On good connected preimages

Topology and its Applications 125 (2002) 489–496

On good connected preimages

Alessandro Fedelia,∗, Attilio Le Donneb

a Department of Mathematics, University of L’Aquila, Italyb Department of Mathematics, University of Rome “La Sapienza”, Italy

Received 30 March 2001; received in revised form 8 November 2001

Abstract

In this paper we investigate when a connected space has a connected preimage with someadditional properties. In particular we completely characterize the continuous and the quotientimages of metric connected spaces. Our results answer several problems posed by V.V. Tkachuk. 2001 Elsevier Science B.V. All rights reserved.

MSC: 54C05; 54D05; 54D55; 54E35

Keywords: Connected space; Metric space; Quotient mapping; Preimage; Finer connected topology

1. Introduction

The study of those connected spaces which have nice connected preimages has notreceived a systematic attention till a recent paper of Tkachuk [13]. In his paper the authorshows, among other things, that a connected Tychonoff space with countable network neednot be represented as a continuous image of a connected sequential Hausdorff space, andthat every connected Tychonoff space is an open continuous image of a connected leftseparated strictlyσ -discrete Tychonoff space.

Motivated by some problems posed in [13], we concentrate our attention to some classesof connected spaces which are continuous or quotient images of metric (or paracompact)connected spaces, moreover we give an example of a sequential connectedT4-space withcountable network which has no finer first countable connected (even dense in itself)topology.

* Corresponding author.E-mail addresses: [email protected] (A. Fedeli), [email protected] (A. Le Donne).

0166-8641/01/$ – see front matter 2001 Elsevier Science B.V. All rights reserved.PII: S0166-8641(01)00294-2

490 A. Fedeli, A. Le Donne / Topology and its Applications 125 (2002) 489–496

Recall that a subset of a space is said to be sequentially closed if together with anysequence it contains all its limits. A space in which every sequentially closed subset isclosed is called sequential.

We refer the reader to [4] for notations and terminology not explicitly given.

2. Spaces which are continuous or quotient images of metric (or paracompact)connected spaces

In this section we answer affirmatively to the following questions (see [13, Prob-lems 3.1–3.3]):

(a) LetX be a Tychonoff sequential (first countable) connected space. IsX a continuousor a quotient image of a connected metric space?

(b) Is any Hausdorff (Tychonoff) connected space a continuous image of a normal(paracompact) connected space?

Theorem 2.1. A space is a quotient image of a connected metric space if and only if it isconnected and sequential.

Proof. Since sequential spaces are precisely the quotient images of metric spaces [6], letus show that every connected sequential spaceX is a quotient image of a metric connectedspace.

Let q : (M,d)→X be a quotient mapping, where(M,d) is a metric space.It is straightforward to show that the following functionρ defined onM × [0,1]:

ρ((y1, t1), (y2, t2)

) = d(y1, y2)+ t1 + t2 if y1 �= y2,

ρ((y1, t1), (y2, t2)

) = |t2 − t1| otherwise,

is a metric.LetZ be the quotient set ofM×[0,1] obtained identifying the points(y1,1) and(y2,1)

wheneverq(y1)= q(y2), and letπ :M × [0,1] → Z be the canonical projection.Now define the following metricσ onZ:

σ(π(y1, t1),π(y2, t2)

) = ρ((y1, t1), (y2, t2)

)if q(y1) �= q(y2),

σ(π(y1, t1),π(y2, t2)

) = min{ρ((y1, t1), (y2, t2)

),2− (t1 + t2)

}otherwise.

Now let p : (M × [0,1], ρ)→ X be the mapping defined byp(y, t) = q(y) for every(y, t) ∈M × [0,1], and letf : (Z,σ )→X be the mapping given byf ◦ π = p.

Sincep−1(G)= q−1(G)× I for everyG ⊂X, q is a quotient mapping and a setA isopen in(M,d) if and only ifA× I is open in(M × [0,1], ρ), it follows thatp is quotient.Moreoverπ , as a map from(M × [0,1], ρ) onto(Z,σ ), is continuous.

We claim thatf is a quotient mapping.Let A be a subset ofX such thatf−1(A) is open in(Z,σ ), sincep is quotient and

p−1(A) = π−1(f−1(A)) is open in(M × [0,1], ρ) it follows thatA is open inX. Now

A. Fedeli, A. Le Donne / Topology and its Applications 125 (2002) 489–496 491

let us show thatf is continuous. First observe thatπ(B × [0,1[ ) andπ(q−1(x)×]0,1])are open in(Z,σ ) for every open subsetB of (M,d) and everyx ∈X. Now let us take anopen subsetA of X, sinceq−1(A) is open in(M,d) andf−1(A)= π(q−1(A)× [0,1[ )∪⋃x∈Aπ(q−1(x)×]0,1]) it follows thatf is continuous.It remains to show that(Z,σ ) is connected. IfA andB are non-empty open disjoint

subsets of(Z,σ ) such thatZ = A ∪ B, then there is a partition{C,D} of X such thatA= ⋃{f−1(x): x ∈ C} andB = ⋃{f−1(x): x ∈D}.

SinceA= f−1(C), B = f−1(D) andf is quotient, it follows thatC andD are open,soX is disconnected, a contradiction.✷

A spaceX is called sequentially connected if there are no non-empty disjointsequentially closed subsetsA andB of X such thatA ∪ B = X. Clearly every connectedsequential space and every pathwise connected space is sequentially connected.

Theorem 2.2. A space is a continuous image of a connected metric space if and only if itis sequentially connected.

Proof. Let M be a connected metric space and letf :M → (X, τ) be a continuoussurjection. Let us show that(X, τ) is sequentially connected.

Call τseqthe coarsest topology onX in which every sequentially closed subset of(X, τ)

is closed. Clearly if(X, τseq) is connected, then(X, τ) is sequentially connected. Letσ bethe quotient topology onX induced byf . Now (X,σ) is a connected sequential space andτseq⊂ σ , therefore(X, τ) is sequentially connected.

Now let us show that every sequentially connected spaceX is a continuous image of aconnected metric space. We may assume thatX is infinite.

LetS be the set of all convergent sequences inX formed by distinct points, and let(Y, d)be the hedgehog([0,1] × S)/∼ of spininess|S|.

SetYx = {x} × Y for everyx ∈X andZ = ⋃{Yx : x ∈X}.Let ρ be the metric onZ given by:(a) ρ((x, y1), (x, y2))= d(y1, y2) for everyx ∈X andy1, y2 ∈ Y ;(b) ρ((x, t, s), (y, t ′, s′))= 1

n+ 2 − (t + t ′) wheneverx �= y, s = s′ and the following

condition holds:(x = sn ∈ s ∧ y ∈ lim s)∨ (y = sn ∈ s ∧ x ∈ lim s);(c) ρ(z1, z2)= 1, otherwise.For example, ifs = (xn)n ∈ S converges to somex, thenρ((x,1, s), (xn,1, s))= 1

nfor

everyn.(i) (Z,ρ) is connected. Suppose not, and letA andB be non-empty disjoint closed

subsets of(Z,ρ) such thatA ∪ B = Z. Then there are two non-empty disjoint subsetsC

andD of X such thatA= ⋃{Yx : x ∈C}, B = ⋃{Yx : x ∈D}. ClearlyC ∪D =X.We will reach a contradiction if we show thatC andD are sequentially closed inX.Let s = (xn)n ⊂ C and suppose thatx ∈ lim s (we may assume thats ∈ S). Since

ρ((x,1, s), (xn,1, s))= 1n

for everyn, it follows that(x,1, s) ∈ A= A, thereforex ∈ C,andC is sequentially closed. SimilarlyD is sequentially closed.


(ii) X is a continuous image of(Z,ρ). Let f : (Z,ρ) → X be the mapping given byf (Yx)= {x} for everyx ∈X. We claim thatf is continuous.

LetA be closed inX. It is enough to show thatf−1(A) is sequentially closed in(Z,ρ).Take a convergent sequence(zn)n ⊂ f−1(A) and letz= lim zn.

So there arex ∈ X andxn ∈ A such thatz = (x, t, s) andzn = (xn, tn, sn) for everyn.We may assume that:(xn)n ∈ S, x �= xn andρ(z, zn) < 1 for everyn.

Clearlys = sn andxn ∈ s for everyn, andxn → x. Sox ∈A=A andz ∈ f−1(A). ✷Remark 2.3. A spaceX is called (T2-)subsequential if it has a sequential (T2-)extension(see, e.g., [7]). One may ask if there is some relationship between a sequentiallyconnected space and a connected subsequential space. As we shall see, they turn out tobe independent.

(i) A connected T2-subsequential space which is not sequentially connected.For everyn ∈ ω let Xn be the subspace{n} × [0,1] of the Euclidean plane and set

X= ⊕{Xn: n ∈ ω} ⊕ (ω+ 1). Let ∼ be the equivalence relation onX given by:(a) (n,0)∼ n for everyn ∈ ω,(b) (m,1)∼ (n,1) for everym,n ∈ ω.Now callZ the quotient spaceX/∼ and setY =Z \ {[n]: n ∈ ω}.ClearlyZ is a Hausdorff space andY is connected.MoreoverZ is sequential (it is a quotient of the metrizable spaceX), thereforeY is

T2-subsequential.SinceY \{[ω]} and{[ω]} are sequentially closed inY , it follows thatY is not sequentially

closed.(ii) A sequentially connected T5-space which is not subsequential.LetX = ([0,1]× (ω1+1))/∼ be the hedgehog of spininessℵ1,X is a connected metric

space. LetZ be the quotient space ofX⊕ (ω1 + 1) obtained identifying[(1, α)] with α foreveryα ∈ ω1 + 1.

ClearlyZ is a hereditarily normal space which is a continuous image ofX, althoughZ is not subsequential (Z contains a copy of the ordinal spaceω1 + 1, which is notsubsequential).

A spaceX is countably tight (or has countable tightness) if for everyA⊂ X and everyx ∈ A there is a countableB ⊂ A such thatx ∈ B. It is clear that every subsequentialspace has countable tightness. Clearly the space given in Remark 2.3(ii) has not countabletightness, moreover observe that for everym> ℵ0, Im is another example of a (compact)sequentially connectedT2-space which is not countably tight.

It is known that under Proper Forcing Axiom (PFA) every compact Hausdorff spaceof countable tightness is sequential ([1], see also [2,3]), although, under the Jensen’sprinciple♦, there are nonsequential countably tight compact Hausdorff spaces [5,10]. Onthe other hand it is an open question, in ZFC, whether every (countably) compact Hausdorffsubsequential space is sequential ([7], see also [9]).

Therefore it is natural to pose the following


Question 1. LetX be a subsequential continuum. IsX a continuous (quotient) image of aconnected metric space?

Remark 2.4. Another property which should be considered isσ -sequentiality. Let usrecall that a spaceX is σ -sequential ifX = ⋃{Xn: n ∈ ω} where eachXn is a closed andsequential subspace ofX (see, e.g., [11]). Clearly everyσ -sequential space has countabletightness. However the space constructed in Remark 2.3(ii) andIm for everym > ℵ0 areexamples of sequentially connected spaces which are notσ -sequential. Moreover there isa countable connected Hausdorff space (henceσ -sequential) which is not a continuousimage of a connected metric space, i.e., it is not sequentially connected [13, Example 2.5].

Question 2. LetX be aσ -sequential continuum. IsX a continuous (quotient) image of aconnected metric space?

Remark 2.5. Supercompact spaces, introduced by De Groot [8], are spacesX whichpossess an open subbaseS such that every cover ofX consisting of members ofS hasa subcover of at most 2 members. It is well known that every compact metric space and allTychonoff cubesIm are supercompact, nevertheless the quotient of the sum of two copiesV × {1} andV × {2} of the long segmentV obtained identifying(ω1,1) with (ω1,2) is aconnected supercompact space which is not sequentially connected.

Question 3. Let X be a connected supercompactT2-space of countable tightness. IsX acontinuous (quotient) image of a connected metric space?

Remark 2.6. Observe that, under PFA, the answer to the above problems is positive. Onthe other hand, it is worth noting that the cone over an Ostaszewski’s (or Fedorchuk’s)space is a consistent example of a sequentially connected compact Hausdorff space ofcountable tightness which is not sequential.

Theorem 2.7. Every connected Hausdorff space is a quotient image of a paracompactconnected space.

Proof. Let X be a connected Hausdorff space and letK be the setX ×X endowed withthe topology in which:

(i) every point(x, x ′), with x �= x ′, is isolated;(ii) {V × {x}: V neighbourhood ofx in X} is a neighbourhood base forK at (x, x) for

everyx ∈X.SetY =K × [0,1] and letσ be the topology onY generated from the product topology

by the addition of all sets of the form{κ}× ]0,1] whereκ ∈K.Let Z be the quotient space ofY obtained identifying the set{x} ×X × {1} to a point

for everyx ∈X, and letπ :Y →Z be the quotient mapping.For anya = (x, x ′, t) ∈ Y , with t �= 1, we will make the assumption, for the sake of

simplicity, thatπ(a)= a.


ClearlyZ is a Hausdorff space.(1) Z is connected. ClearlyZx = π(Yx), whereYx = {x} × X × [0,1], is (pathwise)

connected. Now let{A,B} be an open partition ofZ, then there is a partition{C,D} ofX such thatA = ⋃{Zx : x ∈ C} andB = ⋃{Zx: x ∈ D}. We reach a contradiction byshowing thatC andD are open inX.

Takex ∈ C, then(x, x,0) ∈ A, then there is a neighbourhoodV of x in X such thatV × {x} × {0} ⊂A, thereforeV ⊂ C andC is open. By symmetryD is open too.

(2) Z is paracompact. SinceZ = π(K × [0, 12]) ∪ π(K × [1

2,1]), it is enough to showthat both these closed pieces are paracompact subspaces ofZ. Clearly (K × [1

2,1]) isparacompact, so let us show the paracompactness of(K × [0, 1

2]).SetFx = X × {x} × [0,1] for everyx ∈ X, and observe thatY = ⊕{Fx : x ∈ X} ∼=

K × [0, 12] ∼= π(K × [0, 1

2]). Let us note that{x} × {x} × [0,1] is a closed and open subsetof Fx for everyx �= x.

Let x ∈ X and letU be an open cover ofFx . ChooseUx ∈ U such that( x, x,0) ∈ Uxand letV be a neighbourhood ofx in X such thatV × {x} × [0, ε] ⊂Ux for some positivenumberε. Now choose, for everyx, a finite open refinementUx of {U ∩ ({x} × {x} ×[0,1]): U ∈ U}.

SetVx = Ux if x /∈ V , and setVx = {U ∩ ({x} × {x}× ]ε,1]): U ∈ Ux} otherwise.It is clear that

⋃x Vx ∪ {Ux} is an open locally finite refinement ofU .

Now it remains to show that the mappingf :Z→X defined byf (x, x ′, t)= x for every(x, x ′, t) ∈Z, is quotient. First observef−1(C)= ⋃{Yx : x ∈C} for everyC ⊂X. Clearlyf is continuous. Moreover iff−1(C) is open, thenC is open by the same argument givento show the connectedness ofZ. ✷

3. A counterexample

Recall that a familyN of subsets of a spaceX is called a network forX if any opensubset ofX is union of some subfamily ofN .

The following example gives a negative answer to the following problem (see [13, Prob-lem 3.4]): does every Tychonoff sequential connected space with countable network havea finer Tychonoff second countable connected topology?

Example 3.1. A sequential connectedT4-space with countable network which has no finerfirst countable dense in itself topology.

Proof. Let p be the open filter onR generated by{⋃n�κ ]n− εn,n+ εn[ : κ ∈ N, εn ∈]0,+∞[ }, and letσ be the topology onR ∪ {p} in which the points ofR have the usualneighbourhoods and basic neighbourhoods ofp are of the formA∪ {p} with A ∈ p.

SetX = (R ∪ {p}, σ ).ClearlyX is aT3-space, moreoverR is dense inX, soX is connected.(i) X is sequential. LetA be a non-closed subset ofX, we may assume thatA=A∪{p}.

SinceA ∩ N must be infinite (otherwise there is someκ ∈ N such thatn /∈ A for every


n� κ , so for everyn� κ there exists someεn > 0 such that]n− εn,n+ εn[∩A= ∅, andp /∈A, a contradiction), it follows thatA contains a sequence converging top.

(ii) X has countable network. LetB be a countable base forR, thenN = B ∪ {B ∪{p}: B ∈ B} is a countable network forX.

(iii) X is aT4-space. LetC1 andC2 be two disjoint closed subsets ofX, we may assumethatp /∈ C1. SinceC1 andC2 \ {p} are disjoint closed subsets ofR there are two disjointopen subsetsU andV of R such thatC1 ⊂ U andC2 \ {p} ⊂ V . SinceX is regular thereare two disjoint open subsetsG andH ofX such thatp ∈G andC1 ⊂H . Now it is enoughto observe thatU ∩H andG∪V are two disjoint open subsets ofX containingC1 andC2

respectively.(iv) There is no dense in itself first countable topologyτ onR ∪ {p} such thatσ ⊂ τ .

Suppose thatτ is such a topology and let{An: n ∈ N} be a nested neighbourhood basefor (R ∪ {p}, τ ) at p. If for every n there is somexn ∈ An \ (N ∪ {p}) with xn > n,thenA = R \ {xn: n ∈ N} ∈ p, so A ∪ {p} is a neighbourhood ofp in (R ∪ {p}, τ )which does not contain anyAn, a contradiction. Therefore there is somen ∈ N such thatAn ∩ [n,+∞) ⊂ N . SinceV =]n,+∞[∪{p} ∈ σ ⊂ τ , there is somem � n such thatAm ⊂ V , soAm ⊂ [An ∩ [n,+∞)] ∪ {p} ⊂N ∪ {p}. SinceAm �= {p}, it follows that thereexistsκ ∈N ∩Am, therefore]κ − 1

2, κ + 12[ ∩Am = {κ} ∈ τ , a contradiction. ✷

Remark 3.2. The above example should be compared with the following result: afirst countable infinite connected Tychonoff space admits a strictly finer connected firstcountable Tychonoff topology [12, Corollary 5].

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