The argument showing that the membership graph of a model of ZFC is a simple graph does depend on Foundation, as we saw. In ZFA, we have sets x with \(x\in x\), giving loops in the graph; and pairs x, y of distinct sets with \(x\in y\in x\), giving double edges.
The random loopy graph is obtained with probability 1 if we choose a graph on a countable vertex set by choosing edges (including loop edges) from pairs of not necessarily distinct vertices independently with probability 1/2. It is homogeneous, and is the Fraïssé limit of the class of finite loopy graphs. The relevant version of the Alice’s Restaurant property characterises it as the countable graph such that, for any two finite disjoint sets U and V of vertices, there are vertices \(z_1\) and \(z_2\), where \(z_1\) is loopless and \(z_2\) has a loop, each joined to all vertices in U and to none in V. The proof is very similar to the usual proof for the random graph, and we will not give it here.
Theorem 2
The membership graph of a countable model of ZFA, ignoring multiple edges but keeping loops, is isomorphic to the random loopy graph.
Proof
We begin with some preliminaries. In a model of a subset of ZF including at least Selection, there is no set whose members are all sets. For, if such a set S exists, then Selection would give Bertrand Russell’s set \(R=\{x\in S:x\notin x\}\), whose existence leads to a contraction on examining whether \(R\in R\) or not.
It follows that, if the Union axiom also holds, there is no set \(S'\) which contains all the p-element sets, for a fixed positive natural number p: for the union of \(S'\) would be S. In particular, if T is any set, then there is a set of cardinality p which is not a member of T.
Now let \(\Gamma \) be the membership graph of a countable model of ZFA, with loops but no double edges. We will show that \(\Gamma \) satisfies the loopy version of ARP. For this let \(\{u_1,\ldots ,u_m\}\) and \(\{v_1,\ldots ,v_n\}\) be disjoint sets, and let \(z_1 = \{x, u_1, \ldots , u_m \}\) and \(z_2 = \{z_2, x, u_1, \ldots , u_m\}\), where x is a vertex satisfying the following conditions:
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x is not equal to any of the \(v_j\);
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x is not contained in any of the \(v_j\) or the \(u_i\);
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x is not contained in any of the sets contained in any of the \(v_j\).
Such an x exists, since otherwise the union of V, \(\bigcup U\), \(\bigcup V\), and \(\bigcup \bigcup V\) would contain every set, a contradiction.
Furthermore, we may assume that \(|x|=m+3\), since by our earlier remarks there is a set of this cardinality not a member of the “forbidden set” \(V\cup (\bigcup U)\cup (\bigcup V)\cup (\bigcup \bigcup V)\) above.
We remark that the existence of \(z_1\) follows simply from Pairing and Union; for \(z_2\), we invoke Anti-Foundation, letting \(z_2\) be the unique solution of the equation
$$\begin{aligned} z = \{z, x, u_1, \ldots , u_m\}. \end{aligned}$$
Both \(z_1\) and \(z_2\) are joined to all the vertices \(u_i\); and by construction, there is a loop on \(z_2\). We claim that there is no loop on \(z_1\). For such a loop would imply one of the following:
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\(z_1=u_i\) for some i. Then we have \(x\in u_i\), contradicting our choice of x.
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\(z_1=x\). But we chose x with \(|x|=m+3\), whereas \(|z_1|\le m+1\). (Note that in the same way we see that \(z_2\ne x\).)
Finally we have to show that \(z_1\) and \(z_2\) are not joined to any \(v_j\). We cannot have any \(v_j\) contained in \(z_1\) or \(z_2\); for the \(v_j\) are distinct from the \(u_i\) by hypothesis, not equal to x by choice of x, and not equal to \(z_2\) since if so then x would be a member of \(v_j\), again contrary to our choice of x. Also we cannot have \(z_1\) or \(z_2\in v_j\), since if so then x belongs to a member of \(v_j\), again contrary to our choice of x. \(\square \)
Remark
As in the case of ZFC, it is interesting to note which axioms are actually used in the proof. The Empty Set, Pairing and Union axioms are once again used; of course, the Anti-Foundation Axiom is used; and as well, we use the Selection Axiom.
What happens if we keep the multiple edges? We cannot describe all graphs that can arise, but we note that there is no such graph which has the nice properties of \(\aleph _0\)-categoricity and homogeneity which hold in the random and random loopy graphs. This will follow from the theorem of Engeler, Ryll-Nardzewski and Svenonius (see [9, Theorem 6.3.1]), according to which a countable first-order structure is \(\aleph _0\)-categorical (that is, determined uniquely by its first-order theory and the property of countability) if and only if its theory has only finitely many n-types for all n. An n-type of a first order theory T is a set \(\Phi (\bar{x})\) of formulas with at most n free variables in the language of T such that \(T\cup \{\exists \bar{x}\wedge \phi \}\) is consistent for every finite subset \(\phi (\bar{x})\) of \(\Phi \). In a sense, an n-type describes how a set of up to n elements might behave in a model of T.
Theorem 3
The membership graph of a countable model of ZFA, keeping double edges, is not \(\aleph _0\)-categorical.
Proof
Take a countable model of ZFA. Let \(a_n\) be distinct well-founded sets for \(n\in \mathbb {N}\), for example, the natural numbers. For every natural number n, consider the equations
$$\begin{aligned} y= & {} \{x_0,\ldots ,x_{n-1}\},\\ x_i= & {} \{y,a_i\} \hbox { for }i=0,\ldots ,n-1. \end{aligned}$$
By AFA, these equations have a unique solution in the model. We have \(x_i\in y\) and \(y\in x_i\), so all the edges \(\{x_i,y\}\) are double. (These sets are all distinct, by extension.) There are no further double edges on y, since if \(\{y,z\}\) is a double edge then \(z\in y\) and so \(z=x_i\) for some i.
Thus, for every natural number n, there is a set lying on exactly n double edges. Since this property of the set is expressible in first-order logic there are infinitely many 1-types in the graph, and so the graph cannot be \(\aleph _0\)-categorical, by the theorem of Engeler, Ryll-Nardzewski and Svenonius.
Moreover, we can take the infinite set of equations
$$\begin{aligned} y= & {} \{x_n:n\in \mathbb {N}\}, \\ x_n= & {} \{y,a_n\} \hbox { for }n\in \mathbb {N}. \end{aligned}$$
A solution to these equations will be a point lying on infinitely many double edges. \(\square \)
Another natural reduct is obtained by keeping only the double edges. The double-edge graph of a model of ZFA has as vertices the sets and as edges all pairs \(\{x,y\}\) with \(x\in y\) and \(y\in x\) (allowing \(x=y\)). Thus, it includes loops and double edges but omits all “conventional” instances of the membership relation (where \(a\in b\) but \(b\notin a\)).
Theorem 4
Let D be the double-edge graph of a countable model of ZFA. Then, for any finite connected loopy graph \(\Gamma \), D has infinitely many connected components isomorphic to \(\Gamma \). It also has at least one infinite component.
Proof
An example will illustrate the general proof. Let \(\Gamma \) be the 4-cycle with edges \(\{v_0,v_1\}\), \(\{v_1,v_2\}\), \(\{v_2,v_3\}\) and \(\{v_3,v_0\}\), together with a loop at \(v_0\). Take any four well-founded sets \(a_0,a_1, a_2,a_3\) (for example, the first four natural numbers), and consider the equations
$$\begin{aligned} y_0= & {} \{a_0,y_0,y_1,y_3\},\\ y_1= & {} \{a_1,y_0,y_2\},\\ y_2= & {} \{a_2,y_1,y_3\},\\ y_3= & {} \{a_3,y_0,y_2\}. \end{aligned}$$
The unique solution gives an induced subgraph isomorphic to \(\Gamma \). Note that there are no other double edges meeting these vertices: if, say, \(\{y_1,x\}\) were a double edge, then \(x\in y_1\), and so by Extensionality, \(x=a_1\) or \(x=y_0\) or \(x=y_2\); the first is impossible since \(y_1\notin a_1\) by assumption. So the given set is a connected component.
Since there are infinitely many possible choices of \(a_0,\ldots ,a_3\), there are infinitely many such connected components.
We saw earlier that there is a vertex with infinite valency; it lies in an infinite component of D. \(\square \)
This leaves a few questions which we have not been able to answer in this current work.
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1.
Is it true that the first-order theory of the membership graph of a countable model of ZFA has infinitely many countable models?
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2.
Is it true that there are infinitely many non-isomorphic graphs which are membership graphs of countable models of ZFA?
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3.
Can more be said about infinite connected components of the double-edge graph?
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4.
What about models of ZFA where the Axiom of Infinity is replaced with its negation?
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5.
Is it true that, if two countable multigraphs are elementarily equivalent, and one is the membership graph of a model of ZFA, then so is the other?
However, in a sequel by Adam-Day et al. [2] the first two questions are answered affirmatively and a characterisation of the connected components of double-edge graphs is given, thus answering the third question. The analogue to Question 5 for double-edge graphs is shown to be negative, and indeed it is shown that, for any double-edge graph, there is an elementarily equivalent countable structure which is not itself a double-edge graph.