Characteristics of Eulerian Graph

Theorem
A finite graph is Eulerian if and only if it is connected and each vertex is even.

Proof
First, we prove that any Eulerian graph is connected and has only even vertices.

Suppose that $$G$$ is Eulerian.

Then $$G$$ contains a circuit that uses each vertex and passes through each edge exactly once.

Since a circuit must be connected, $$G$$ is connected.

Beginning at a vertex $$v$$, follow the Eulerian circuit through the graph.

As the circuit passes through each vertex, it uses two edges: one going to the vertex and another leaving.

Each edge is used exactly once, so each of the vertices must be even. Since the circuit must also end at $$v$$, so $$v$$ is also even.

To prove the converse, suppose that a graph $$G$$ is connected and its vertices all have even degree.

If there is more than one vertex in the graph, then each vertex must have degree greater than 0.

Begin at a vertex $$v$$.

Since the graph is connected, there must be an edge $$\{v,v_1\}$$ for some vertex $$v_1\neq v$$.

Since $$v_1$$ has even degree greater than 0, there is an edge $$\{v_1,v_2\}$$.

These two edges make a trail from $$v$$ to $$v_2$$.

Continue this trail, leaving each vertex on an edge that was not previously used, until returning to $$v$$.

Call the circuit formed by this process $$C_1$$.

If $$C_1$$ covers all the edges of $$G$$, then we are done.

Otherwise, remove $$C_1$$ from the graph, leaving the graph $$G_o$$.

The remaining vertices are still even, and since $$G$$ is connected there is some vertex $$u$$ in both $$G_o$$ and $$C_1$$.

Repeat the same process as before, beginning at $$u$$.

The new circuit, $$C_2$$, can be added to $$C_1$$ by starting at $$v$$, moving along $$C_1$$ to $$u$$, travelling around $$C_2$$ back to $$u$$ and then along the remainder of $$C_1$$ back to $$v$$.

Repeat this process, adding each new circuit found to create a larger circuits.

Since $$G$$ is finite, this process must end at some point, and the resulting circuit will be an Eulerian circuit.