Conflicting definition of eulerian graph and finite graph? I'm reading Van Lint and Wilson: _A Course in Combinatorics_. There is one part of the book where he defines a finite graph: > A graph is nite when both E(G) and V (G) are nite sets. > > **Theorem 1.1.** A nite graph G has an even number of vertices with odd valency. Reading it further, there is: > **Theorem 1.2.** A nite graph G with no isolated vertices (but possibly with multiple edges) is Eulerian if and only if it is connected and every vertex has even degree. It was first said that a finite graph has an odd valency at each vertex, now a finite graph with every vertex with even degree is presented. How should I proceed? Should I drop the condition given in the first theorem (odd valency)? Does Eulerian graph mean a finite graph without this condition? I believe it is, but I want to be sure.

The confusion is that you read the theorem the following way:

> A nite graph G has an even number of vertices, each with odd valency.

What the theorem states actually is:

> A nite graph G has an [even number of vertices with odd valency].

Or, to make it more clear:

**Theorem** In a finite graph, the number of vertices which has an odd valency is even.