Let $G = (V, E)$ be a simple, undirected graph with $|V|>2$, and let $S\subseteq V$ be a set with more than $1$ element. By $G/S$ we denote the graph obtained by collapsing $S$ to one point. More formally:

  • $V(G/S) = (V \setminus S) \cup \{S\}$, and
  • $E(G/S) = \big\{\{v,w\} \in E: v, w \notin S \big\}\cup \big\{\{v, \{S\}\}: v\in (V\setminus S)$ and there is $s \in S$ with $\{v,s\}\in E\big\}$.

We call a connected graph $G=(V,E)$ collapse-resistant $|V|> 2$ and if for all $S\subseteq V$ with $1<|S|<|V|$ we have $G \cong G/S$.

Clearly every infinite complete graph is collapse-resistant.

Question. Is every infinite connected collapse-resistant graph complete?

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    $\begingroup$ Not a complete answer, but this is at least false if one restricts least false if one restricts only to contractions involving finitely many vertices. Construction: Given a graph $G$ let $C(G)$ be the union of all finite collapses of $G$ Then given a graph $G$, consider $G \cup C(G) \cup C(C(G)) \cdots $. This does not work in the obvious way if one allows countable contractions since one can take a contraction that involves taking a vertex from each level of the construction, but a similar argument then taking a union up to the smallest uncountable ordinal will then work. $\endgroup$
    – JoshuaZ
    Commented Jan 14 at 16:52
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    $\begingroup$ Wouldn't an infinite graph with no edges be collapse-resistant, for the same reason that an infinite complete graph is? $\endgroup$
    – paste bee
    Commented Jan 14 at 17:16
  • $\begingroup$ Sorry I have to add "connected" graph $\endgroup$ Commented Jan 14 at 17:40
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    $\begingroup$ Thanks @JoshuaZ - can you maybe make this into an answer? $\endgroup$ Commented Jan 14 at 17:42
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    $\begingroup$ I would have called it contraction invariant, or possibly finite contraction invariant, since one could consider $\kappa$-contraction invariance, allowing the contraction of fewer than $\kappa$ many vertices. The concept makes sense in any relational structure, not just graphs. Also there is a weak and a strong form, depending on whether the new relation instances occur because of a single instance in $S$, as you defined, or whether you require all the instances with $S$ to hold. Paste bee's solution with the countable random graph is both weak and strong contraction invariant. $\endgroup$ Commented Jan 16 at 12:40

3 Answers 3


No. Let $G$ be the Rado graph (which is infinite, connected, and not complete), and $S$ a finite subset of the vertices of $G$ (because the Rado graph is countable). $G/S$ still has the extension property: for sets $U$ and $V$, there is a point in $G$ adjacent to all points in $U$, not adjacent to any points in $V$, and that is also not in $S$. (If either $U$ or $V$ contains the point $S$, we require the point in $G$ to be adjacent/not adjacent to all points in $S$). This point is adjacent in $G/S$ to all points in $U$, and not adjacent to any of the points in $V$. Any countable graph with this extension property is the Rado graph, so $G \cong G/S$.


No. Let $V = \mathbb{N}$ and $E = (0, i)$ for $i$ in $\mathbb{N}^*$ (a "star" graph where every vertex is connected to $0$).

If you collapse a subset containing $0$, the collapsed vertex can be mapped to $0$ and the remaining ones can be mapped to $\mathbb{N}^*$ as they are still all connected to $0$.

If you collapse a subset not containing $0$, you can map it to $1$ as it will only be connected to $0$. You can then map $0$ to $0$ and all the remaining vertices to $\mathbb{N} \setminus \{0, 1\}$

$G = (V, E)$ is collapse-resistant.

  • $\begingroup$ This was the simplest I could think of, I started with a covering tree because we want a connected graph, and then I realised the diameter of the graph will not grow when collapsing a subset. Turns out a tree of diameter 2 will do the trick ! $\endgroup$
    – Vincent
    Commented Jan 16 at 8:25
  • $\begingroup$ I am trying to think of other graphs of diameter 2, but it is starting to look complicated as soon as you have vertices of degree different from 1 and infinity. Actually the Rado graph is of diameter 2 (according to wikipedia) so maybe you have to jump from the star graph directly to the Rado graph ? $\endgroup$
    – Vincent
    Commented Jan 16 at 8:31

I add this answer to provide some variety, I believe this construction is collapse-resistant but the proof is not as easy and elegant as the other answers. Please comment if you find an error or gap in the proof !

Let $\mathbb{G}_f$ be the set of all finite graph. For $g \in \mathbb{G}_f$ a finite graph, let $\hat{g}$ be the union of a countable number of copy of $g$. Now let $G$ be the union of all $\hat{g}$ for $g \in \mathbb{G}_f$.

$G$ is not connected but it is collapse-resistant. When we collapse a finite subset of $G$, the resulting vertex will be connected to a finite number of vertices and be part a finite graph. It can be mapped to the corresponding finite graph from $G$. There will be a countable number of every finite graph remaining (not taking part in the collapse), so the result is still a countable number of every finite graph.

To get a connected graph, let us construct $\bar{G}$ by adding a vertex $o$ to $G$, connected to every other vertex.

$\bar{G}$ is connected and also collapse-resistant. If we collapse a subset $S$ containing $o$, the resulting vertex will be connected to all other vertices and can be mapped to $o$. For the remaining vertices ($\bar{G} \setminus \{o\}$), it is just as if we deleted $S \setminus \{o\}$, it will still result in a countable number of every finite graph. If we collapse a subset not containing $o$, it will work just the same as for $G$ (but with an added vertex $o$ connected to everything)

Note that this graph $\bar{G}$ is not the Rado graph, for example it has a vertex connected to every other vertex.


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