Adding edge between 2 trees gives a tree
Dependencies:
Let $G = (V, E)$ be an undirected graph which has 2 connected components, $T_1 = (V_1, E_1)$ and $T_2 = (V_2, E_2)$, which are trees. Let $u \in V_1$ and $v \in V_2$. Let $G'$ be the graph obtained by adding $(u, v)$ to $G$. Then $G'$ is a tree.
Proof
Since $T_1$ is a tree, all vertices in $V_1$ have a path to all other vertices in $V_1$. Since $T_2$ is a tree, all vertices in $V_2$ have a path to all other vertices in $V_2$. Adding the edge $(u, v)$ would create a path from all vertices in $V_1$ to all vertices in $V_2$ (the path goes from a vertex in $V_1$ to $u$, from $u$ to $v$ and from $v$ to a vertex in $V_2$). Therefore, $G'$ is connected.
Suppose adding $(u, v)$ created a cycle in $G'$. This cycle includes $(u, v)$. Therefore, there must exist multiple paths from $u$ to $v$. This is a contradiction, since $[u, v]$ is the only path from $u$ to $v$. Therefore, $G'$ is acyclic.
Since $G'$ is connected and acyclic, it is a tree.
Dependency for:
Info:
- Depth: 3
- Number of transitive dependencies: 6
Transitive dependencies:
- /sets-and-relations/equivalence-classes
- /sets-and-relations/relation-composition-is-associative
- /sets-and-relations/equivalence-relation
- Graph
- Path in Graph
- Tree