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Topology and its Applications 148 (2005) 33–38

www.elsevier.com/locate/topo

On a problem related to a non-squeezing theore

Alessandro Fedelia,∗, Attilio Le Donneb

a Dipartimento di Matematica, Universitá dell’Aquila, 67100 l’Aquila, Italyb Dipartimento di Matematica, Universitá di Roma “La Sapienza”, 00100 Roma, Italy

Received 16 June 2004; received in revised form 25 July 2004; accepted 25 July 2004

Abstract

In this paper we give a solution to a problem of Kulpa about the interior of the image of ccontinuous mapsf :X → Rn whereX is a compact subset ofRn with non-empty interior. Moreovewe show that the image of every continuous mapf :X → R2 whereX is a non-empty compacsubset ofR2 and diamf −1f (x) <

√3aX for everyx ∈ FrX, has non-empty interior.

2004 Elsevier B.V. All rights reserved.

MSC:54C05; 54E40; 54E45

Keywords:Compact; Interior; Diameter

Let d be the euclidean metric onRn. For each (non-empty)A ⊂ Rn, x ∈ Rn andε > 0let us setd(x,A) = inf{d(x, a): a ∈ A}, diamA = sup{d(x, y): x, y ∈ A}, D(x, ε) = {p ∈Rn: d(x,p) < ε} and�D(x, ε) = {p ∈ X: d(p,x) � ε}. The boundary ofA will be denotedby FrA.

For each non-empty compact setX ⊂ Rn let aX = sup{d(x,FrX): x ∈ X}.The reader is referred to [3,4] for notations and terminology not explicitly given. In

note we will give a complete solution to the following problem of Kulpa.

Problem [6]. Let f :X → Rn be a continuous map from a compact subsetX ⊂ Rn withnon-empty interior, such that diamf −1(f (x)) < 2aX.

Is Intf (X) �= ∅?

* Corresponding author.E-mail addresses:afedeli@univaq.it (A. Fedeli), ledonne@mat.uniromal.it (A. Le Donne).

0166-8641/$ – see front matter 2004 Elsevier B.V. All rights reserved.

doi:10.1016/j.topol.2004.07.010

34 A. Fedeli, A. Le Donne / Topology and its Applications 148 (2005) 33–38

The problem above is related to various results of Kulpa [6,7], e.g., the following non-

at

see,

nt

squeezing theorem (which is a strengthening of a classical result of Brouwer [1]):Let a > 0 and let In = [−a, a]n; if f : In → Rn is a continuous map such th

diamf −1(f (x)) < 2a for eachx ∈ FrIn, then Intf (In) �= ∅.A mappingf :X → Y between topological spaces is called (linearly) Darboux iff (C)

is connected wheneverC is (pathwise) connected. Clearly every Darboux function (e.g., [2,5]) is linearly Darboux.

Now let us show that the problem above has a positive answer forn = 1 and a negativeone forn > 1.

Theorem 1.

(i) Letf :X → R be a linearly Darboux function from a non-empty compact subsetX ⊂Rn such thatdiamf −1(f (x)) < 2aX for everyx ∈ FrX.ThenIntf (X) �= ∅.

(ii) For everyn > 1 there is a continuous mapf :X → Rn from a compact subsetX ⊂ Rn

with non-empty interior such thatIntf (X) = ∅ anddiamf −1(f (x)) < 2aX for everyx ∈ X.

Proof. (i) Let U = IntX. ClearlyU �= ∅. Since the mappingψ :X → R given byψ(x) =d(x,FrX) for everyx ∈ X is continuous andX is compact, we can take somep ∈ X suchthatψ(p) = aX. Clearly such ap belongs toU .

We claim thatD(p,aX) ⊂ U . Let q ∈ D(p,aX) and letIpq be the closed line segmejoining p andq . SinceaX = d(p,FrX), it follows thatIpq ∩ FrX = ∅.

SoIpq = (Ipq ∩U)∪ (Ipq ∩ (Rn \X)). SinceIpq is connected, it follows thatIpq ⊂ X.Thereforeq ∈ Ipq ⊂ U .

Now let φ : FrX → R be the mapping given byφ(x) = d(p,x). By the continuity ofφ and the compactness of FrX, it follows that there is somey ∈ FrX such thatφ(y) =minφ(FrX), i.e.,d(p,y) = aX.

Let g : �D(p,aX) → R be the restriction off to �D(p,aX). Clearly g is a linearlyDarboux function. Moreoverg is not constant, otherwise�D(p,aX) = g−1(g(y)) ⊂f −1(f (y)), so 2aX = diam�D(p,aX) � diamf −1(f (y)), a contradiction (recall thaty ∈FrX).

Sog(�D(p,aX)) is a non-degenerate interval contained inf (X).Therefore Intf (X) �= ∅.(ii) Let 0 = (0,0) andX = {p ∈ R2: d(p, 0) � 1}. For eachκ ∈ {0, . . . ,5} setAκ =

(cosπ3 κ,sin π

3 κ), set alsoA6 = A0.For everyκ ∈ {0, . . . ,5} let Dκ be the sector ofX bounded by the arcAκAκ+1 and the

two radii 0Aκ and0Aκ+1.For everyκ ∈ {0,2,4} let gκ be the projection ofDκ onto the segment0Aκ along

0Aκ+1.For everyκ ∈ {1,3,5} let gκ be the projection ofDκ onto the segment0Aκ+1 along

0Aκ .Clearly every mappinggκ :Dκ → R2 is continuous.

A. Fedeli, A. Le Donne / Topology and its Applications 148 (2005) 33–38 35

Since{D0, . . . ,D5} is a finite closed cover ofX and{g0, . . . , g5} is a family of compat-

gative

d for a

dges

ts

r-

s

axis,

ch

e

ible mappings, it follows that the combinationf = ∇κgκ :X → R2 is continuous.Sincef (X) = 0A0 ∪ 0A2 ∪ 0A4, it follows that Intf (X) = ∅.SinceaX = 1, it remains to show that diamf −1(f (x)) < 2 for everyx ∈ X. Let P ∈

f (X). If P = 0, thenf −1(P ) = 0A1 ∪ 0A3 ∪ 0A5 and diamf −1(P ) = d(A1,A3) = √3.

If P ∈ 0A0 \ {0}, thenf −1(P ) = PQ ∪ PQ′ for someQ ∈ D0 andQ′ ∈ D5 such thatPQ‖ 0A1 andPQ′ ‖ 0A5.

So diamf −1(P ) = d(Q,Q′) < d(A1,A5) = √3.

In a similar way it can be shown that diamf −1(P ) <√

3 for everyP ∈ (0A2 ∪ 0A4) \{0}.

With suitable modifications one can show, in the same vein, that the answer is nefor everyn � 2. �

The spaceX constructed in Theorem 1 is such that diamf −1f (x) �√

3aX for everyx ∈ FrX. The next theorem shows that this is, in the plane, the best possible bouncounterexample.

Theorem 2. LetX be a non-empty compact subspace ofR2 andf :X → R2 a continuousmap such thatdiamf −1f (x) <

√3aX for everyx ∈ FrX. ThenIntf (X) �= ∅.

Proof. ClearlyaX > 0, so IntX �= ∅. Moreover we may assume thataX = 1,B((0,0),1) ⊂X andX is connected.

Let us suppose that Intf (X) = ∅.

Claim. We may assume thatX is the union of finitely many closed rectangles whose eare parallel to the axis and do not contain the origin.

Proof. Let Dn = {x ∈ X: d(x,FrX) � 1n} for everyn ∈ N . Let us show that there exis

somem such that diamf −1f (x) <√

3 for everyx ∈ Dm. If not, there existxn, yn ∈ Dn

such thatf (xn) = f (yn), d(xn, yn) �√

3 for everyn ∈ N . So, by eventually consideing a subsequence, we may assume thatxn → x, yn → y, for somex, y ∈ X. Therefored(x, y) �

√3 andx, y ∈ FrX.

So, by compactness ofDm, b = 1√3

sup{diamf −1f (x): x ∈ Dm} < 1. Sinceb+12 < 1,

we may take somen > m such thatB((0,0), b+12 ) ⊂ X \ Dn. SinceX \ Dn ⊂ IntX, we

may cover it by the unionV of finitely many open squares contained inX, whose edgeare parallel to the axis and do not contain the origin. SoX may be replaced withY = �V ,in fact Y is the union of finitely many closed rectangles, with edges parallel to theand diamf −1f (x) <

√3b <

√3aY for everyx ∈ FrY (observe that FrY ⊂ Dn andb <

b+12 � aY ).

So, by the claim, there is a gridG (i.e., a union of finitely many lines in the plane, eaparallel to thex-axis or they-axis, see, e.g., [4, p. 85]) such thatX is the union of someminimal compact rectanglesR1, . . . ,Rm for the gridG and no line of the grid contains thorigin. �

36 A. Fedeli, A. Le Donne / Topology and its Applications 148 (2005) 33–38

A grid H is said to be a refinement of a gridG if G ⊂ H. A node of a gridH is the

t

e

nt

at

at

f

intersection of two non-parallel lines ofH. N (G) will denote the set of all nodes ofG.Moreover, for every non-empty setA, let us denote byF(A) the free Abelian group onA.

So FrX can be seen as the closed rectangular 1-chain∂X = ∂R1 + ∂R2 +· · ·+ ∂Rm forG (for R = [a, b]× [c, d] with a � b andc � d , the boundary∂R is the 1-chainγ1 + γ2 −γ3 − γ4 whereγ1 is the straight path from(a, c) to (b, c), γ2 from (b, c) to (b, d), γ3 from(a, d) to (b, d), γ4 from (a, c) to (a, d), sof∗∂R = f ◦ γ1 + f ◦ γ2 − f ◦ γ3 − f ◦ γ4, see,e.g., [4, Chapter 6]). Since eachRi is contained inX, it follows that∂X is a 1-boundary inX (see, e.g., [4, p. 82]). Moreover, sincef :X → f (X) is continuous, it follows also thaf∗∂X = f∗∂R1 + f∗∂R2 + · · · + f∗∂Rm is a 1-boundary inf (X) (see [4, Chapter 6]).

Now let us take three half-linesl1, l2, l3 starting from the origin, not parallel to thaxis, and dividing the plane in three equal anglesA1, A2, A3, with Ai ∩ Ai+1 = li+1 fori = 1,2,3, A4 = A1 andl4 = l1.

Now setSi = Ai ∩ FrX andYi = f (Si) for i = 1,2,3. Note that eachYi is compact.Since diamf −1f (x) <

√3 for everyx ∈ FrX and B((0,0),1) ⊂ X, it follows that

Y1 ∩ Y2 ∩ Y3 = ∅.Let us callM(H) the family of all minimal compact rectangles for a gridH.SinceY1 ∩Y2 is disjoint fromY3, we may find a gridG′ such thatf (X) ⊂ ⋃

M(G′) andmoreoverK = ⋃{C ∈ M(G′): C ∩ (Y1 ∩ Y2) �= ∅} andS = ⋃{C ∈ M(G′): C ∩ Y3 �= ∅}are disjoint.

SinceY1 \ Int(K ∪ S) is disjoint from Y2 \ Int(K ∪ S), we can, up to a refinemeof the gridG′, assume that the setsT = ⋃{C ∈ M(G′): C ∩ (Y1 \ (K ∪ S)) �= ∅} andU = ⋃{C ∈ M(G′): C ∩ (Y2 \ (K ∪ S)) �= ∅} are disjoint.

So we havef (FrX) ⊂ Int(K ∪ S ∪ T ∪U) andK ∩ S = T ∩U = ∅. Moreover observethat IntK ∩ IntT = IntT ∩ IntS = IntU ∩ IntK = IntU ∩ IntS = ∅.

SetM(G′) = {Q1, . . . ,Qκ} and putM = ⋃{FrQi : 1 � i � κ}.Since we are assuming thatf (X) has empty interior we may pick someqi ∈ Int(Qi) \

f (X) for everyi. Let g :⋃{Qi \ {qi}: 1 � i � κ} → M be the continuous map such th

g|Qi\{qi} is the radial projection fromqi to FrQi . Note thatg(K) ⊂ K, g(S) ⊂ S, g(T ) ⊂ T

andg(U) ⊂ U .Seth = g ◦ f . Thenh∗∂X = g∗f∗∂X is a 1-boundary (inM). Since we assumed th

the half-linesl1, l2, l3 are not parallel to the axis, it follows thatL = Fr(X) ∩ (l1 ∪ l2 ∪ l3)

is finite. So we may assume, up to a refinement of the gridG′, that the points ofL go underf to some nodes ofG′. Henceg is the identity onL.

Let S be the family of all straight pathsσab from a to b (i.e., σab(t) = a + t (b − a)),wherea, b is a pair of distinct adjacent nodes of the gridG′. ∂σab = b − a is an element oF(N (G′)).

For everyµ = ∑{nσ σ : σ ∈ S} ∈ F(S), nσ ∈ Z, let us put Imµ = ⋃{Imσ : nσ �= 0},∂µ = ∑{nσ ∂σ : σ ∈ S} ∈ F(N (G′)).

Now observe that∂X = ∑ni=1 γi , whereγi is either a path fromai to bi and Im(γi) ∩

L = {ai, bi} or it is a closed path whose image is disjoint fromL.Now for eachi ∈ {1, . . . , n} let us fix a memberµi of F(S) homologous toh∗γi onM,

so that Imµi ⊂ Imh∗γi and∂µi = ∂(h∗γi) = ∂(f∗γi) = ∂(f ◦ γi) = f∗(∂γi) = f (bi) −f (ai) ∈ F(N (G′)) (where the last equality holds whenever∂γi = bi − ai ). Note that ifγi

is a closed path then∂µi = 0.

A. Fedeli, A. Le Donne / Topology and its Applications 148 (2005) 33–38 37

Observe that a memberµ = ∑{nσ σ : σ ∈ S} of S is a 1-boundary inM (i.e., homol-

-

s

ce

er

ogous to zero onM) if and only if nσ = nτ whenever∂σ = −∂τ (i.e., σ is the oppositestraight path ofτ ).

Moreover ifµ = ∑{nσ σ : σ ∈ S} is a 1-boundary inM, thenµ|B = ∑{nσ σ : Imσ ⊂B} is a 1-boundary inM for everyB ⊂ R2.

Now let :F(N (G′)) → Z be the group homomorphism given, for eachx ∈ N (G′), by(x) = 1 wheneverx ∈ K, (x) = 0 otherwise.

Let us setµ = ∂µ for eachµ ∈ F(S) andγi = µi for everyi ∈ {1, . . . , n}. Observethat, if ∂γi = bi − ai , thenγi is equal to 1 wheneverai /∈ l1 andbi ∈ l1, it is equal to−1wheneverai ∈ l1 andbi /∈ l1, and it is equal to 0 otherwise. Observe thatγi = f (bi) −f (ai) if ∂γi = bi − ai , in factf (x) ∈ K if and only if x ∈ l1 for everyx ∈ L.

Now let us show that∑n

i=1 µi |T is not a 1-boundary inM, thus reaching a contradiction, in fact

∑ni=1 µi is homologous to

∑ni=1 h∗γi = h∗∂X which is a 1-boundary.

Observe that ifµ is a 1-boundary onM, then∂µ is the identity element ofF(N (G′)),soµ = 0.

We will show that(∑n

i=1 µi |T ) �= 0.

Claim. (µi |T ) is equal toµi wheneverImγi ⊂ A1, it is 0 otherwise.

Proof. First observe thatσ = 0 whenever Imσ ⊂ K or Imσ ∩ K = ∅.If Im γi ⊂ A3, then Imf∗(γi) ⊂ S. Hence Imh∗γi ⊂ S, so Imµi ∩ K = ∅. Therefore

Im(µi |T ) ∩ K = ∅, so(µi |T ) = 0.If Im γi ⊂ A2, then

µi =∑

{nσ σ : Imσ ⊂ U} +∑

{nσ σ : Imσ ⊂ K \ U} +∑

{nσ σ : Imσ ⊂ S \ U}.Soµi |T = ∑{nσ σ : Imσ ⊂ (K \ U) ∩ T } + ∑{nσ σ : Imσ ⊂ (S \ U) ∩ T }.Hence(µi |T ) = 0.If Im γi ⊂ A1, then

µi =∑

{nσ σ : Imσ ⊂ T } +∑

{nσ σ : Imσ ⊂ K \ T } +∑

{nσ σ : Imσ ⊂ S \ T }.Soµi = (µi |T ). �Therefore we have(

∑ni=1 µi |T ) = ∑n

i=1 (µi |T ) = ∑{µi : Imγi ⊂ A1} =∑{γi: Imγi ⊂ A1} = (∑{γi : Imγi ⊂ A1}).

Now if R is a rectangle inR2, we can write the 1-chain∂R as a finite sum of pathη1, . . . , ηr whose images are contained in one ofA1, A2 andA3.

Then we set∂R|A1 = ∑{ηl : Imηl ⊂ A1} (this definition is independent from the choiof the pathsη1, . . . , ηr ).

Now (∑{γi : Imγi ⊂ A1}) = (

∑ni=1 γi |A1) = ∂X|A1 =

∑mj=1 ∂Rj |A1 =∑m

j=1 ∂Rj |A1 = ∂Rκ|A1 = 1, whereκ is the only index for which(0,0) ∈ Rκ .This completes the proof.�We conclude the paper with the following observation. Letr be the greatest real numb

such that for each compact subsetX of Rn and for eachx ∈ FrX, diamf −1(f (x)) < raX

38 A. Fedeli, A. Le Donne / Topology and its Applications 148 (2005) 33–38

implies Intf (X) �= ∅. As pointed out by the referee, it should be interesting to investigate

r and its dependence on the dimensionn.

References

[1] L.E.J. Brouwer, Beweis der Invarianz der Dimensionalzahl, Math. Ann. 70 (1911) 61–65.[2] K. Ciesielski, Set theoretic real analysis, J. Appl. Anal. 3 (2) (1997) 143–190.[3] R. Engelking, General Topology, Heldermann, Berlin, 1989.[4] W. Fulton, Algebraic Topology: A First Course, Springer, New York, 1995.[5] R.G. Gibson, T. Natkaniec, Darboux like functions, Real Anal. Exchange 22 (2) (1996–1997) 453–492.[6] W. Kulpa, The Poincaré–Miranda theorem, Amer. Math. Monthly 104 (6) (1997) 545–550.[7] W. Kulpa, On extensions of the domain invariance theorem, Topology Appl. 130 (2003) 253–258.