Page 61 - Textos de Matemática Vol. 34

MINIMAL UNIVERSALITY AND MICHAELLIKE SPACES 51
Proposition 3.2. Let G be a homeomorphism group of R and [a] ∈ R/G. If A = p−1([a]) is open and, for f ∈ G, f(A) = A, then R/G and RA/G are homeomorphic.
Proof. The equivalence classes will be denoted by [x] and [x]A. Define φ : R/G → RA/G by φ([x]) = [x]A.
Since the identity map from RA to R is continuous it follows that φ is open. We
show next that its inverse is also open. Let UA = φ(U) be open in RA/G. Then
p−1(UA)=A1∪U1,withA1 ⊂AandU1 openinR.
A
IfA1 ̸=∅then[a]A ∈UA andA⊂A1∪U1. ThereforeA∪U1 =A1∪U1 and p−1(U)=p−1(UA)=A∪U1 isopenandU isopenininR/G.
A
The result above is false if A is not open. Take, for instance, the homeomorphism group G = {f : R → R | f(0) = 0}. Then R/G is the Sierpinski space while R{0}/G is a 2-point discrete space.
It is known [1] that the Davey space can be obtained as a quotient R/G. Con- sequently we can use Proposition 3.2 to conclude
Theorem 3.3. There exists an open set A of R and a homeomorphism group G of RA such that RA/G and the Davey space are homeomorphic.
Only case 3) remains to be dealt with.
Theorem 3.4. For X as in 3), there is a set A and a group of homeomorphisms G of RA such that X and RA/G are homeomorphic.
Proof. Take A = [−1, 1] \ {0}, B = {0} and let C(resp.D) be the set of rational (resp. irrational) numbers greater than 1 or smaller than −1. Let G be formed by the homeomorphisms f : RA → RA such that f(0) = 0,f(A) = A,f(C) = C,f(D) = D. Then RA/G is homeomorphic to X.
Since reflection in the origin O is a homeomorphism of RA the only problem which may arise is, given irrationals i1 , i2 greater than 1, to find a homeomorphism from (1, +∞) to itself which maps i1 to i2 and preserves irrationals. In [2] we give a way of doing that.
Let (xi)i∈N (resp. (yi)i∈N) be an increasing (resp. decreasing) sequence of ra- tional numbers greater than 1 converging from below (resp. above) to i1. Similarly let now (zi)i∈N , (wi)i∈N be analogous sequences but converging to i2 this time. We define f : (1,+∞) → (1,+∞) by
IfA1 =∅thenp−1(U)=p−1(UA)=U1 andU isopeninR/G. A