Page 60 - Textos de Matemática Vol. 34

50 SHEILA CARTER AND F. J. CRAVEIRO DE CARVALHO
Proof. The condition is necessary:
First observe that any two points y1, y2 in Y are related by the homeomorphism
which interchanges them and is the identity outside {y1,y2}. Also no point z ∈/ Y isrelatedwithapointinY. Infactifwehadf :RY →RY suchthatf(y)=z we would conclude that {z} would be open in R. Since RY /G is discrete RY \ Y is open in RY and, consequently, in R. Therefore Y is closed.
The condition is sufficient:
Assume that Y is closed in R. It is enough to show that any two points z1, z2 ∈/ Y are related. Let C1, C2 be the connected components of RY \Y which contain z1 and z2, respectively. Observe that the subset topology for RY \Y is the same whether we consider it as a subspace of RY or R. Therefore C1, C2 are open intervals and there isahomeomorphismh:C1 →C2 whichmapsz1 toz2. Usingh,h−1 ifC1 ̸=C2,and the identity outside C1 ∪ C2 we obtain a global homeomorphism f : RY → RY such that f(z1) = z2. Therefore there are two equivalence classes in RY /G with Y and RY \Y as inverse images by the projection pY . Consequently, RY /G is a discrete 2- -point space.
If instead of working with R we chose to work with a topological n-manifold M we would have a similar result. That is, MY /G is discrete if and only if Y is closed. In that situation the equivalence classes would have as inverse images Y and the unions of the homeomorphic components of M \ Y .
3. Universal Spaces
Definition 3.1. A topological space X is universal if any topological space is home-
omorphic to a subspace of some topological power of X.
In [3] it was shown that X is universal if and only if it contains a copy of one of
the following spaces, where τ stands for the topology:
1) X = {a,b,c}, τ = {∅,X,{a,b}}.
2) X = {a, b, c}, τ = {∅, X, {a}}, this is the so-called Davey space. 3) X = {a,b,c,d}, τ = {∅,X,{a},{a,b},{c,d},{a,c,d}}.
These spaces can then in a way be considered minimal universal spaces. Can they be homeomorphic to a space RY /G?
Theorem 3.1. For X as in 1), there is no homeomorphic space of the form RY /G.
Proof. We just have to point out that, for y ∈ Y , {y} is open and therefore its projectionisopeninRY/GforpY isanopenmap.Thismeansthatsuchaquotient always has an open singleton.
Let us now consider case 2), the Davey space case.
We start with the following observation:
Iff :R→Riscontinuousandf(R\Y)⊂R\Y thenf :RY →RY iscontinuous.