Locally Finite Connected Graph is Countable

Theorem
Let $G = \struct {V, E}$ be a graph which is connected and locally finite.

Then $G$ has countably many vertices and countably many edges.

Proof
We first show that $V$ is countable.

If $V$ is finite, then it is surely countable.

Suppose instead that $V$ is infinite.

Choose an arbitrary vertex $q \in V$.

Recursively define a sequence $\sequence {S_n}$:


 * Let $S_0 = \set q$.


 * Let $S_{n + 1}$ be the set of all vertices that are adjacent to some element of $S_n$ but not adjacent to any element of $S_k$ for $k < n$.


 * That is, $S_n$ is the set of vertices whose shortest path(s) to $q$ have $n$ edges.

Since $G$ is connected, $V = \ds \bigcup_{n \mathop \in \N} S_n$.

Furthermore, $S_n$ is finite and non-empty for each $n$.

But by Countable Union of Finite Sets is Countable, $\ds \bigcup_{n \mathop \in \N} S_n$ is countable, so $V$ is countable.

By Cartesian Product of Countable Sets is Countable, $V \times V$ is countable.

Let $E' = \set {\tuple {a, b}: \set {a, b} \in E}$.

By Subset of Countable Set is Countable, $E'$ is countable.

Let $g: E' \to E$ be defined by $\map g {a, b} = \set {a, b}$.

Then $g$ is surjective.

Thus by Surjection from Natural Numbers iff Countable and Composite of Surjections is Surjection, $E$ is countable as well.