(VERY) INFORMAL NOTES ON THE RANDOM GRAPH
MIKE COHEN
Definition. By a countably infinite graph G = (V, E) we mean a countably infinite
collection V of vertices, together with a subset E ⊆ V × V of edges. We require that
E does not have any loops, i.e. (v, v) ∈
/ E for v ∈ V , and that E is not directed, i.e.
(v1 , v2 ) ∈ E ↔ (v2 , v1 ) ∈ E. An isomorphism of graphs f : G1 = (V1 , E1 ) → G2 =
(V2 , E2 ) is a bijective function which satisfies (v, w) ∈ E1 ↔ (f (v), f (w)) ∈ E2 ). Since
we are being informal, we will usually identify the graph G with its set of vertices
V , i.e. if we wish to refer to a particular vertex v we will probably say v ∈ G rather
than v ∈ V .
Remark. We will wish to refer to the space of countably infinite graphs by
identifying all such graphs with the Cantor space 2ω in the following way: If G is
a graph, let ω represent some fixed enumeration of the distinct pairs of vertices in
G. Then let xG ∈ 2ω be such that xG (n) = 1 if the n-th pair is joined by an
edge; xG (n) = 0 if not. Clearly all possible graphs on a given set of vertices can be
represented in this way. This representation will allow us to talk about both Baire
category and measure in the space of countably infinite graphs, where our canonical
measure is the Lebesgue (coin-flipping) measure λ.
Now let us define a few countably infinite graphs of interest.
Definition (The Rado Graph R). Let V = ω = N. For each
P n ∈ N, let (n0 , n1 , ...) ∈
2N denote the unique binary expansion of n, i.e. n = i=0 2i · ni . For n, m ∈ N,
declare that (n, m) ∈ E ↔ (nm = 1 ∨ mn = 1).
Example. The neighbors of 0 in the Rado graph are exactly the odd numbers. The
neighbors of 1 are 0 and the integers congruent to 2 or 3 modulo 4.
Definition (The Gilton/Holshouser Graph GH). Let M be a countable transitive
model of ZFC. Let V = M , and for x, y ∈ M , declare that (x, y) ∈ E ↔ (x ∈ y∨y ∈ x)
(i.e. we symmetrize the distinguished set membership relation).
Definition (The RTG Scholarship Graph RT G$). Recall that a set S ⊆ N is called
universal if, when we regard S as an infinite string of 0’s and 1’s in 2N , then every
finite string of 0’s and 1’s occurs as a substring of S. Fix any such universal set
S ⊆ Z. Let V = Z, and for n, m ∈ Z, declare that (n, m) ∈ E ↔ |n − m| ∈ S.
Theorem (Main Theorem). R ∼
= GH ∼
= RT G$. Moreover, if X is the isomorphism
class of R, then X has full Lebesgue measure and X is comeager.
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2
MIKE COHEN
To prove this fact the following definition and lemmas are helpful.
Definition. If G is a countably infinite graph, we will say that G satisfies property
(*) if the following holds: Whenever U and V are finite disjoint sets of vertices in G,
then there exists a vertex w ∈ G − (U ∩ V ) such that w is connected by an edge to
everything in U , but not connected by an edge to anything in V .
Lemma. If G and H are graphs which both satisfy property (*), then G and H are
isomorphic.
Proof. We wish to construct an isomorphism f : G → H. Fix enumerations G =
{g0 , g1 , g2 , ...} and H = {h0 , h1 , h2 , ...}. We will simultaneously define f and f −1 recursively, in such a way that at each step f and f −1 both define partial isomorpisms.
We proceed as follows: Let f (g0 ) = h0 , so f −1 (h0 ) = g0 .
Now suppose f (g0 ), ..., f (gn ) and f −1 (h0 ), ..., f −1 (hn ) have already been defined
for some particular n, and both are partial isomorphisms. We must define f (gn+1 )
and f −1 (hn+1 ). If f (gn+1 ) has already been defined, i.e. gn+1 = f −1 (hi ) for some
i ≤ n, then we are done. Otherwise, let W = {g1 , ..., gn , f −1 (h0 ), ..., f −1 (hn )}. Let
U = {g ∈ W : g is connected by an edge to gn+1 and set V = W − U . Then f (U )
and f (V ) are finite disjoint sets in H, and since H satisfies property (*), we may
find some least integer m > n for which hm is connected to everything in f (U ) but
nothing in f (V ). Set f (gn+1 ) = hm .
Now if f −1 (hn+1 ) is already defined, then we’re done. Otherwise, repeat the
process above in the reverse direction: let W = {h0 , ..., hn , f (g0 ), ..., f (gn+1 )}, let
U = {h ∈ W : h is connected to hn+1 and let V = W − U . Since f −1 (U ) and
f −1 (V ) are finite and disjoint in G, we may use property (*) in G to find a least
m > n for which gm is connected to everything in f −1 (U ) and nothing in f −1 (V ). Set
f −1 (hn+1 ) = gm . This completes our recursive definition.
Now f is our isomorphism!
Lemma. If G is a graph which satisfies property (*), then every finite or countably
infinite graph A embeds into G as an induced subgraph.
Proof. Suppose G satisfies property (*), and A = (a0 , a1 , ...) is a countable graph.
Define f : A → G recursively as follows: Let f (a0 ) be arbitrary. If f (a0 ), ..., f (an )
have already been defined for some particular n, then let U = {f (ai ) : 0 ≤ i ≤ n ∧ ai
and an+1 are connected by an edge}, and let V = {f (a0 ), ..., f (an )} − U . Then use
property (*) to find a w ∈ G which connects to everything in U but nothing in V .
Set f (an+1 ) = w. This clearly defines an embedding.
Lemma. Suppose G is a graph which satisfies property (*). Let (A, B) be any partition of G, i.e. A ∪ B = R and A ∩ B = ∅. Then either A ∼
= G or B ∼
= G.
(VERY) INFORMAL NOTES ON THE RANDOM GRAPH
3
Proof. Suppose by way of contradiction there were some partition (A, B) such that
neither component is isomorphic to R. Then neither one satisfies property (*), and
hence we may find U1 , V1 ⊆ A and U2 , V2 ⊆ B, all finite, which do not admit witnesses
to property (*). Then there is no w ∈ R which is connected to all the points in U1 ∪U2
but none in V1 ∪ V2 , contradicting property (*) in R.
Corollary. Suppose G1 is a graph which satisfies property (*), and we obtain G2 from
G1 by any of the following operations:
(1) deleting a finite number of vertices;
(2) deleting or adding a finite number of edges;
(3) switching edge connections between a finite subset and its complement.
Then G1 ∼
= G2 .
Proof of Main Theorem. In light of our previous lemma it suffices for the first part of
the theorem to show that R, GH, and RT G$ all satisfy property (*).
For R: Let U = {u0 , ..., un } and V = {v0 , ..., vm } be disjoint sets in N. Let
M > max(log2 (v0 ), ..., log2 (vm )), and set w = 2u0 + ... + 2u+n + 2M . We claim that w
witnesses property (*). It is clear that the ui -th digit in the binary expansion of w is
1, and hence w is connected to ui for all ui ∈ U . Since V is disjoint from U , we also
see that the vi -th digit of w is not 1, so we need only check that the w-th digit of vi
is not 1. Our choice of M guarantees this, for 2w ≥ 2M > vi .
For GH: Let U = {u0 , ..., un } and V = {v0 , ..., vm } be disjoint sets in M . By
the Pairing Axiom we have V ∈ M . Again using Pairing, set w = {u0 , ..., un , V }.
Obviously w is connected to everything in U . If w were connected to any vi , then we
would have either vi ∈ w or w ∈ vi . If w ∈ Vi , then vi = V , in which case vi ∈ vi ,
violating the Foundation Axiom. If w ∈ vi , then w ∈ vi ∈ V ∈ w, again violating
Foundation. So w witnesses property (*) for GH.
For RT G$: Let U = {u0 , ..., un } and V = {v0 , ..., vm } be disjoint sets in Z. Let
a = min(U, V ) and b = max(U, V ), and build a map g : {a, ..., b} → 2 such that
g(z) = 1 whenever z ∈ U and g(z) = 0 whenever z ∈ V . This gives a finite string
of 0’s and 1’s of length b − a + 1. By the definition of universality, there exists some
(positive or negative) integer w such that, whenever s ∈ {a − w, ..., b − w}, we have
s ∈ S ↔ g(s + w) = 1. So for any ui ∈ U ⊆ {a, ..., b}, we have g((ui − w) + w) =
g(ui ) = 1 ⇒ ui − w ∈ S ⇒ ui and w are connected by an edge. On the other hand
for any vi ∈ V ⊆ {a, ..., b}, we have g((vi − w) + w) = g(vi ) = 0 → vi − w ∈
/ S ⇒ vi
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MIKE COHEN
and w are not connected by an edge. So w witness property (*) in RT G$. Thus our
three graphs of interest are indeed isomorphic.
It remains to show that the isomorphism class X of R is comeager, and that
λ(X) = 1. For any finite sets U = {u0 , ..., un } and V = {v0 , ..., vm } in R and any vertex w ∈ R−(U ∪V ), let j0 , ..., jn denote the indeces in ω for each pair (w, u0 ), ..., (w, un )
respectively, and let k0 , ..., km denote the indeces in ω for each pair (w, v0 ), ..., (w, vn )
respectively. Write XU,V,w = {x ∈ 2ω : x(j0 ) = ... = x(jn ) = 1 ∧ x(k0 ) = ... = x(km ) =
0}, so XU,V,w is clopen in 2ω . Now since X is exactly the set of all graphs which
satisfy property (*), we may write

X=
\ \

[

U ⊆R V ⊆R
XU,V,w 
w∈R−(U ∩V )
This displays X as a Gδ set in 2ω . Moreover X is dense in 2ω , for given any finite
string of initial edge conditions x(0), ..., x(n), we can modify R finitely to find R0
which satisfies the conditions. Then by our corollary above, R0 ∼
= R, so X is indeed
dense. This implies X is comeager!
Recall now that any countable intersection of full measure sets has full measure. So
in order to show that X has full Lebesgue measure, weSneed only check that for any
fixed U = {u0 , ..., un } and V = {v0 , ..., vm }, we have λ( w∈R−(U ∩V ) XU,V,w ) = 1. It is
T
equivalent to check that λ(Y ) = 0, where Y = w∈R−(U ∩V ) [XU,V,w ]c . But λ(XU,V,w ) =
( 21 )(n+1)+(m+1) by its definition, and the family of events {XU,V,w : w ∈ R − (U ∩ V )}
Q
is clearly independent. So we have λ(Y ) ≤ w∈R−(U ∩V ) (1 − ( 21 )n+m+2 ). Since the
latter product is infinite and each factor is bounded below 1, we have λ(Y ) = 0 as
required.
Remark. The Main Theorem justifies our use of the definite article when we refer to
”the random graph R.” R is sometimes called the universal countable graph. The
random graph was first discovered by Paul Erdos and Alfred Renyi in 1963, where
they proved that there is a graph whose isomorphism class has full Lebesgue measure
in 2ω . In 1964 Richard Rado independently gave the first explicit construction of R,
which is the one included in these notes.
Definition. Let AutR denote the group of automorphisms of the random graph,
i.e. permutations of R which preserve edge relations, endowed with the topology of
pointwise convergence. Then R is a non-locally compact simple Polish group.
Proposition. AutR embeds as a closed subgroup of S∞ .
(VERY) INFORMAL NOTES ON THE RANDOM GRAPH
5
Remark. J. K. Truss has completely classified the cycle-types of AutR in 1985. We
will mention a few simple theorems along these lines below.
Proposition. AutR contains permutations which consist of a single infinite cycle.
Proof. In RT G$, the map z 7→ z + 1 is such a permutation.
Proposition. AutR does not contain any permutations which consist only of a single
fixed point and a single infinite cycle.
Proof. Exercise!
Proposition. R is ultrahomogenous, i.e. if A, B are finite subsets of R and p : A →
B is a graph isomorphism, then there exists a π ∈ AutR which extends p.