aa r X i v : . [ m a t h . D S ] O c t CONTINUOUS IMAGES OF CANTOR’S TERNARY SET
F. DREHER ∗ AND T. SAMUEL † A bstract . The Hausdor ff –Alexandro ff Theorem states that any compact metric space isthe continuous image of Cantor’s ternary set C . It is well known that there are compactHausdor ff spaces of cardinality equal to that of C that are not continuous images of Cantor’sternary set. On the other hand, every compact countably infinite Hausdor ff space is acontinuous image of C . Here we present a compact countably infinite non-Hausdor ff spacewhich is not the continuous image of Cantor’s ternary set.
1. I ntroduction .One of the simplest sets that is widely studied by and most important to many mathe-maticians, in particular analysts and topologists, is Cantor’s ternary set (also referred to asthe middle third Cantor set), introduced by H. Smith [7] and by G. Cantor [2]. Cantor’sternary set is generated by the simple recipe of dividing the unit interval [0 ,
1] into threeparts, removing the open middle interval, and then continuing the process so that at eachstage, each remaining subinterval is similarly subdivided into three and the middle openinterval removed. Continuing this process ad infinitum one obtains a non-empty set con-sisting of an infinite number of points. We now formally define Cantor’s ternary set inarithmetic terms.
Definition 1.1.
Cantor’s ternary set is defined to be the set C : = X n ∈ N ω n n : ω n ∈ { , } for all n ∈ N . Throughout, C will denote Cantor’s ternary set equipped with the subspace topologyinduced by the Euclidean metric. Below we list some of the remarkable properties ofCantor’s ternary set. For a proof of Property (1), more commonly known as the Hausdor ff –Alexandro ff Theorem, we refer the reader to [8, Theorem 30.7]. Both F. Hausdor ff [4] andP. S. Alexandro ff [1] published proofs of this result in 1927. For Properties (2) to (8), werefer the reader to [3] or [6, Counterexample 29]; for a proof of Properties (9) and (10), thedefinition of the Lebesgue measure, Hausdor ff dimension and that of a self-similar set, werefer the reader to [3]. For basic definitions of topological concepts see [5] or [8].(1) Any compact metric space is the continuous image of C .(2) The set C is totally disconnected.(3) The set C is perfect.(4) The set C is compact.(5) The set C is nowhere dense in the closed unit interval [0 , C is Hausdor ff .(7) The set C is normal.(8) The cardinality of C is equal to that of the continuum.(9) The one dimensional Lebesgue measure and outer Jordan content of C are both zero.(10) The set C is a self-similar set and has Hausdor ff dimension equal to log(2) / log(3).In this note we are interested in whether the Hausdor ff –Alexandro ff Theorem can bestrengthened. To avoid any misunderstandings regarding the compactness condition that
Date : October 7, 2013. ∗ [email protected] - Fachbereich 3 - Mathematik, Universit¨at Bremen, 28359 Bremen, Germany. † [email protected] - Fachbereich 3 - Mathematik, Universit¨at Bremen, 28359 Bremen, Germany. might stem from di ff erent naming traditions, we shall explicitly define what we mean bycompact. Note that a compact space does not have to be Hausdor ff . Definition 1.2 (Compact) . Given a subset A ⊆ X of a topological space ( X , τ ), an opencover of A is a collection of open sets whose union contains A . An open subcover is a sub-collection of an open cover whose union still contains A . We call a subset A of X compact if every open cover has a finite open subcover.It is known that for compact Hausdor ff spaces the properties (i) metrizability, (ii) second-countability and (iii) being a continuous image of Cantor’s ternary set are equivalent. Thisfollows from the fact that a compact Hausdor ff space is metrizable if and only if it issecond-countable (see for instance [5, p. 218]) the Hausdor ff –Alexandro ff Theorem, andthe fact that the continuous image of a compact metric space (in our case the Cantor set C )in a compact Hausdor ff space is again a compact metrizable space ([8, Corollary 23.2]).Obviously the cardinality of a space that is the continuous image of C cannot exceedthat of the continuum. This restriction on cardinality is a necessary, but not a su ffi cientcondition because there are compact Hausdor ff spaces with cardinality equal to that ofthe continuum which are not second-countable, for instance the Alexandro ff one-pointcompactification of the discrete topological space ( R , P ( R )), where P ( R ) denotes the powerset of R .Restricting the cardinality even further leads to a su ffi cient condition. If we only look atcountably infinite target spaces, we can deduce that for compact countably infinite spacesthe Hausdor ff property already implies metrizability. In a countably infinite Hausdor ff space, every point is a G δ point which together with compactness implies that the spaceis first-countable [8, Problem 16.A.4]; since the space is countably infinite it follows thatit is also second-countable. Hence, we have a space which is compact Hausdor ff andsecond-countable and so, using the above mentioned equivalence, metrizable. Therefore,any compact countably infinite Hausdor ff space is the continuous image of C . Is this strongrestriction on the space’s cardinality a su ffi cient condition for the Hausdor ff –Alexandro ff Theorem, i.e. is every compact countably infinite topological space the continuous imageof C ? This is precisely the question we address and, in fact, show that the answer is no byexhibiting a counterexample. Theorem 1.3.
There exists a compact countably infinite topological space ( T , τ ) which isnot the continuous image of C. In order to prove this result we require an auxiliary result, Lemma 2.1. In the proof ofthis result we rectify an error in [6, Counterexample 99].The main idea behind the proof of Theorem 1.3 is to choose a specific non-Hausdor ff space and to show that if there exists a continuous map from the Cantor set into this space,then the continuous map must push-forward the Hausdor ff property of Cantor’s ternary set,which will be a contradiction to how the target space was originally chosen.2. P roof of T heorem Lemma 2.1.
There exists a countable topological space ( T , τ ) with the following proper-ties:(a) ( T , τ ) is compact,(b) ( T , τ ) is non-Hausdor ff , and(c) every compact subset of T is closed with respect to τ .Proof. This proof is based on [6, Counterexample 99]. We define T : = ( N × N ) ∪ { x , y } ,namely the Cartesian product of the set of natural numbers N with itself unioned withtwo distinct arbitrary points x and y . We equip the set T with the topology τ whose baseconsists of all sets of the form:(i) { ( m , n ) } , where ( m , n ) ∈ N × N , ONTINUOUS IMAGES OF CANTOR’S TERNARY SET 3 (ii) T \ A , where A ⊂ ( N × N ) ∪ { y } contains y and is such that the cardinality of the set A ∩ { ( m , n ) : n ∈ N } is finite for all m ∈ N ; that is, the set A contains at most finitelymany points on each row (these sets are the open neighbourhoods of x ) and(iii) T \ B , where B ⊂ ( N × N ) ∪ { x } contains x and is such that there exists an M ∈ N , sothat if ( m , n ) ∈ B ∩ ( N × N ), then m ≤ M ; that is B contains only points from at mostfinitely many rows (these sets are the open neighbourhoods of y ).Property (a) follows from the observation that any open cover of T contains at least oneopen neighbourhood U ⊇ T \ A of x and one open neighbourhood V ⊇ T \ B of y with A and B as given above. The points not already contained in these two open sets are contained in T \ ( U ∪ V ) ⊆ T \ (( T \ A ) ∪ ( T \ B )) = A ∩ B which, by construction, is a finite set. In thisway a finite open subcover can be chosen and hence the topological space ( T , τ ) is compact.To see why ( T , τ ) has Property (b), consider open neighbourhoods of x and y . An openneighbourhood of x contains countably infinitely many points on each row of the lattice N × N ; an open neighbourhood of y contains countably infinitely many full rows. It followsthat there are no disjoint open neighbourhoods U ∋ x and V ∋ y and thus T is non-Hausdor ff .We use contraposition to prove Property (c). Suppose that E ⊂ T is not closed. Notethat we may assume that E is a strict subset of T since T itself is closed by the fact that ∅ ∈ τ . By construction of the topology, a set that is not closed cannot contain both x and y . Also, there needs to be at least one point in the closure E of E , but not already in E ;this has to be one of the points x or y , because singletons { ( m , n ) } which are subsets of thelattice N × N are open. Thus the point ( m , n ) cannot be a limit point of E . We shall nowcheck both cases, that is, (i) if x ∈ E \ E and (ii) if y ∈ E \ E .(i) If x ∈ E \ E , then every open neighbourhood of x has a non-empty intersection with E . It follows that there is at least one row in N × N that shares infinitely many pointswith E . Denote this row by B . Then the open cover { T \ B } ∪ {{ b } : b ∈ B ∩ E } of E cannot be reduced to a finite open subcover, and therefore, E is not compact.(ii) If y ∈ E \ E , then, similar to (i), we have that E contains points from infinitely manyrows. Take one point from each of these rows and call the resulting set A . Then theopen cover { T \ A } ∪ {{ a } : a ∈ A ∩ E } of E cannot be reduced to a finite open subcoverand hence E is not compact. (cid:3) Proof of 1.3.
Assume that there exists a surjective continuous map f : C → T . We willshow that this implies that ( T , τ ) is Hausdor ff which contradicts Lemma 2.1(b).Choose two distinct points u , v ∈ T . Since singletons are compact, Lemma 2.1(c) impliesthat the sets { u } and { v } are closed in T with respect to τ . Therefore, their pre-images under f are non-empty closed subsets of C and have disjoint open neighbourhoods U ( u ) and U ( v ), as C is normal. The complements U ( u ) c and U ( v ) c of these open neighbourhoodsare compact subsets of C . Thus, their images under f are compact in T because f iscontinuous and they are closed because of Lemma 2.1(c). Therefore, V ( u ) : = f ( U ( u ) c ) c and V ( v ) : = f ( U ( v ) c ) c are open neighbourhoods of u and v respectively. We claim thatthese sets are disjoint: V ( u ) ∩ V ( v ) = f (cid:0) U ( u ) c (cid:1) c ∩ f (cid:0) U ( v ) c (cid:1) c = (cid:0) f (cid:0) U ( u ) c (cid:1) ∪ f (cid:0) U ( v ) c (cid:1)(cid:1) c = (cid:0) f (cid:0) U ( u ) c ∪ U ( v ) c (cid:1)(cid:1) c = (cid:0) f (cid:0) ( U ( u ) ∩ U ( v )) c (cid:1)(cid:1) c = (cid:0) f (cid:0) ∅ c (cid:1)(cid:1) c = ∅ . Hence we have separated the points u and v by open neighbourhoods. Since u , v ∈ T werechosen arbitrarily, we conclude that ( T , τ ) is Hausdor ff , giving the desired contradiction. (cid:3) R eferences [1] P. S. Alexandro ff , ¨Uber stetige Abbildungen kompakter R¨aume, Math. Annalen (1927) 55- 571.[2] G. Cantor, Grundlagen einer allgemeinen Mannigfaltigkeitslehre, Math. Annalen (1883) 545-591. F. DREHER AND T. SAMUEL [3] K. J. Falconer,
Fractal Geometry. second edition. John Wiley and Sons, Chichester, West Sussex, 2003.[4] F. Hausdor ff , Mengenlehre, Zweite, neubearbeitete Auflage . Walter de Gruyter & Co., 1927.[5] J. R. Munkres,
Topology. second edition. Prentice-Hall, Upper Saddle River, NJ, 2000.[6] L. R. Steen and J. A. Seebach (Jr.),
Counterexamples in Topology.