PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 139, Number 8, August 2011, Pages 2987–2993
S 0002-9939(2011)10876-1
Article electronically published on March 22, 2011
A CIRCLE IS NOT THE GENERALIZED INVERSE LIMIT
OF A SUBSET OF [0, 1]2
ALEJANDRO ILLANES
(Communicated by Alexander N. Dranishnikov)
Abstract. In this paper we show that the simple closed curve cannot be
obtained as the inverse limit of an upper semi-continuous multivalued function
from [0, 1] into [0, 1].
1. Introduction
A continuum is a compact connected metric space with more than one point.
We denote the Hilbert cube by [0, 1]∞ , and we consider [0, 1]∞ with the metric
d((x1 , x2 , . . .), (y1 , y2 , . . .)) =
∞
|xi − yi |
i=1
2i
.
Given a closed subset M of the square [0, 1]2 , it is possible to define the generalized inverse limit of the set M as
lim M = {(x1 , x2 , . . .) ∈ [0, 1]∞ : for each i ∈ N, (xi+1 , xi ) ∈ M }.
←
When M is the graph of a continuous function, lim M is the usual inverse limit.
←
In this case, it is well known that the continua that can be expressed as lim M
←
are exactly the chainable continua and, even for very simple maps, they can be
extremely complicated; see for example [1].
Generalized inverse limits were introduced by W. S. Mahavier in [4], where he
proved some basic properties and offered several interesting and illustrative examples. Generalized inverse limits are the natural notion of an inverse limit (or, in
category theoretic terms, the limit) in the category whose objects are topological
spaces, and where an arrow between X and Y is a closed subset of X × Y , both of
whose projections are onto. Even when these objects are very natural, they do not
seem to be specifically studied from the categorical point of view.
An important problem in this area is:
Problem. What spaces can be obtained as lim M , for a closed subset M of [0, 1]2 ?
←
Received by the editors October 8, 2009 and, in revised form, March 26, 2010.
2010 Mathematics Subject Classification. Primary 54C60; Secondary 54F15.
Key words and phrases. Inverse limits, Hilbert cube, simple closed curve, unit square, upper
semi-continuous, set valued functions.
The author wishes to thank Gabriela Sanginés for her technical help during the preparation of
this paper.
c
2011
American Mathematical Society
2987
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2988
ALEJANDRO ILLANES
In order to illustrate the variety of spaces that can be constructed with this
technique, consider the following:
(a) If M = [0, 1] × { 12 }, then lim M is the singleton {( 12 , 12 , . . .)}.
←
(b) If M = {(x, x) : x ∈ [0, 1]}, then lim M is an arc.
←
(c) If M = ([0, 1] × {0, 1}), then lim M is a Cantor set.
←
(d) If M = [0, 1]2 , lim M is the Hilbert cube.
←
(e) If M = {(x, x) : x ∈ [0, 1]} ∪ {(x, 1 − x) : x ∈ [0, 1]}, then lim M is the cone
←
over the Cantor set (Example 4 of [4]).
Of course, more interesting examples can be found in the references.
Given a closed subset M of [0, 1]2 , in general it is very difficult to visualize
the space lim M . This originated the comment of W. S. Mahavier at the end of
←
[4]: “All of our examples yield inverse limits that are either infinite dimensional
or 1-dimensional continua. We suspect that this is true in general and that lim M
←
must either contain a Hilbert cube or be 1-dimensional.” The answer to Mahavier’s
remark was given independently by A. Peláez in [5] and T. Ingram and W. S.
Mahvier in [2]. They showed that for each positive integer m it is possible to find
a closed subset M of [0, 1]2 such that lim M is m-dimensional.
←
In his talk in the Spring Topology and Dynamics Conference, 2009, Van C. Nall
remarked that “It is not known if there is a one dimensional continuum that is
not homeomorphic to an inverse limit with a single set valued function from [0, 1]
to [0, 1]. We will discuss some facts about a leading candidate, a simple closed
curve.” Later, in the Third Workshop in Continua and Its Hyperspaces, celebrated
in the city of Toluca, in México (July, 2009), Van C. Nall gave a series of talks
in which he showed several interesting examples and discussed his ideas about the
problem of determining if a simple closed curve can be obtained as lim M , for some
←
closed subset of [0, 1]2 . This paper is the result of the discussions we had in this
workshop, and the author wishes to thank all the participants. In particular, he
wants to thank Professor Van C. Nall for transmitting his experience and expertise
on this topic.
In this paper we prove that there is no closed subset M of [0, 1]2 such that the
projection of M in both coordinates is onto and lim M is a simple closed curve.
←
A finite graph is a continuum G which is the union of a finite number of arcs,
each two of which intersect in a finite set. The following problem is open.
Problem. Does there exist a finite graph G, different from an arc, such that G is
homeomorphic to lim N for some closed subset N of the square [0, 1]2 ?
←
2. Two auxiliary results
In this section we present two results that belong to the area of uniformization
of functions. The ideas of the proofs come from the paper [6].
Let N be the set of positive integers. Given n ∈ N, let Pn = {0, 1, . . . , n}.
Lemma 1. Let f, g : [0, 1] → [0, 1] be continuous functions such that f (0) = 0 =
g(0) and f (1) = 1 = g(1) and let δ > 0. Then there exist n ∈ N and functions
α, β : Pn → [0, 1] such that f ◦ α = g ◦ β, α(0) = 0 = β(0), α(n) = 1 = β(n), and,
for each i ∈ {0, . . . , n − 1}, |α(i + 1) − α(i)| , |β(i + 1) − β(i)| < δ.
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THE GENERALIZED INVERSE LIMIT OF A SUBSET OF [0, 1]2
2989
Proof. Let A = {(x, y) ∈ [0, 1]2 : f (x) = g(y)}. The set A is a compact subset of
[0, 1]2 such that (0, 0), (1, 1) ∈ A. Let L be the component of A such that (0, 0) ∈ L.
First, we prove that (1, 1) ∈ L. Suppose to the contrary that (1, 1) ∈
/ L. Then
there exist compact disjoint subsets H and K of [0, 1]2 such that A = H ∪ K,
(0, 0) ∈ H and (1, 1) ∈ K. By [3, §57, III, Theorem 2, p. 438], there exists a
separator C between (0, 0) and (1, 1) which is a locally connected continuum disjoint
from A. Thus there exist points (x, y) ∈ C ∩ (({0} × [0, 1]) ∪ ([0, 1] × {1})) and
(u, v) ∈ C ∩ (([0, 1] × {0}) ∪ ({1} × [0, 1])). Notice that f (x) ≤ g(y) and f (u) ≥ g(v).
Since C is connected, there exists a point (t, s) ∈ C such that f (t) = g(s). This
implies that (t, s) ∈ C ∩ A, a contradiction. Hence (1, 1) ∈ L.
For each (x, y) ∈ L, let U (x, y) = (x − δ2 , x + δ2 ) × (y − δ2 , y + 2δ ). Consider
the open cover U = {U (x, y) : (x, y) ∈ L} of L. By the connectedness of L, there
exist n ∈ N and points (x1 , y1 ), . . . , (xn , yn ) ∈ L such that (0, 0) ∈ U (x1 , y1 ),
(1, 1) ∈ U (xn , yn ) and U (xi , yi ) ∩ U (xi+1 , yi+1 ) ∩ L = ∅ for each i ∈ {1, . . . , n − 1}.
For each i ∈ {1, . . . , n − 1}, choose a point (ui , vi ) ∈ U (xi , yi ) ∩ U (xi+1 , yi+1 ) ∩ L.
Let α, β : Pn → [0, 1] be given by α(0) = 0, α(1) = u1 , . . ., α(n − 1) = un−1 ,
α(n) = 1, β(0) = 0, β(1) = v1 , . . ., β(n − 1) = vn−1 and β(n) = 1. Clearly, α and
β have the required properties.
Lemma 2. Let f, g : [0, 1] → [0, 1] be continuous functions such that f (0) = 0 and
f (1) = 1 and let δ > 0. Then there exist n ∈ N and functions α, β : Pn → [0, 1]
such that f ◦ α = g ◦ β, β(0) < δ, 1 − β(n) < δ and, for each i ∈ {0, . . . , n − 1},
|α(i + 1) − α(i)| , |β(i + 1) − β(i)| < δ.
Proof. Define h : [0, 1] → [0, 1] by
⎧
3tg(0),
⎪
⎪
⎨
g(3t − 1),
h(t) =
⎪
⎪
⎩
(3 − 3t)(g(1)) + 3t − 2,
if t ∈ [0, 13 ],
if t ∈ [ 13 , 23 ],
if t ∈ [ 23 , 1].
Notice that h is a well defined continuous function such that h(0) = 0 and
h(1) = 1. Thus we can apply Lemma 1 and obtain m ∈ N and functions α1 , β1 :
Pm → [0, 1] such that f ◦ α1 = h ◦ β1 , α1 (0) = 0 = β1 (0), α1 (m) = 1 = β1 (m) and,
for each i ∈ {0, . . . , m − 1}, |α1 (i + 1) − α1 (i)| , |β1 (i + 1) − β1 (i)| < min{ 19 , δ3 }.
Let i0 = max{i ∈ {0, . . . , m} : β1 (i) ∈ [0, 13 ]}. Then 0 < i0 < m. Let j0 =
min{j ∈ {i0 , . . . , m} : β1 (j) ∈ [ 23 , 1]}. Then 0 < i0 + 1 < j0 − 1 < j0 < m and,
for each i ∈ {i0 + 1, . . . , j0 − 1}, β1 (j) ∈ [ 13 , 23 ]. Let n = j0 − i0 − 2. Define
α, β : Pn → [0, 1] by α(i) = α1 (i + i0 + 1) and β(i) = 3β1 (i + i0 + 1) − 1. Given
i ∈ {0, 1, . . . , n}, i + i0 + 1 ∈ {i0 + 1, . . . , j0 − 1}, so β1 (i + i0 + 1) ∈ [ 13 , 23 ] and
f ◦ α(i) = f ◦ α1 (i + i0 + 1) = h ◦ β1 (i + i0 + 1) = g(3β1 (i + i0 + 1) − 1) = g ◦ β(i).
Thus f ◦ α = g ◦ β.
Since β(0) = 3β1 (i0 + 1) − 1, by the definition of i0 , β1 (i0 + 1) ∈ [ 13 , 13 + 19 ),
so β(0) ∈ [0, δ). Similarly, β(n) ∈ (1 − δ, 1]. It is easy to check that, for each
i ∈ {0, . . . , n − 1}, |α(i + 1) − α(i)| , |β(i + 1) − β(i)| < δ.
3. Main result
For each i ∈ {1, 2} let ρi : [0, 1]2 → [0, 1] be the i-th projection.
Throughout this section, M will denote a closed subset of [0, 1]2 such that
ρi (M ) = [0, 1] for each i ∈ {1, 2}.
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2990
ALEJANDRO ILLANES
Given x = (x1 , x2 , . . .) ∈ [0, 1]∞ , A ⊂ [0, 1]∞ and ε > 0, let B(ε, x) be the εneighborhood around x in I ∞ and N (ε, A) = {x ∈ [0, 1]∞ : there exists a ∈ A such
that d(a, x) < ε}.
Theorem 1. There is no closed subset M of [0, 1]2 such that lim M is a simple
←
closed curve.
Proof. Denote S = lim M . Suppose that S is a simple closed curve. For each
←
i ∈ N, let πi : [0, 1]∞ → [0, 1] be the i-th projection. Since ρ2 (M ) = [0, 1] , π1 (S) =
[0, 1] = π2 (S) and the projection π1,2 : S → M given by π1,2 ((x1 , x2 , . . .)) = (x1 , x2 )
is onto, we obtain that M is connected.
Let Δ be the diagonal of [0, 1]2 ; that is, Δ = {(x, x) ∈ [0, 1]2 : x ∈ [0, 1]}. Given
x ∈ [0, 1], we denote x = (x, x, . . .) ∈ [0, 1]∞ . Notice that if (x, x) ∈ Δ ∩ M , then x
∈ S. Given a map σ : [0, 1] → [0, 1]∞ and m ∈ N, let σm : [0, 1] → [0, 1]m be the
map given by σ(t) = (π1 (σ(t)), . . . , πm (σ(t))). We consider the space [0, 1]m with
|xi −yi |
.
the metric dm ((x1 , . . . , xm ), (y1 , . . . , ym )) = m
i=1
2i
Claim 1. If there exist p, q ∈ S and x ∈ [0, 1] such that π1 (p) = 0, π1 (q) = 1 and x
∈ S, then x ∈ {p, q}.
Suppose, contrary to Claim 1, that there exist p, q ∈ S such that π1 (p) = 0,
π1 (q) = 1 and x ∈ S − {p, q}, where x ∈ [0, 1]. Let σ, γ : [0, 1] → S be one-toone continuous functions such that σ(0) = p = γ(0), σ(1) = q = γ(1), σ( 12 ) = x,
Im σ ∩ Im γ = {p, q} and Im σ ∪ Im γ = S.
Let ε > 0 be such that B(ε, x) ∩ Im γ = ∅, N (ε, Im σ) ∩ N (ε, γ([ 13 , 23 ])) =
∅, N (ε, σ([0, 13 ]) ∪ γ([0, 13 ])) ∩ N (ε, σ([ 12 , 1]) ∪ γ([ 23 , 1])) = ∅ and N (ε, σ([0, 12 ]) ∪
1
γ([0, 13 ])) ∩ N (ε, σ([ 23 , 1]) ∪ γ([ 23 , 1])) = ∅. Let m ∈ N be such that 2m−1
< ε.
Since Im σ is homeomorphic to [0, 1], there exists λ > 0 such that λ < ε and, if
ε
. Since the
d(σ(s), σ(t)) < λ, where t, s ∈ [0, 1], then diameter (σ([s, t])) < 2m−1
m
functions σm : [0, 1] → [0, 1] and γ are continuous, there exists δ > 0 such that, if
t, s ∈ [0, 1] and |s − t| < δ, then dm (σm (s), σm (t)) < λ2 and d(γ(s), γ(t)) < λ2 .
Since π1 (γ(0)) = π1 (p) = 0 and π1 (γ(1)) = π1 (q) = 1, we can apply Lemma 2 to
the maps π1 ◦ γ and πm ◦ σ and the number δ, so there exist n ∈ N and functions
α, β : Pn → [0, 1] such that π1 ◦ γ ◦ α = πm ◦ σ ◦ β, β(0) < δ, 1 − β(n) < δ and, for
each i ∈ {0, . . . , n − 1}, |α(i + 1) − α(i)| , |β(i + 1) − β(i)| < δ.
Consider the function ϕ : Pn → S given by
ϕ(i) = (π1 (σ(β(i))), . . . , πm (σ(β(i))), π2(γ(α(i))), π3 (γ(α(i))), . . .).
Given i ∈ Pn , we have that πm (σ(β(i))) = π1 (γ(α(i))), then ϕ(i) ∈ S. Notice
that σ(β(i)) and ϕ(i) coincide in the first m coordinates, so d(σ(β(i)), ϕ(i)) ≤
1
/ γ([ 31 , 23 ]). Let J = γ([0, 13 ]) ∪ Im σ ∪ γ([ 23 , 1]).
2m−1 < ε. By the choice of ε, ϕ(i) ∈
Then J is a subarc of S with end points γ( 13 ) and γ( 23 ) such that ϕ(Pn ) ⊂ J.
Consider the natural order < on J such that γ( 13 ) < γ( 23 ).
Since 0 ≤ β(0) < δ, we have that dm (σm (0), σm (β(0))) < λ2 < 2ε . Since
ϕ(0) = (π1 (σ(β(0))), . . . , πm (σ(β(0))), π2(γ(α(0))), π3 (γ(α(0))), . . .), we have that
d(σ(0), ϕ(0)) < ε. This implies that ϕ(0) ∈
/ σ([ 12 , 1]) ∪ γ([ 32 , 1]). That is, ϕ(0) <
x. Similarly, x < ϕ(n). Thus there exists i0 = max{i ∈ {0, 1, . . . , n} : ϕ(i) ≤ x}.
Notice that i0 < n.
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THE GENERALIZED INVERSE LIMIT OF A SUBSET OF [0, 1]2
2991
Given i ∈ {0, 1, . . . , n − 1}, |α(i + 1) − α(i)| , |β(i + 1) − β(i)| < δ. This implies that dm (σm (β(i)), σm (β(i + 1))) < λ2 and d(γ(α(i)), γ(α(i + 1))) < λ2 . Thus
d(ϕ(i), ϕ(i + 1)) < λ < ε.
/ σ([ 23 , 1])∪γ([ 32 , 1]).
Since ϕ(i0 ) ∈ γ([0, 13 ])∪σ([0, 12 ]), by the choice of ε, ϕ(i0 +1) ∈
1
2
1
Hence σ( 2 ) < ϕ(i0 + 1) < σ( 3 ). Similarly, σ( 3 ) < ϕ(i0 ) ≤ σ( 21 ). In particular,
ϕ(i0 ), ϕ(i0 + 1) ∈ Im σ. Let s, t ∈ [0, 1] be such that ϕ(i0 ) = σ(s) and ϕ(i0 + 1) =
ε
σ(t). Then s ≤ 12 < t. By the choice of λ, diameter(σ([s, t])) < 2m−1
. Thus
ε
d(ϕ(i0 ), x) < 2m−1 . Hence,
1
d(γ(α(i0 )), x) =
2m−1
≤
1
2m−1
<
i=1
2i
∞
|πi (γ(α(i0 ))) − x|
i=1
=
∞
|πi (γ(α(i0 ))) − x|
2m−1+i
+
m−1
i=1
|πi (σ(β(i0 ))) − x|
2i
d(ϕ(i0 ), x)
ε
.
2m−1
Therefore, d(γ(α(i0 ), x)) < ε. This contradicts the choice of ε and proves that
Claim 1 holds.
Claim 2. (0, 0) ∈
/ M or (1, 1) ∈
/ M.
Suppose, contrary to Claim 2, that (0, 0), (1, 1) ∈ M . Then 0, 1 ∈ S. Let
σ, γ : [0, 1] → S be one-to-one continuous functions such that σ(0) = 0 = γ(0),
σ(1) = 1 = γ(1), Im σ ∩ Im γ = {0, 1} and Im σ ∪ Im γ = S.
Let ε > 0 be such that N (ε, Im σ) ∩ N (ε, γ([ 13 , 23 ])) = ∅, N (ε, σ([0, 13 ]) ∪ γ([0, 13 ]))
∩N (ε, σ([ 12 , 1]) ∪ γ([ 23 , 1])) = ∅, and N (ε, σ([0, 12 ]) ∪ γ([0, 13 ])) ∩ N (ε, σ([ 23 , 1]) ∪
γ([ 23 , 1])) = ∅, and if d(γ(s), γ(t)) < ε, then |t − s| < 13 . Let m ∈ N be such
1
that 2m−1
< 2ε . Let L = Im σ ∪ γ([0, 13 ]) ∪ γ([ 23 , 1]). Then J is an arc. We consider
J with the natural order < for which γ( 13 ) < γ( 23 ). Given points w, z ∈ J, denote
by J(w, z) the subarc of J joining them if w = z, and let J(w, z) = {w} if w = z.
Then there exists λ > 0 such that λ < ε and if z, w ∈ J and d(w, z) < λ, then
ε
. Let δ > 0 be such that δ < 13 , and if s, t ∈ [0, 1] and
diameter(J(w, z)) < 2m−1
|s − t| < δ, then d(γ(s), γ(t)) < 2λm and d(σ(s), σ(t)) < 2λm .
Since π1 (γ(0)) = π1 (σ(0)) = πm (σ(0)) = 0 and π1 (γ(1)) = π1 (σ(1)) = πm (σ(1))
= 1, we can apply Lemma 1 to the pairs of maps (π1 ◦ γ, πm ◦ σ) and (π1 ◦ σ, πm ◦ σ)
and to the number δ, so there exist n, k ∈ N and functions α, β : Pn → [0, 1] and
ζ, η : Pk → [0, 1] such that π1 ◦ γ ◦ α = πm ◦ σ ◦ β, π1 ◦ σ ◦ ζ = πm ◦ σ ◦ η,
α(0) = β(0) = ζ(0) = η(0) = 0, and α(n) = β(n) = ζ(k) = η(k) = 1, for
each i ∈ {0, . . . , n − 1}, |α(i + 1) − α(i)| , |β(i + 1) − β(i)| < δ and, for each j ∈
{0, . . . , k − 1}, |ζ(j + 1) − ζ(j)| , |η(j + 1) − η(j)| < δ.
Consider the functions ϕ : Pn → S and ψ : Pk → S given by
ϕ(i) = (π1 (σ(β(i))), . . . , πm (σ(β(i))), π2(γ(α(i))), π3 (γ(α(i))), . . .)
and
ψ(j) = (π1 (σ(η(j))), . . . , πm (σ(η(j))), π2(σ(ζ(j))), π3(σ(ζ(j))), . . .).
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2992
ALEJANDRO ILLANES
Given i ∈ Pn , we have that πm (σ(β(i))) = π1 (γ(α(i))); then ϕ(i) ∈ S. Notice
that σ(β(i)) and ϕ(i) coincide in the first m coordinates, so d(σ(β(i)), ϕ(i)) ≤
1
/ γ([ 13 , 23 ]). Thus ϕ(Pn ) ⊂ J. Similarly,
2m−1 < ε. By the choice of ε, ϕ(i) ∈
ψ(j) ∈ S, for each j ∈ Pk and ψ(Pk ) ⊂ J.
Given i ∈ {0, 1, . . . , n − 1}, |α(i + 1) − α(i)| , |β(i + 1) − β(i)| < δ. This implies that d(σ(β(i)), σ(β(i + 1))) < 2λm and d(γ(α(i)), γ(α(i + 1))) < 2λm . Thus
λ
ε
d(ϕ(i), ϕ(i + 1)) < 2m−1
≤ λ. Hence diameter(J(ϕ(i), ϕ(i + 1))) < 2m−1
. Similarly,
ε
for each j ∈ {0, 1, . . . , k − 1}, diameter(J(ψ(j), ψ(j + 1))) < 2m−1 .
Since ϕ(0) = 0 = ψ(0) and ϕ(n) = 1 = ψ(k), we can define i0 = max{i ∈ Pn :
ϕ(i) ∈ γ([0, 13 ])}, i1 = min{i ∈ {i0 , . . . , n} : ϕ(i) ∈ γ([ 23 , 1])}, j0 = max{j ∈ Pk :
ψ(j) ∈ γ([0, 13 ])} and j1 = min{j ∈ {j0 , . . . , k} : ψ(j) ∈ γ([ 23 , 1])}. Notice that i0 <
i1 and j0 < j1 . Notice also that, for each i ∈ {i0 +1, . . . , i1 −1}, σ(0) < ϕ(i) < σ(1).
/ σ([ 12 , 1]) ∪ γ([ 32 , 1]), so
Since d(ϕ(i0 ), ϕ(i0 + 1)) < ε, by the choice of ε, ϕ(i0 + 1) ∈
1
1
ϕ(i0 +1) ∈ σ([0, 2 )). Similarly, ϕ(i1 −1) ∈ σ(( 2 , 1]). Thus i0 +1 < i1 −1. Similarly,
ε
j0 + 1 < j1 − 1. Notice that 0 ∈ J(ϕ(i0 ), ϕ(i0 + 1)). Thus d(0, ϕ(i0 + 1)) < 2m−1
.
1
This implies that d(γ(0), γ(α(i0 + 1))) < ε. Hence, α(i0 + 1) < 3 . Similarly,
2
3 < α(i1 − 1).
Given i ∈ {i0 + 1, . . . , i1 − 1}, ψ(j0 ) ≤ σ(0) < ϕ(i) < σ(1) ≤ ψ(j1 ). Thus,
there exists j(i) ∈ {j0 , . . . , j1 } such that ψ(j(i)) ≤ ϕ(i) < ψ(j(i) + 1). Hence
ε
. This implies that
d(ϕ(i), ψ(j(i))) < 2m−1
1
d(γ(α(i)), σ(ζ(j(i)))) =
2m−1
1
2m−1
∞
|πl (γ(α(i))) − πl (σ(ζ(j(i))))|
l=1
2l
≤ d(ϕ(i), ψ(j(i)))
ε
.
<
2m−1
Hence, d(γ(α(i)), σ(ζ(j(i)))) < ε and γ(α(i)) ∈ N (ε, Im σ). By the choice of ε,
α(i) ∈ [0, 13 ] ∪ [ 23 , 1].
Therefore, α(i0 + 1) ∈ [0, 13 ], α(i1 − 1) ∈ [ 23 , 1], α(i) ∈ [0, 13 ] ∪ [ 23 , 1] for each
i ∈ {i0 + 1, . . . , i1 − 1} and |α(i + 1) − α(i)| < δ < 13 for each i ∈ {i0 + 1, . . . , i1 − 2}.
Clearly, this is absurd. We have shown that Claim 2 holds.
We are ready to obtain a final contradiction. Since π1 (S) = [0, 1] = π2 (S), there
exist points p, q ∈ S such that π1 (p) = 0 and π1 (q) = 1. Since M is connected
and ρi (M ) = [0, 1] for each i ∈ {1, 2}, we have that M ∩ Δ = ∅. Let x ∈ [0, 1]
be such that (x, x) ∈ M ∩ Δ. Then x ∈ S . By Claim 1, x ∈ {p, q}. If x = p,
then 0 = π1 (p) = π1 (x), so x = 0 and p = 0. Similarly, if x = q, then x = 1
= q. Therefore, x ∈ {0, 1}. This shows that M ∩ Δ ⊂ {(0, 0), (1, 1)}. By Claim 2,
(0, 0) ∈
/ M or (1, 1) ∈
/ M . Thus M ∩ Δ = {(0, 0)} or M ∩ Δ ⊂ {(1, 1)}. We may
assume that M ∩ Δ = {(0, 0)}; the other case is similar. Notice that 0 ∈ S.
Let T = {r ∈ S : π1 (r) ≥ π2 (r)} and R = {r ∈ S : π1 (r) ≤ π2 (r)}. Then
T and R are closed subsets of S such that S = T ∪ R and {0} ⊂ T ∩ R. If
r = (r1 , r2 , r3 , . . .) ∈ T ∩R, then r1 = r2 . Thus (r2 , r1 ) ∈ M ∩Δ. Hence r1 = 0 = r2 .
If r = 0, we can take m = min{i ∈ N : ri = 0}. Then m ≥ 3; the point
p0 = (rm−1 , rm , rm+1 , . . .) belongs to S and π1 (p0 ) = 0. Applying Claim 1 to
p0 , q and 0, we obtain a contradiction. Therefore, r = 0. We have shown that
T ∩ R = {0}.
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THE GENERALIZED INVERSE LIMIT OF A SUBSET OF [0, 1]2
2993
Since ρi (M ) = [0, 1] for each i ∈ {1, 2}, we have that T −{0} = ∅ and R−{0} = ∅.
Thus 0 is a cut point of S. This is impossible since S is a simple closed curve. This
finishes the proof of the theorem.
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Instituto de Matemáticas, Universidad Nacional Autónoma de México, Circuito Exterior, Ciudad Universitaria, México 04510, D.F.
E-mail address: [email protected]
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