Aren't two infinite asymmetric graphs always identical?

Question

Suppose you have an infinite graph $G$. I assume $G$ to be cubic and planar. No further conditions, so it will be asymmetric, maybe in the sense of cubic planar version of Rado's graph: Every possible combination of faces exists somewhere in this graph (like in normal numbers).

Now you are given a second one of such a graph $G'$. They might look different at first glance, but aren't they actually identical?

In simple words, you are asking whether or not the theory of cubic and planar graphs with the requirement that the graph is infinite is $\aleph_0$ categorical? — Asaf Karagila, Nov 24 '13 at 21:57
I don't know... If "just cubic and planar" imply random (in the sense of Rado graph) then yes. — Asaf Karagila, Nov 24 '13 at 22:01
@Asaf What else could? So and do you mean that $G$ is the planar cubic subgraph of Rado's graph? — draks ..., Nov 24 '13 at 22:04
@Asaf Brian's answer seems to answer your reformulation of my problem: Negative. You need more than cubic planar. What do you think? — draks ..., Nov 24 '13 at 22:31
I think that I don't know enough terminology in graph theory to give a comprehensive answer (otherwise I would have), and I trust Brian enough to be certain of his knowledge of graph theory. I'd go and fiddle around the definitions, but all my brain resources have been allocated to four problems as it is... :-) — Asaf Karagila, Nov 24 '13 at 22:33
@Asaf Thanks for your point of view and good luck with the four... — draks ..., Nov 24 '13 at 22:35

score 8 · Answer 1 · answered Nov 24 '13 at 22:05

8

These are not isomorphic: the second has bridges, and the first has none.

          --o----o----o----o--  
            |    |    |    |  
          --o----o----o----o--

            o          o          o  
           /|\        /|\        /|\  
          / | \      / | \      / | \  
       --o  |  o----o  |  o----o  |  o--  
          \ | /      \ | /      \ | /  
           \|/        \|/        \|/  
            0          o          o

answered Nov 24 '13 at 22:05

Brian M. Scott

635,418

...that's not what I thought about. I meant something random like Rado's graph (see Asaf's comment above). – draks ... Nov 24 '13 at 22:07
@draks...: They’re cubic planar graphs. Perhaps you need to add some conditions? – Brian M. Scott Nov 24 '13 at 22:07
Yes they are. Formally you're right: It answer what I've written, but it doesn't answer what I thought. Thanks anyway... – draks ... Nov 24 '13 at 22:09
@draks...: You’re welcome. I now have a general idea of what you had in mind, but I don’t immediately see how to make it precise. – Brian M. Scott Nov 24 '13 at 22:10
I thought irregular was enough to express my thoughts. Do you have any suggestions? – draks ... Nov 24 '13 at 22:10
@draks...: Do you perhaps want to require that it contain a copy of every finite $3$-regular planar graph? – Brian M. Scott Nov 24 '13 at 22:10
2

Does this work? How could such a graph even contain $K_4$? – draks ... Nov 24 '13 at 22:14
@draks...: Good point; it couldn’t, if it’s connected. In fact the same problem arises with any finite $3$-regular graph. – Brian M. Scott Nov 24 '13 at 22:18
I think Asaf's right and I'm asking whether or not the theory of cubic and planar graphs with the requirement that the graph is infinite is $ℵ_0$ categorical? – draks ... Nov 24 '13 at 22:22
@draks...: My examples answer that in the negative. You need more than just cubic and planar. – Brian M. Scott Nov 24 '13 at 22:25
Right you are... – draks ... Nov 24 '13 at 22:28
Do you think I should say something about connectedness? $G$ should be connected in such a way that after leaving out some edges no two infinite components remain. Finite ones should be fine... – draks ... Nov 25 '13 at 06:50
@draks...: You mean require that $G$ be $n$-edge-connected for every finite $n$? Is that actually possible for a $3$-regular graph? – Brian M. Scott Nov 25 '13 at 06:55
No, I rather try to exclude examples like yours, by requiring the following: If you remove $k$ edges from $G$, you will get a finite and an infinite component. Your example 2 could give two infinite components by removing one edge...Concerning your question: I don't know. How can $G$ be $1-$ and $2-$connected at the same time...? – draks ... Nov 25 '13 at 07:04
@draks...: And the other can leave two infinite components when you remove two edges. – Brian M. Scott Nov 25 '13 at 07:05

Aren't two infinite asymmetric graphs always identical?

1 Answers1