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Given an infinite cardinal $\kappa$, is there a graph on $\kappa$ vertices that contains $2^\kappa$ pairwise non-isomorphic induced subgraphs?

Dominic van der Zypen
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    In Universal graphs and universal functions, Acta Arith. 9 (1964), R. Rado constructed a graph with denumerably many nodes (points) in which every graph with at most denumerably many nodes can be embedded. It seems to me that this affirmatively answers your question in the case $\kappa=\aleph_0$, right? – Francesco Polizzi Jul 16 '18 at 12:44
  • Thanks @FrancescoPolizzi for pointing me to this paper! That's correct, that would be an affirmative answre for $\kappa = \aleph_0$. Can this be generalized to higher cardinals? – Dominic van der Zypen Jul 16 '18 at 12:51
  • The Rado graph seems like overkill for the case $\kappa = \aleph_0$, why wouldn't one just use the complete graph on $\aleph_0$ vertices? And then, for arbitrary $\kappa$, why wouldn't one just use the complete graph on $\kappa$ vertices? – Lee Mosher Jul 16 '18 at 13:54
  • @LeeMosher The complete graph doesn't work, as all the induced subgraphs are also complete. There are only $\aleph_0$ pairwise distinct induced subgraphs of the complete graph on $\aleph_0$ vertices. – Dominic van der Zypen Jul 16 '18 at 13:59
  • Ah, okay, I did not know what "induced subgraph" meant (although you even supplied a nice link, I see now). – Lee Mosher Jul 16 '18 at 14:12

1 Answers1

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You don't need something as complicated as the Rado graph.

Take the disjoint union of $K_n$ (the complete graph on $n$ vertices) for all $n$. This has countably many vertices. For any subset of the natural numbers $S \subseteq \mathbb{N}$ consider the induced subgraph consisting of $K_i$ for $i \in S$. For distinct S these subgraphs are non-isomorphic. There are uncountably many such subsets, so you are done. This is not a very interesting example of course. Perhaps you want connected induced subgraphs? But then you can just take the complement of this example.

This only answers the question for $\kappa = \aleph_0$ of course.

David Roberson
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    But all you need to generalize to arbitrary $\kappa$ is a family of $\kappa$ many pairwise nonisomorphic graphs, each with at most $\kappa$ vertices. Which is easy. – Nik Weaver Jul 16 '18 at 14:41
  • On second thought, not as easy as I just thought ... Still, surely not too hard. – Nik Weaver Jul 16 '18 at 14:43
  • @NikWeaver Yeah, I thought maybe some sort of generalization of this might work. But I am so familiar with infinite graphs or infinite cardinals, so I just stopped here. – David Roberson Jul 16 '18 at 15:16
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    Well, it seems to me that you can encode any relation on $\kappa$ by a graph with $\kappa$ vertices. In particular you can encode every smaller ordinal, which does the trick. The encoding is a little too complicated to describe in a comment though. – Nik Weaver Jul 16 '18 at 15:33
  • Why not connect kappa many non isomorphic small connected graphs to an extra point? That allows you lots of disconnected subgraphs. Gerhard "Let Topology Be Your Helper" Paseman, 2018.07.16. – Gerhard Paseman Jul 16 '18 at 15:42
  • @GerhardPaseman: the problem isn't connectivity, it's just finding $\kappa$ many nonisomorphic small graphs. But as I said above, this isn't really hard. – Nik Weaver Jul 16 '18 at 15:49
  • @NikWeaver For each $\alpha \in \kappa$, $G_\alpha=(\alpha,\in)$ is a graph. If $\alpha \neq \beta$ then $G_\alpha$ and $G_\beta$ are nonisomorphic. – Ramiro de la Vega Jul 18 '18 at 15:14
  • @RamirodelaVega: the membership relation is not symmetric. – Nik Weaver Jul 18 '18 at 16:17
  • @NikWeaver Ok, for some reason a thought we were dealing with directed graphs. On the other hand I think the details of your idea are spelled out in this (https://mathoverflow.net/a/281040/17836) answer to a previous question of Dominic. – Ramiro de la Vega Jul 19 '18 at 19:26