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Applied General Topology

c© Universidad Politécnica 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 T0 topological spaces and denoting by S the
Sierpinski space, if X × S ≈ Y × S then 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 X and Y denote T1 topological spaces.

Proposition 1.1. Let S be a topological space with a unique closed singleton
{p}. If there is a homeomorphism φ : X×S → Y ×S then 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 ∈ Y and q ∈ S \ {p} we have
φ(x,p) = (y,q). Then {(y,q)} is closed and, therefore, (Y × S) \ {(y,q)} is
open.

Let r belong 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 y and 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 Fundação para
a Ciência e Tecnologia, Lisboa, Portugal.



16 S. Carter and F. J. Craveiro de Carvalho

An example of such an S is obtained as follows. Let S be a set with 4
elements at least. Let a,b ∈ S and denote by S1 the complement of the subset
they form. Take then as basis for a topology on S the set {{a},{a,b},S1}. If
S happens to have just 4 points then it is the only minimal, universal space
with such a number of elements [3].

Proposition 1.2. Let S be 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 × S then 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 ∈ Y and q 6= p we 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 s elements. Also, observe

that p /∈ {q}.

Since (x,p) belongs to φ−1({y}×{q}) and this set has s elements, there is an

r in {q} such that (x,r) does not belong to this set. There are then open sets

Ux,Ur, containing x and r, respectively, with Ux×Ur ⊂ (X×S)\φ
−1({y}×{q}).

We have a contradiction since (x,p) ∈ Ux × Ur. �

An example for S can be the following Door space. Let S be a set and fix
p ∈ S. Define U ⊂ S to be open if it is empty or contains p.

2. 3-point spaces

We go on assuming that X,Y are T1 topological spaces though such assum-
ption 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 S can
be cancelled except in the following cases

- S is discrete,
- S has {{a},{b},{a,c}} as a topological basis,
- S is trivial.

If S is discrete the situation is not as simple as one might be led to think.
Let us take the following example. Let S = Z, here Z stands 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} × Z by
φ(x,r) = (0,nr + x). This map is a homeomorphism and however Z cannot be
cancelled.

We can say something when the spaces X,Y have a finite number of con-
nected components.

Proposition 2.1. Let S be a finite discrete space and assume that X has a
finite number of connected components. If X × S ≈ Y × S then X ≈ Y .



Cancellation of 3-Point Topological Spaces 17

Proof. The connected components of X × S or Y × S are of the type X′ ×
{x},Y ′ ×{y}, where X′,Y ′ are components of X and Y , respectively. It follows
that Y has the same number of components as X.

Let us consider in the sets of connected components of X and connected
components of Y the homeomorphism equivalence relation and take an equi-

valence class of components of X, say {X1, . . . ,Xk}. The subspace
k⋃

i=1

Xi × S

has kn components, where n is the cardinal of S. The same happens with

φ(

k⋃

i=1

Xi × S), where φ is a homeomorphism between X × S and Y × S.

Let p ∈ S. For every i = 1, . . . ,k, φ(Xi × {p}) = Yi × {qi}, where the
qi’s belong to S and the Yi’s are components of Y homeomorphic 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 re-

verse 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 × S have the same number of components and it follows that k = l.

From each component class in X choose a representative and use φ to es-
tablish a homeomorphism between that representative and a component in Y .
These homeomorphisms can then be used to conclude that every component
of X is homeomorphic to a component of Y . Since components are closed and
finite in number, X is homeomorphic to Y . �

Proposition 2.2. Let X and Y be topological spaces with the same finite num-
ber of connected components and S be a discrete space. Assume, moreover,
that neither space has two homeomorphic components. If X × S ≈ Y × S then
X ≈ Y .

Proof. Let Xi, i = 1, . . . ,n, be the components of X and fix p ∈ S.
If φ is a homeomorphism between X × S and Y × S then 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 φ induces 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 X and Y . �

Proposition 2.3. Let S have {{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 → S denote 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 ∈ Y we 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 Uy of y such 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 S is trivial.

Above we have an example of a homeomorphism φ : X × S → Y × S which
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 S with also, at least,
2 elements and let ψ : S → S be a fixed point free bijection. Fix x0 ∈ X and
define φ : X × S → X × S by φ(x,s) = (x,s), for x 6= x0, and φ(x0,s) =
(x0,ψ(s)).

Then φ is a bijection and φ({x} × S) = {x} × S, for x ∈ X. Since open sets
in X × S are of the form U × S, U open in X, and φ(U × S) = U × S, φ is a
homeomorphism. Obviously no slice X × {x} is mapped onto a similar slice.

Proposition 2.4. Let S be a finite trivial space. If X × S ≈ Y × S then
X ≈ Y .

Proof. Open (closed) sets in X × S and Y × S are of the form U × S, where U
is open (closed).

We are going to define f : X → Y as follows. Let x ∈ X. Then {x} is
closed and so are {x} × S and φ({x} × S), where φ : X × S → Y × S is a
homeomorphism. Hence φ({x}×S) = C ×S, for some closed set C in Y . Since
S is finite, C is a singleton and we make {f(x)} = C.

This way we obtain an f which is a bijection since we began with a bijective
φ.

If C is closed in X, φ(C × S) = f(C) × S is closed in Y × S. Consequently
f(C) is closed in Y . Therefore f is closed and f−1 is continuous.

Taking φ−1, we would conclude that f is continuous the same way. �

We can now state.

Theorem 2.5. For X and Y T1 topological spaces and S a non-discrete 3-point
topological space, if X × S ≈ Y × S then X ≈ Y .

3. A particular case

We will no longer assume X,Y to be T1 and will suppose that S has 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 S have 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 × S is a homeomorphism then φ(X × {a}) = Y × {a}.

Proof. Let πS : Y × S → S denote 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 ∈ X is 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 Z be a set. Fix p ∈ Z and define U ⊂ Z to be open if U = Z or 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.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