carvalag.dvi @ Applied General Topology c© Universidad Politécnica de Valencia Volume 5, No. 1, 2004 pp. 91- 96 Homeomorphisms of R and the Davey Space Sheila Carter and F. J. Craveiro de Carvalho Abstract. Up to homeomorphism, there are 9 topologies on a three point set {a, b, c} [4]. Among the resulting topological spaces we have the so called Davey space, where the only non-trivial open set is, let us say, {a}. This is an interesting topological space to the extent that every topological space can be embedded in a product of Davey spaces [3]. In this note we will consider the problem of obtaining the Davey space as a quotient R/G, where G is a suitable homeomorphism group. The present work can be regarded as a follow-up to some previous work done by one of the authors and Bernd Wegner [1]. 2000 AMS Classification: 54F65. Keywords: Davey space, homeomorphism group, Cantor set. 1. R/G as the Davey space -necessary conditions We will take the topological space ({a, b, c}, τ), with τ = {∅, {a}, {a, b, c}}, as a model for the Davey space and the real line will be denoted by R. Our purpose is to obtain a group G of homeomorphisms of R whose natural action on R gives rise to the Davey space and we start by establishing a number of observations which guided our quest. Below we assume that the homeomorphism group G is such that R/G is the Davey space and π will stand for the projection from R to R/G. Proposition 1.1. G is not finite. Proof. If G were finite then, for instance, π−1(b) would be finite and, conse- quently, {a, c} would be open. � Proposition 1.2. π−1(a) is bounded neither above nor below. Proof. Assume that π−1(a) is bounded above and let x be its supremum. Then π((x, +∞)) is open in R/G and, consequently must contain a. � 92 Sheila Carter and F. J. Craveiro de Carvalho Proposition 1.3. π−1({b, c}) is bounded neither above nor below. Proof. Assume that x is the supremum of π−1({b, c}). Since this set is closed in R, x belongs to it. Let us suppose that π(x) = b. As we will see below it then follows that | π−1(b) |≤ 2 which, as remarked above, is impossible. Let y, z be points in π−1(b) with y < z < x. There is a homeomorphism f in G such that f(z) = x. If f were increasing then f(x) > x. Therefore f must be decreasing and, since y < z, f(y) > x which, again, is impossible. � Proposition 1.4. π−1(b), π−1(c) are bounded neither above nor below. Proof. Assume that π−1(b) is bounded above and let x be its supremum. Then π((x, +∞)) must be {a} and x is an upper bound for π−1(c). Consequently π−1({b, c}) would be bounded above. � We are now in a position which allows us to conclude Theorem 1.5. The action of G is not free. Proof. Let π−1(a) = ⋃ i∈I Ci, where the Ci’s are the connected components. From above it follows that, for each i, Ci = (ai, bi). Fix an i and choose x, y distinct in Ci. There is an f ∈ G such that f(x) = y. Since f maps [ai, bi] into itself, it must have a fixed point. � It is also clear that π−1({b, c}) is totally disconnected and that every point in it is a limit point of that set. Proposition 1.6. π−1({b, c}) is uncountable. Proof. Write π−1(a) = ⋃ i∈I Ci and choose x, y in different components, with x < y. Then π−1({b, c}) ⋂ [x, y] is a compact, Hausdorff space having all its elements as limits points. Therefore it is uncountable [4]. � 2. An example This section is devoted to the construction of an example of a group G such that R/G is the Davey space. Since they are homeomorphic spaces we will use the open interval (0, 1) instead of R. Let C denote the intersection of the Cantor set [2], [5] with (0, 1) and consider the partition (0, 1) = A∪B∪C, where A = (0, 1)\C is a union of open intervals, the “middle thirds”, B is the set of end-points of the open intervals in A and C = C \ B. The Cantor set can be described in terms of ternary expansions. We then have, for x ∈ (0, 1), that Homeomorphisms of R and the Davey Space 93 x ∈ A if and only if there is n ∈ N such that x = ∞∑ i=1 xi 3i , where, for i < n, xi = 0 or 2, xn = 1 and 0 < ∞∑ i=n+1 xi 3i < 1 3n , x ∈ B if and only if there is n ∈ N such that x = n∑ i=1 xi 3i , where, for i < n, xi = 0 or 2, xn = 1 or 2. x ∈ C if and only if x = ∞∑ i=1 xi 3i , with xi = 0 or 2, and there are arbitrarily large i and j for which xi = 0, xj = 2. Proposition 2.1. The quotient topological space originated by the partition (0, 1) = A ∪ B ∪ C is the Davey space. Proof. Let X = {a, b, c} be the quotient space obtained by identifying A, B, C to points a, b, c, respectively. Since A is open in (0, 1) it follows that {a} is open in X. Let now x ∈ B and suppose that x = n−1∑ i=1 xi 3i + 1 3n , where xi = 0 or 2. For k ≥ n + 1, define yk ∈ C by yk = n−1∑ i=1 xi 3i + k∑ i=n+1 2 3i + ∞∑ j=1 2 3k+2j . Then the sequence (yk) converges to n−1∑ i=1 xi 3i + ∞∑ i=n+1 2 3i , which is x. Similarly if x = n−1∑ i=1 xi 3i + 2 3n , where xi = 0 or 2, for k ≥ n + 1, define yk = x + ∞∑ j=1 2 3k+2j . This sequence also converges to x. Thus every element of B belongs to the closure of C and, since it is an end-point of an open interval in A, it also lies in the closure of A. Hence every open set in X containing b also contains a and c and the only such open set is X itself. Next consider x ∈ C, say x = ∞∑ i=1 xi 3i , where xi = 0 or 2. There exists an arbitrarily large i for which xi = 2. Let xl be the first nonzero term and, for k ≥ l, define yk ∈ B by yk = k∑ i=1 xi 3i . The sequence (yk) converges to x. Thus x lies in the closure of B and, as each yk is in the closure of A, it also lies in the closure of A. So every open set in X containing c also contains a and b and the only such open set is X itself. Therefore X is the Davey space. � 94 Sheila Carter and F. J. Craveiro de Carvalho Let G = {h : (0, 1) → (0, 1) | h is a homeomorphism and h(A) = A}. If h ∈ G then h takes an open interval in A to an open interval in A and, consequently, the end-points to end-points. So h(B) = B and then h(C) = C. If we prove that G acts transitively on A, B and C we may conclude that those subsets are the orbits of the natural action of G on (0, 1) and, by Proposition 2.1, (0, 1)/G is the Davey space. Proposition 2.2. G acts transitively on A. Proof. We start by observing that, given any open interval (α, β) in (0, 1) and x, y ∈ (α, β), there exists a homeomorphism h : (0, 1) → (0, 1) such that h((α, β)) = (α, β), h(x) = y and h | (0, 1) \ (α, β) is the identity function. To prove transitivity on A it is therefore enough to show that, for any open interval (α, β) in A, with α, β ∈ B, there is h ∈ G such that h((α, β)) = (1 3 , 2 3 ). Assume α = 2 3i1 +. . .+ 2 3ik + 1 3n , 1 ≤ i1 < i2 < . . . < ik < n. So β = α+ 1 3n . Let j1, . . . , jl be such that 1 ≤ j1 < . . . jl < n, {j1, . . . , jl}∪{i1, i2, . . . , ik} = {1, 2, . . . , n − 1}, l + k = n − 1. Hence α + 2 3j1 + . . . + 2 3jl + 1 3n = n∑ i=1 2 3i = 1 − 1 3n and, in the construction of h, we will use 1 − ( 2 3j1 + . . . + 2 3jl + 1 3n + s 3n ) = α + t 3n , where s = 1 − t, t ∈ [0, 1]. We define h : (0, 1) → (0, 1) as follows: (0, 2 3i1 ] is mapped to (0, 2 32 ] by h( 2t 3i1 ) = 2t 32 , with t ∈ (0, 1], for r = 2, . . . , k, [ 2 3i1 + . . . + 2 3ir−1 , 2 3i1 + . . . + 2 3ir ] is mapped to [ 2 32 + . . . + 2 3r , 2 32 + . . . + 2 3r+1 ] by h( 2 3i1 + . . . + 2 3ir−1 + 2t 3ir ) = 2 32 + . . . + 2 3r + 2t 3r+1 , with t ∈ [0, 1], [ 2 3i1 +. . .+ 2 3ik , α] is mapped to [ 2 32 +. . .+ 2 3k+1 , 1 3 ] by h( 2 3i1 +. . .+ 2 3ik + t 3n ) = 2 32 + . . . + 2 3k+1 + t 3k+1 , with t ∈ [0, 1], [α, α + 1 3n ] is mapped to [1 3 , 2 3 ] by h(α + t 3n ) = 1 3 + t 3 , with t ∈ [0, 1], [α + 1 3n , α + 2 3n ] is mapped to [2 3 , 2 3 + 1 3l+1 ] by h(1 − ( 2 3j1 + . . . + 2 3jl + s 3n )) = 1 − ( 2 32 + . . . + 2 3l+1 + s 3l+1 ), with s ∈ [0, 1], for r = 2, . . . , l, [1 − ( 2 3j1 + . . . + 2 3jr ), 1 − ( 2 3j1 + . . . + 2 3jr−1 )] is mapped to [1 − ( 2 32 + . . . + 2 3r+1 ), 1 − ( 2 32 + . . . + 2 3r )] by h(1 − ( 2 3j1 + . . . + 2 3 jr−1 + 2s 3jr )) = 1 − ( 2 32 + . . . + 2 3r + 2s 3r+1 ), with s ∈ [0, 1], [1 − 2 3j1 , 1) is mapped to [1 − 2 32 , 1) by h(1 − 2s 3j1 ) = 1 − 2s 32 , with s ∈ (0, 1]. We have then a homeomorphism h : (0, 1) → (0, 1) such that h((α, β)) = (1 3 , 2 3 ). On each interval of its definition, h is of the form h(x) = λx + µ, for some λ, µ ∈ R. Hence it takes middle thirds in (0, 1 3i1 ) to middle terms in Homeomorphisms of R and the Davey Space 95 (0, 1 32 ), middle thirds in ( 2 3i1 + . . . + 2 3 ir−1 , 2 3i1 + . . . + 2 3 ir−1 + 1 3r ) to middle thirds in ( 2 32 + . . . + 2 3r , 2 32 + . . . + 2 3r + 1 3r+1 ), for r = 2, . . . , k, and so on for the other intervals. So h(A) = A and h ∈ G as required. � Proposition 2.3. G acts transitively on B. Proof. The homeomorphism h constructed above maps [α, β] to [1 3 , 2 3 ], with h(α) = 1 3 , h(β) = 2 3 and every element in B is such an α or β. Composing h with the reflection of (0, 1) that sends x to 1 − x gives an element of G that takes β to 1 3 . Hence, for α ∈ B, there exists g ∈ G with g(α) = 1 3 . Therefore G acts transitively on B. � Proposition 2.4. G acts transitively on C. Proof. Since 1 4 ∈ C, it suffices to show that, for γ ∈ C, there is h ∈ G such that h(γ) = ∞∑ n=1 2 32n = 1 4 . Let γ = ∞∑ n=1 2 3in , I = {i1, i2, . . .}, J = N \ I = {j1, j2, . . .}. Define h : (0, 1) → (0, 1) as follows: (0, 2 3i1 ] is mapped to (0, 2 32 ] by h( 2t 3i1 ) = 2t 32 , with t ∈ (0, 1], for n = 2, . . ., [ 2 3i1 +. . .+ 2 3in−1 , 2 3i1 +. . .+ 2 3in ] is mapped to [ 2 32 + 2 34 +. . .+ 2 32n−2 , 2 32 +. . .+ 2 32n ] by h( 2 3i1 +. . .+ 2 3 in−1 + 2t 3in ) = 2 32 + 2 34 +. . .+ 2 32n−2 + 2t 32n , with t ∈ [0, 1], h( ∞∑ n=1 2 3in ) = ∞∑ n=1 2 32n , for n = 2, . . ., [1 − ( 2 3j1 + . . . + 2 3jn ), 1 − ( 2 3j1 + . . . + 2 3jn−1 )] is mapped to [1 − (2 3 + 2 33 + . . . + 2 32n−1 ), 1 − (2 3 + 2 33 + . . . + 2 32n−3 )] by h(1 − ( 2 3j1 + . . . + 2 3jn−1 + 2s 3jn )) = 1 − (2 3 + 2 33 + . . . + 2 32n−3 + 2s 32n−1 ), with s ∈ [0, 1], [1 − 2 3j1 , 1) is mapped to [1 3 , 1) by h(1 − 2s 3j1 ) = 1 − 2s 3 , with s ∈ (0, 1]. We have therefore defined an h ∈ G with h(γ) = 1 4 as required. � We can now conclude with our main result. Theorem 2.5. There is a group G of homeomorphisms of R such that the quotient space R/G is the Davey space. 96 Sheila Carter and F. J. Craveiro de Carvalho Acknowledgements. The authors are very grateful to Alan West for dis- cussions on the topic of this paper which have led to a complete proof of Theorem 2.5. References [1] F. J. Craveiro de Carvalho and Bernd Wegner, Locally Sierpinski spaces as interval quotients, Kyungpook Math. J. 42 (2002), 165-169. [2] Ryszard Engelking, General Topology, Heldermann Verlag, 1989. [3] Sidney A. Morris, Are finite topological spaces worthy of study?, Austral. Math. Soc. Gazette 11 (1984), 563-564. [4] James R. Munkres, Topology, a first course, Prentice-Hall, Inc., 1975. [5] Stephen Willard, General Topology, Addison-Wesley, Inc., 1970. Received December 2002 Accepted April 2003 Sheila Carter (S.Carter@leeds.ac.uk) School of Mathematics, University of Leeds, Leeds LS2 9JT, U. K. F. J. Craveiro de Carvalho (fjcc@mat.uc.pt) Departamento de Matemática, Universidade de Coimbra, 3001 - 454 Coimbra, Portugal