Definition of Tn: Let R [E0] be such that R(N0n ) = T(N0n ). Then
define Tn on Ns,s
as follows:
Tn(x) = TgsT
As gs,TgsT
[E0], clearly Tn [E0]. Also Tn|N0n = R|N0n , Tn(Ns) =
T(Ns). We now check that TngsTn
= TgsT
∀y. If y
T(Ns) T(N0n ),TgsT
= TT
= y and TngsTn
= TnTn
y. If y T(Ns) T(N0n ) this follows from

Since clearly [E0] = N[E0] (e.g., any infinite A N induces an element
of N[E0] \ [E0] via x x + χA(mod 2)), it follows that for any ergodic,
hyperfinite E, Out(E) is not Polish. In fact one can say quite a bit more.
Call Out(E) turbulent if the action of ([E],u) by (right) translation on
N[E] is turbulent. In that case, the equivalence relation induced by the
cosets of [E] in N[E] admits no classification by countable structures (i.e.,
Out(E) cannot be Borel embedded in the isomorphism types of countable
Theorem 7.3 (Kechris). If E is not E0-ergodic, then every element of [E]
(closure of [E] in N[E]) is turbulent for the translation action of ([E],u) on
N[E]. In particular, if also [E] is not closed in N[E], the translation action
of ([E],u) on [E] is turbulent and the associated coset equivalence relation
admits no classification by countable structures.
Corollary 7.4. If E is ergodic, hyperfinite, then Out(E) is turbulent.
Proof. Let G = [E]. Since the periodic g [E] are dense in ([E],u), fix
a countable set {gj} of periodic elements in [E] which is dense in ([E],u).
Then a basic nbhd of g G has the form
U =
{S G :
Also fix a basic nbhd V of 1 in ([E],u).
We will show that O(g, U, V ) is dense in U. Now {gT
: T [E]} is
dense in G, so {gT
U : T [E]} is dense in U. Therefore {ggj
U :
j = 0, 1,... } is dense in U, thus it is enough, for each ρ 0, and h {gj}
such that
U, to show that O(g, U, V ) {S : δu(S,
ρ} = ∅.
Note that for S G:
gS U ∀j
It is thus enough, for each ρ 0, to find a continuous path λ in
([E],u) with h0 = 1 such that δu(hλ
, ∀j n, and δu(h, h1) ρ.
Let α be such that
gj) α, ∀j n. Using the notation of 5.2,
we can find
h, ¯j,j g n, which are in
) E] for some large enough
N and δu(h,
h δ0,δu(gj, ¯j) g δ0, ∀j n, where 0 δ0 ρ, α + 6δ0 .
h g h, ¯j) g 3δ0 + α + δ0 = α + 4δ0. Then, as in the proof of
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