# Graph theory: Proof that if the graph G(V1V2,E1E2) is conntected then the intersection (V1V2) is not empty.

I'm attempting to prove the following with contradiction. Unfortunately i'm not sure if my deduction is flawless in this one.

Given: $G_1=(V_1,E_1),\quad G_2=(V_2,E_2),\quad G=(V_1\cup V_2,E_1\cup E_2)$

Prove that: $(V_1\cap V_2) \neq \emptyset$ if $G$ is connected.

Assume: $(V_1\cap V_2)=\emptyset$

Then $\forall v \in V_1$ and $w \in V_2$ exists a path between every pair of vertices $v$ and $w$. $\implies$ $G$ is connected $\implies$ (I'm not sure if I can already deduce it) $(V_1\cap V_2)\neq \emptyset$

I know the intersection can't be empty if G is connected. But I have no idea how I can express it in formal mathematical proof. Any help appreciated.

Suppose: $V_1\cap V_2= \emptyset$ and $G$ is connected.
Then $\forall v\in V_1, w \in V_2$ exists a path between every pair of vertices $v$ and $w$.
$\implies v,w \in V_1 \wedge v,w \in V_2$
$\implies V_1 \cap V_2 \neq \emptyset$