3

A question asked before (There exists no zero-order or first-order theory for connected graphs) proved that no theory exists such that axiomtizes connected graphs.

Using the same type of argument (Compactness theorem for first order logic), is it possible to prove the disconnected graphs are axiomtizable?

Minnee
  • 33

1 Answers1

2

There is no theory in the language of graphs whose models are exactly the disconnected (by which I assume you mean "not connected") graphs. It is possible to prove this fact simultaneously with the fact that there is no theory of connected graphs: the problem is that we can build graphs that are elementarily equivalent (i.e., have the same theory), but where one is connected and the other disconnected.

Let $G_0$ be the graph whose vertices are the integers, with edges between adjacent integers. Let $L$ be the language of graphs, and $T$ the complete theory of $G_0$ as an $L$ structure. Let $L'$ be $L$ together with two new constant symbols, and let $T'$ be $T$ together with an axiom schema asserting that there is no path of length $n$ connecting the two constant symbols, for any $n$. $T'$ is finitely consistent, using $G_0$ as the graph and interpreting the constant symbols as integers that are sufficiently far apart. By compactness, $T'$ has a model. That model will be a graph $G_1$ that is elementarily equivalent to $G_0$ in the graph language, but the new constant symbols from $L'$ must be interpreted as vertices in different connected components.

So we have $G_0$ elementarily equivalent to $G_1$, with $G_0$ connected and $G_1$ disconnected, so you can't tell whether a graph is connected by looking at its theory. Consequently, you can't axiomatize either "connected graphs" or "disconnected graphs" (in the language of graphs).

  • Hi Mike, thanks for your answer! I understand everything up to "By compactness, T' has a model", but I am confused about how we derive graph G_1. By construction G_0 is connected. Do we not need to construct another graph G_1 to use in our proof, or can we just state that G_0 and G_1 are elementarily equivalent? – Minnee May 31 '17 at 06:33
  • @Minnee The overall sketch of my construction in paragraph two is 1) build a theory $T'$, 2) Show it is finitely consistent, 3) Use compactness to conclude it is consistent, 4) The definition of consistent means that $T'$ has a model. Call that model $G_1$. Compactness is letting us do something very sneaky, here: when we're showing that $T'$ is finitely consistent, the model for every finite part of $T'$ is just going to be the graph $G_0$ and the work is finding interpretations for the constants. But the graph we get from step 4 is not isomorphic to $G_0$. –  May 31 '17 at 06:38
  • @Minnee If you've seen the classic compactness proof that there is a model of the theory of $\mathbb{R}$ with an infinite element, it's very similar to what's happening there. –  May 31 '17 at 06:39
  • I see! So the final step is stating G_1 is disconnected, but as T is the theory whose models are connected graphs, we arrive at a contradiction? – Minnee May 31 '17 at 06:47
  • @Minnee The way I framed it, $T$ was just the complete theory of $G_0$ in the graph language, and the contradiction is that, if a class of structures is axiomatizable, you can't have a structure in the class model the exact same sentences as one not in the class. I'm pretty tired; sorry if I'm not being very clear. –  May 31 '17 at 06:55