TITLE: Conflicting definition of eulerian graph and finite graph?
QUESTION [0 upvotes]: 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.

REPLY [3 votes]: 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.
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