(G)$. Now let $G_{1}$ be some arbitrary component of $G - v$. Clearly there is a vertex $u \in G_{1}$ such that $uv \in E(G)$.
Notice that $G_{1}$ is connected; for if it were otherwise, then $G$ would be disconnected to begin with; which contradicts our assumption.
Now, let $H = G[V(G_{1}) \cup \{v\}]$. Then $H$ is an induced subgraph such that every pair of distinct vertices is adjacent in $H$ $\iff$ they are adjacent in $G$.
It follows that since $G_{1}$ is connected and $uv \in E(G)$, then $uv \in E(H)$.
Hence $H$ is connected. $\square$