n-Colorability is Equivalent to Finite n-Colorability (A Formal Logic Proof)

Warning: this proof requires a bit of familiarity with the terminology of propositional logic and graph theory.

Problem: Let G be an infinite graph. Show that G is n-colorable if and only if every finite subgraph G_0 \subset G is n-colorable.

Solution: One of the many equivalent versions of the Compactness Theorem for the propositional calculus states that if \Sigma \subset \textup{Prop}(A), where A is a set of propositional atoms, then \Sigma is satisfiable if and only if any finite subset of \Sigma is satisfiable. (Recall that a set of propositions is satisfiable if for some truth valuation t:A \to \left \{ 0,1 \right \} the unique extension \hat{t}:\Sigma \to \left \{ 0,1 \right \} satisfies \hat{t}(p) = 1 for all p \in \Sigma. The function t is called a model for \Sigma). This is equivalent to the Completeness Theorem, which says that if one can use \Sigma to prove p, then every model of \Sigma satisfies \hat{t}(p) = 1. Both are fundamental results in logic.

And so we will convert this graph coloring problem into a logical set of propositions, and use the Compactness Theorem against it. We want a set of propositions \Sigma which has a model if and only if the corresponding graph is n-colorable. Then we will use the n-colorability of finite subgraphs, to show that all finite subsets of \Sigma have models, and this implies by the compactness theorem that \Sigma has a model, so the original infinite graph is n-colorable.

We may think of a coloring of a graph G as a function on the set of vertices: c:V \to \left \{ 1, 2, \dots, n \right \}. Define our set of propositional atoms as A = V \times \left \{ 1, 2, \dots, n \right \}. In other words, we identify a proposition p_{v,i} to each vertex and possible color. So we will define three sets of propositions using these atoms, which codify the conditions of a valid coloring:

  • \left \{ p_{v,1} \vee p_{v,2} \vee \dots \vee p_{v,n} : v \in V \right \} i.e. every vertex must have some color,
  • \left \{ \lnot (p_{v,i} \wedge p_{v,j}) : i,j = 1, \dots, n, i \neq j, v \in V \right \} i.e. no vertex may have two colors, and
  • \left \{ \lnot (p_{v,i} \wedge p_{w,i}) : \textup{whenever } (v,w) \textup{ is an edge in } G \right \} i.e. no two adjacent vertices may have the same color.

Let \Sigma be the union of the above three sets. Take \Sigma_0 \subset \Sigma to be any finite set of the above propositions. Let V_0 be the finite subset of vertices of G which are involved in some proposition of \Sigma_0 (i.e., p_{v,i} \in \Sigma_0 if and only if v \in V_0). Since every proposition involves finitely many atoms, V_0 is finite, and hence the subgraph of vertices of V_0 is n-colorable, with some coloring c: V_0 \to \left \{ 1, 2, \dots, n \right \}. We claim that this c induces a model on \Sigma_0.

Define a valuation t:A \to \left \{ 0,1 \right \} as follows. If v \notin V_0, then we (arbitrarily) choose t(p_{v,i}) = 1. If v \in V_0 and c(v) = i then t(p_{v,1}) = 1. Finally, if v \in V_0 and c(v) \neq i then t(p_{v,i} = 0).

Clearly each of the possible propositions in the above three sets is true under the extension \hat{t}, and so \Sigma_0 has a model. Since \Sigma_0 was arbitrary, \Sigma is finitely satisfiable. So by the Compactness Theorem, \Sigma is satisfiable, and any model s for \Sigma gives a valid graph coloring, simply by choosing i such that the proposition p_{v,i} satisfies s(p_{v,i}) = 1. Our construction forces that such a proposition exists, and hence G is n-colorable. _\square.

Note that without translating this into a logical system, we would be left with the mess of combining n-colorable finite graphs into larger n-colorable finite graphs. The number of cases we imagine encountering are mind-boggling. Indeed, there is probably a not-so-awkward graph theoretical approach to this problem, but this proof exemplifies the elegance that occurs when two different fields of math interact.

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4 thoughts on “n-Colorability is Equivalent to Finite n-Colorability (A Formal Logic Proof)

      • Yeah, two topics that I enjoy merged together. It’s definitely a great concept. The only frustrating part is that I’ve lost more knowledge than I’d like from my minor classes so I don’t usually get all of the details.

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      • Well if you ever have a question on a particular detail, don’t hesitate to ask! Nobody really asks me for clarification so I assume I’m writing at the correct level.

        Like

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