Proving that the Union of Two Paths with a Connected Intersection Contains at Least One Circuit

The question of how to prove that if the intersection of two paths in a graph is a connected subgraph, then the union of the two paths contains at least one circuit is a fundamental concept within graph theory.

Definitions and Key Concepts

Before delving into the proof, let's establish some key definitions:

Path: A path in a graph is a sequence of vertices such that each consecutive pair of vertices in the sequence is connected by an edge. Intersection: The intersection of two paths, denoted as ( P_1 cap P_2 ), is the set of vertices that are common to both paths. Circuit: A circuit or cycle is a path that starts and ends at the same vertex without retracing any edges.

In the context of this problem, we have two paths ( P_1 ) and ( P_2 ) in a graph ( G ) with a connected intersection ( P_1 cap P_2 ).

Proof Outline

To prove that if ( P_1 cap P_2 ) is a connected subgraph, then the union ( P_1 cup P_2 ) contains at least one circuit, we can follow these steps:

Paths and Intersection

Since ( P_1 ) and ( P_2 ) are paths, they are both connected by definition. The intersection ( P_1 cap P_2 ) is given to be connected, meaning there exists a path within ( P_1 cap P_2 ) that connects any two vertices in this intersection.

Connected Components

The connectedness of ( P_1 cap P_2 ) implies there is a path between any two vertices in this intersection. This is a crucial property for our proof.

Vertices in the Union

Consider the vertices in the union ( P_1 cup P_2 ). Since both paths intersect, they share at least one vertex, say ( v ).

Exploring Paths

Take any two vertices ( u ) and ( v ) in ( P_1 cap P_2 ). From ( u ) to ( v ) we can travel along ( P_1 ). From ( v ) to ( u ) we can travel along ( P_2 ).

Forming a Circuit

By traveling from ( u ) to ( v ) along ( P_1 ) and returning from ( v ) to ( u ) along ( P_2 ), we form a closed loop:

This path ( u to v to u ) constitutes a circuit.

Since we can find at least one circuit formed through the connected intersection of the two paths, we conclude that the union ( P_1 cup P_2 ) contains at least one circuit.

Summary

Therefore, if the intersection of two paths is a connected graph, then their union must contain at least one circuit as demonstrated by the ability to traverse between common vertices through both paths.

Further Considerations

It is important to note that the question is slightly ambiguous. If the intersection of two paths is disconnected, it does not imply the existence of a circuit. For example, consider a path ( P_8 ) with vertices labeled ( v_1 ) to ( v_8 ) taken anticlockwise. Let ( P_1 v_2 - v_3 - v_4 ) and ( P_2 v_4 - v_5 - v_6 ). The intersection of ( P_1 ) and ( P_2 ) is just ( v_4 ), which is trivially disconnected, but their union does not contain any cycle.

However, if the intersection contains more than one vertex, say ( ab ), then you can follow ( P_1 ) from ( a ) to ( b ) in the union graph and then follow ( P_2 ) from ( b ) to ( a ) to form a cycle.

If the intersection contains more than two vertices, you need to choose ( ab ) such that both vertices lie in different components in the intersection.

Understanding these nuances is crucial for grasping the intricacies of circuit formation in graph theory and the implications of path intersections.