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 $k < n$.
Since $G$ is connected, $V = \ds \bigcup_{n \mathop \in \N} S_n$.
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Furthermore, $S_n$ is finite and non-empty for each $n$.
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But by Countable Union of Finite Sets is Countable, $\ds \bigcup_{n \mathop \in \N} S_n$ is countable, so $V$ is countable.
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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.
$\blacksquare$
Axiom:Axiom of Countable Choice for Finite Sets
This theorem depends on Axiom:Axiom of Countable Choice for Finite Sets, by way of Countable Union of Finite Sets is Countable.
Although not as strong as the Axiom of Choice, Axiom:Axiom of Countable Choice for Finite Sets is similarly independent of the Zermelo-Fraenkel axioms.
As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.