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Show that there doesn't exist a (countable) universal graph for each of the following classes of graphs (that is, no universal graph for class 1, and unrelatedly, no universal graph for class 2):

  1. All graphs that do not contain the countable complete graph.

  2. All graphs where the degree of each vertex is finite.

So a proof would involve showing that for every graph $G$ in one of these class, there is a graph in that same class $G'$ that is not isomorphic to some induced subgraph of $G$. How would I go about doing this? I know of the Rado random graph, which contains every countable graph as an induced subgraph, what fails in the two classes above (as opposed to all countable graphs) to make it so that there is no universal graph?

ctlaltdefeat
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2 Answers2

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Hint for #1: Suppose that $G_0$ is the purported graph, which contains any graph $H$ unless $H$ in turn contains $K_\omega$, the complete countable graph. Make a new graph, $G'_0$, by appending a new vertex $v_0$ to $G_0$ and adding edges from $v_0$ to every vertex of $G_0$. Then $G'_0$ does not contain $K_\omega$ (you prove this), so by hypothesis there must be a copy of $G'_0$ strictly inside of $G_0$, which we can call $G'_1$, and $G'_1$ must have an analogous vertex $v_1\ne v_0$.

diagram of graph

Now you fill in the rest.

MJD
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  • This vertex $v_1$ has to be connected to every vertex of the copy $G_1'$. How does this lead to a contradiction? Maybe I should continue this process to construct a countable sequence of $v_n's$ that are all connected with one another? How will that work, if we don't even know that $v_0$ and $v_1$ are necessarily connected? – ctlaltdefeat Jul 22 '13 at 17:56
  • Are you sure you don't know that $v_0$ and $v_1$ are connected? Think about it a little more. – MJD Jul 22 '13 at 17:57
  • Ah silly me of course $v_1$ is connected to $v_0$. Then in the next step, we let $G_1$ = $G_1'$ without $v_1$ and its edges (is that the correct definition?). Now $G_1$ is isomorphic to $G_0$ (right?), and that way we continue. Please correct me if I made a mistake. – ctlaltdefeat Jul 22 '13 at 18:11
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    @ctlaltdefeat: Your reasoning is fine, and ${v_k:k\ge 1}$ ... – Brian M. Scott Jul 23 '13 at 00:25
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For the second problem, suppose that $G$ is a universal countable locally finite graph, and let $V=\{v_n:n\in\omega\}$ be the vertex set of $G$. For each $n\in\omega$ define a function

$$f_n:V\to\omega:v\mapsto|\{w\in V:d_G(v,w)\le n\}|\;,$$

where $d_G(v,w)$ is the length of the shortest path from $v$ to $w$ in $G$. (Verify that the definition makes sense.) Then define

$$f:\omega\to\omega:n\mapsto f_n(v_n)+1\;.$$

Now build a countable locally finite graph $H$ as follows. Let $H_0$ be a copy of $K_1$, with vertex $u$. For $n\ge 1$ let $H_n$ be a copy of $K_{f(n)}$. Form $H$ by taking the disjoint union of the graphs $H_n$, $n\in\omega$, and for each $n\in\omega$ adding an edge from each vertex of $H_n$ to each vertex of $H_{n+1}$.

Without loss of generality assume that $H$ is an induced subgraph of $G$; clearly $u=v_n$ for some $n\in\omega$. Find a lower bound for $f_n(u)$ by considering how many vertices $w$ of $H$ satisfy $d_H(u,w)\le n$, and use this and the definition of $f$ to get a contradiction.

Brian M. Scott
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  • Did you think of that yourself, or did you see it somewhere? – MJD Jul 23 '13 at 03:13
  • @MJD: I’ve seen it somewhere. I think that it’s originally due to de Bruijn. It’s a kind of unexpected use of the diagonal argument. – Brian M. Scott Jul 23 '13 at 03:18
  • The number of vertices with $d_H(u,w) \le n$ is $f(n)$ from your construction, which is equal to $f_n(v_n)+1=f_n(u)+1$. But the distance in $G$ is <= to the dist in $H$, which means this is a lower bound for $f_n(u)$, a contradiction because $f_n(u)+1 > f_n(u)$. Question: why is it necessary for the $H_n$'s to be complete graphs? From what I can tell, can't they be empty graphs (with only vertices), as long as each $H_n$ is connected to all of $H_{n+1}$? – ctlaltdefeat Aug 10 '13 at 21:03