Cancellation of 3-Point Topological Spaces

@Applied General Topology c Universidad Polit´ecnica de Valencia Volume 9, No. 1, 2008 pp. 15-19 Cancellation of 3-Point Topological Spaces S. Carter and F. J. Craveiro de Carvalho∗ Abstract. The cancellation problem, which goes back to S. Ulam [2], is formulated as follows: Given topological spaces X, Y, Z, under what circumstances does X× Z≈Y×Z(≈meaning homeomorphic to) imply X≈Y? In [1] it is proved that, for T0topological spaces and denoting by Sthe Sierpinski space, if X×S≈Y×Sthen X≈Y. This note concerns all nine (up to homeomorphism) 3-point spaces, which are given in [4]. 2000 AMS Classification: 54B10 Keywords: Homeomorphism, Cancellation problem, 3-point spaces. 1. Two cancellation results Below Xand Ydenote T1topological spaces. Proposition 1.1. Let Sbe a topological space with a unique closed singleton {p}. If there is a homeomorphism φ:X×S→Y×Sthen φ(X×{p}) = Y×{p}. Proof. We shall show that φ(X×{p})⊂Y×{p}which, using similar arguments, will be enough to prove that φ(X× {p}) = Y× {p}and, consequently, that X≈Y. Let us suppose that for some x∈X, y ∈Yand q∈S\ {p}we have φ(x, p) = (y, q). Then {(y, q)}is closed and, therefore, (Y×S)\ {(y, q)}is open. Let rbelong to the topological closure of {q}, r 6=q. Then (y, r)∈(Y×S)\ {(y, q)}and we must have open sets Uy, Ur, containing yand r, respectively, such that Uy×Ur⊂(Y×S)\ {(y, q)}. We reach a contradiction since (y, q) belongs to Uy×Ur. ∗The second named author gratefully acknowledges financial support from Funda¸c˜ao para a Ciˆencia e Tecnologia, Lisboa, Portugal. 16 S. Carter and F. J. Craveiro de Carvalho An example of such an Sis obtained as follows. Let Sbe a set with 4 elements at least. Let a, b ∈Sand denote by S1the complement of the subset they form. Take then as basis for a topology on Sthe set {{a},{a, b}, S1}. If Shappens to have just 4 points then it is the only minimal, universal space with such a number of elements [3]. Proposition 1.2. Let Sbe a topological space with a dense, open singleton {p}and such that, for every q∈S\ {p}, the topological closure of {q}is finite. If there is a homeomorphism φ:X×S→Y×Sthen φ(X× {p}) = Y× {p}. Proof. Let {p}be an open, dense singleton in S. We will show that φ(X× {p}) = Y× {p}which, as observed before, is enough to conclude that X≈Y. Assume that for some x∈X, y ∈Yand q6=pwe have φ(x, p) = (y, q). Consider the closed set {y}×{q}, the bar denoting closure, its image φ−1({y}× {q}), which is also closed, and suppose that {q}has selements. Also, observe that p /∈ {q}. Since (x, p) belongs to φ−1({y}×{q}) and this set has selements, there is an rin {q}such that (x, r) does not belong to this set. There are then open sets Ux, Ur, containing xand r, respectively, with Ux×Ur⊂(X×S)\φ−1({y}×{q}). We have a contradiction since (x, p)∈Ux×Ur. An example for Scan be the following Door space. Let Sbe a set and fix p∈S. Define U⊂Sto be open if it is empty or contains p. 2. 3-point spaces We go on assuming that X, Y are T1topological spaces though such assumption is not used in Propositions 2.1 and 2.2 below. If we now consider S={a, b, c}to be one of the 3-point spaces [4], we see that Propositions 1.1 and 1.2 of §1 allow us to deduce immediately that Scan be cancelled except in the following cases -Sis discrete, -Shas {{a},{b},{a, c}} as a topological basis, -Sis trivial. If Sis discrete the situation is not as simple as one might be led to think. Let us take the following example. Let S=Z, here Zstands for the integers with the discrete topology, and consider the discrete spaces X={0,1,...,n− 1}, n ≥2, Y ={0}. Now define φ:{0,1,...,n−1} × Z→ {0} × Zby φ(x, r) = (0, nr +x). This map is a homeomorphism and however Zcannot be cancelled. We can say something when the spaces X, Y have a finite number of connected components. Proposition 2.1. Let Sbe a finite discrete space and assume that Xhas a finite number of connected components. If X×S≈Y×Sthen X≈Y. Cancellation of 3-Point Topological Spaces 17 Proof. The connected components of X×Sor Y×Sare of the type X′× {x}, Y ′×{y}, where X′, Y ′are components of Xand Y, respectively. It follows that Yhas the same number of components as X. Let us consider in the sets of connected components of Xand connected components of Ythe homeomorphism equivalence relation and take an equivalence class of components of X, say {X1,...,Xk}. The subspace k [ i=1 Xi×S has kn components, where nis the cardinal of S. The same happens with φ( k [ i=1 Xi×S), where φis a homeomorphism between X×Sand Y×S. Let p∈S. For every i= 1,...,k,φ(Xi× {p}) = Yi× {qi}, where the qi’s belong to Sand the Yi’s are components of Yhomeomorphic to the Xi’s. Assume that the equivalence class to which the Yi’s belong is {Y1,...,Yl}. Then φ( k [ i=1 Xi× {p})⊂ l [ j=1 Yj×S. Consequently, also φ( k [ i=1 Xi×S)⊂ l [ j=1 Yj×S. Using the inverse homeomorphism φ−1, we are led to conclude that the reverse inclusion holds and, therefore, φ( k [ i=1 Xi×S) = l [ j=1 Yj×S. So k [ i=1 Xi×S and l [ j=1 Yj×Shave the same number of components and it follows that k=l. From each component class in Xchoose a representative and use φto establish a homeomorphism between that representative and a component in Y. These homeomorphisms can then be used to conclude that every component of Xis homeomorphic to a component of Y. Since components are closed and finite in number, Xis homeomorphic to Y. Proposition 2.2. Let Xand Ybe topological spaces with the same finite number of connected components and Sbe a discrete space. Assume, moreover, that neither space has two homeomorphic components. If X×S≈Y×Sthen X≈Y. Proof. Let Xi, i = 1,...,n, be the components of Xand fix p∈S. If φis a homeomorphism between X×Sand Y×Sthen there are qi∈ S, i = 1,...,n, such that φ(Xi× {p}) = Yi× {qi}, i = 1,...,n, where, due to our assumption on the non-existence of homeomorphic components, the Yi’s are the components of Y. Hence φinduces a homeomorphism φi:Xi→Yi, i = 1,...,n. Again, since the number of components is finite and they are closed, the φi’s can be used to obtain a homeomorphism between Xand Y. Proposition 2.3. Let Shave {{a},{b},{a, c}} as basis. If φ:X×S→Y×S is a homeomorphism then φ(X× {b}) = Y× {b}. 18 S. Carter and F. J. Craveiro de Carvalho Proof. Let πS:Y×S→Sdenote the standard projection. The image πS(φ(X× {b})) is open and, therefore, it is either {b}or contains a. Assume that for some x∈X, y ∈Ywe have φ(x, b) = (y, a). The subset {(x, b)}is closed and, consequently, the same happens with {(y, a)}. Hence (Y×S)\ {(y, a)}is open and contains (y, c). We must then have an open neighbourhood Uyof ysuch that Uy× {a, c} ⊂ (Y×S)\ {(y, a)}. Again we have a contradiction and φ(X× {b}) = Y× {b}. To conclude the proof that a non-discrete 3-point space can be cancelled it only remains to deal with the case where Sis trivial. Above we have an example of a homeomorphism φ:X×S→Y×Swhich does take a slice X×{x}onto a slice Y×{y}. More examples can be obtained. Take X=Y, with at least 2 elements, a trivial space Swith also, at least, 2 elements and let ψ:S→Sbe a fixed point free bijection. Fix x0∈Xand define φ:X×S→X×Sby φ(x, s) = (x, s), for x6=x0, and φ(x0, s) = (x0, ψ(s)). Then φis a bijection and φ({x} × S) = {x} × S, for x∈X. Since open sets in X×Sare of the form U×S,Uopen in X, and φ(U×S) = U×S,φis a homeomorphism. Obviously no slice X× {x}is mapped onto a similar slice. Proposition 2.4. Let Sbe a finite trivial space. If X×S≈Y×Sthen X≈Y. Proof. Open (closed) sets in X×Sand Y×Sare of the form U×S, where U is open (closed). We are going to define f:X→Yas follows. Let x∈X. Then {x}is closed and so are {x} × Sand φ({x} × S), where φ:X×S→Y×Sis a homeomorphism. Hence φ({x}×S) = C×S, for some closed set Cin Y. Since Sis finite, Cis a singleton and we make {f(x)}=C. This way we obtain an fwhich is a bijection since we began with a bijective φ. If Cis closed in X,φ(C×S) = f(C)×Sis closed in Y×S. Consequently f(C) is closed in Y. Therefore fis closed and f−1is continuous. Taking φ−1, we would conclude that fis continuous the same way.  We can now state. Theorem 2.5. For Xand Y T1topological spaces and Sa non-discrete 3-point topological space, if X×S≈Y×Sthen X≈Y. 3. A particular case We will no longer assume X, Y to be T1and will suppose that Shas a unique isolated point a. Moreover, the singleton {a}will be assumed to be closed. That is, for instance, the case where S={a, b, c}and {{a},{b, c}} is an open basis. Cancellation of 3-Point Topological Spaces 19 Proposition 3.1. Let Shave a unique isolated point a. Assume that {a}is closed. For X, Y connected with, at least, an isolated point each, if φ:X×S→ Y×Sis a homeomorphism then φ(X× {a}) = Y× {a}. Proof. Let πS:Y×S→Sdenote the standard projection, as before. The image πS(φ(X×{a})) is open and connected. Therefore it is either {a} or some open, connected subset of S, which naturally does not contain a. Let the latter be the case. If x∈Xis an isolated point then {(x, a)}is open and the same happens to its image under πS◦φ. This is impossible because {a}is the unique open singleton of S. Examples of spaces satisfying the conditions of Proposition 3.1 are, again, some Door spaces. Let Zbe a set. Fix p∈Zand define U⊂Zto be open if U=Zor p /∈U. References [1] B. Banaschewski and R. Lowen, A cancellation law for partially ordered sets and T0 spaces, Proc. Amer. Math. Soc. 132 (2004). [2] R. H. Fox, On a problem of S. Ulam concerning cartesian products, Fund. Math. 27 (1947). [3] K. D. Magill Jr, Universal topological spaces, Amer. Math. Monthly 95 (1988). [4] J. R. Munkres, Topology, a first course, Prentice-Hall, Inc., 1975. Received July 2006 Accepted November 2006 S. Carter (s.ca[email protected]) School of Mathematics, University of Leeds, Leeds LS2 9JT, U.K. F. J. Craveiro de Carvalho ([email protected]) Departamento de Matem´atica, Universidade de Coimbra, 3001-454 Coimbra, PORTUGAL