Let $ G_1 = (V_1, E_1), G_2 = (V_2, E_2)$ such that...
- G1 is connected
- G2 is connected
- $V_1\cap V_2={v_o}$
I know since the intersection of both vertex sets only contains a single vertex, implying any path including vertices from $G_1$ and $G_2$ will not include duplicates. How do I formalize this notion to show every vertex in $G_3$ is connected to every other vertex (i.e.: there exists a path between every other vertex).
Let $u$ and $v$ in $V_3=V_1 \cup V_2$. If $u,v$ are both in $V_1$, then there is a path from $u$ to $v$ in $(V_1,E_1)$ and this is still a path from $u$ to $v$ in $(V_3,E_3)$ (as $E_1 \subseteq E_3$).
If $u,v$ are both in $V_2$, mutatis mutandis the same argument applies in $(V_2,E_2)$.
Finally we have (WLOG) $u \in V_1$, $v \in V_2$. Then there is a path from $u$ to $v_0$ in $(V_1,E_1)$ and a path from $v_0$ to $v$ in $(V_2,E_2)$. Combining these paths we have a path in $(V_3,E_3)$ from $u$ to $v$.