Recall that f is an automorphism of A iff f is an isomorphism of A
onto itself and recall that a structure A is rigid iff the only
automorphism of A is the identity function from A onto A.  We define
G_n to be the set of simple graphs with node set {1,...,n} and G to be
the union (for all finite n) of G_n.  Let R contained in G be the set
of rigid simple graphs and let R_n = R (intersect) G_n. Now recall
that the infinitary logic L_(infinity,w) extends FO with infinitary
conjunctions and disjunctions.  L_(infinity,w)^k uses at most k
variables in this logic. The finite variable infinitary logic is the
union of L_(infinity,w)^k for all natural numbers k. The question is
to show that R is not finite variable infinitary logic-definable over
G. It may be useful to use the result lim(n->infinity) |R_n|/|G_n| = 1
(this can be used w/o proof).

Desired timeframe: 5 days