Conflicting definition of eulerian graph and finite graph?

Question

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 finite when both E(G) and V (G) are finite sets.

Theorem 1.1. A finite graph G has an even number of vertices with odd valency.

Reading it further, there is:

Theorem 1.2. A finite 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.

Answer 1

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

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

What the theorem states actually is:

A finite 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.

Answer 2

Where does it say that a finite graph has an odd valency at each vertex? I see it saying the number of such vertices is even. Note that zero is an even number. There is no contradiction here.