arXiv:1703.06862v4 [math.GN] 16 May 2017
Quasi-components in widely-connected spaces
David Lipham
Department of Mathematics and Statistics
Auburn University
Auburn, AL 36849, United States
E-mail: [email protected]
Abstract
An unanswered question in general topology asks whether the Stone-Čech
compactification of a widely-connected space is necessarily an indecomposable
continuum. We describe a property of X that is necessary and sufficient in
order for βX to be indecomposable, and examine the case when X is a separable metric space. We also construct two widely-connected sets in R3 – one is
contained in a composant of each of its compactifications, and the other fails
to be indecomposable upon the addition of a single limit point.
1
Introduction
A connected topological space W is widely-connected if every non-degenerate connected subset of W is dense in W . This paper addresses a collection of open problems
concerning widely-connected spaces. They are contained in the following multi-part
question from the 2004 Spring Topology and Dynamics Conference report, due to
Jerzy Mioduszewski. Part (a) of the question is also posed in [1] by David Bellamy,
who conjectured a positive answer to part (b).
Question 1 (Mioduszewski; 23 in [9]). Let W be a widely-connected space.
(a) Is βW an indecomposable continuum?
(b) If W is metrizable and separable, does W have a metric compactification which
is an indecomposable continuum?
(c) If W is metrizable and separable, does W have a metric compactification γW
such that for every composant P of γW , W ∩ P is (i) totally disconnected?
(ii) finite? (iii) a singleton?
2010 Mathematics Subject Classification: Primary 54D05; Secondary 54F15, 54E50, 54G15.
Key words and phrases: widely-connected, indecomposable, quasi-component
1
2
D. Lipham
γW is a compactification of W if γW is a compact Hausdorff space in which
W is densely embedded; βW denotes the Stone-Čech compactification of W ;
P is a composant of γW if P is the union of all proper subcontinua of γW
that contain a given point.
A connected compact Hausdorff space (a continuum) is indecomposable if it is
not the union of any two of its proper subcontinua. If X is a continuum, then X is
indecomposable if and only if every proper subcontinuum of X is nowhere dense in
X ([7] §48 V Theorem 2). Thus, the term indecomposable can be consistently defined
for arbitrary topological spaces – say that X is indecomposable if every connected
subset of X is either dense or nowhere dense in X.
The process of compactifying a space can destroy indecomposability. In fact, by
a result of Philip Bowers [2] every widely-connected separable metric space has a
compactification equal to the Hilbert cube. There is also a widely-connected subset
of Euclidean 3-space that fails to be indecomposable when a single limit point is
added to it – Section 3.1. The existence of such an example was suggested by Mary
Ellen Rudin [10], who proved that this scenario is impossible in the plane but never
published her example. The special limit point glues together many disjoint closed
subsets of the original space, and is therefore not representative of a point in the
Stone-Čech compactification.
Using properties of quasi-components, we prove that Questions 1(a) and 1(b) are
equivalent when W is metrizable and separable. In the language of continuum theory
they are equivalent to: If W is a connected separable metric space that is irreducible
between every two of its points, then does W have an irreducible compactification?
Perhaps surprisingly, W may not have a compactification that is irreducible between
two points in W . There is a widely-connected W ⊆ R3 such that if γW is any
compactification of W then there is a composant P of γW such that W ⊆ P –
Section 3.2. This provides a complete (negative) answer to Question 1(c), improving
one of our previous results in Section 4 of [8].
2
A property of quasi-components
If X is a topological space, then X is strongly indecomposable means that for every
two non-empty disjoint open sets U and V there are two closed sets A and B such
that X = A ∪ B, A ∩ U 6= ∅, B ∩ U 6= ∅, and A ∩ B ⊆ V .
Remark. X is strongly indecomposable if and only if the quasi-components of
proper closed subsets of X are nowhere dense in X.
Perfect totally disconnected ⇒ strongly indecomposable ⇒ indecomposable.
Quasi-components
3
Lemma 1. If Y is strongly indecomposable and X ⊆ Y such that X = Y , then X
is strongly indecomposable.
Proof. Let U and V be non-empty disjoint open subsets of X. Let U 0 and V 0 be open
subsets of Y such that U = U 0 ∩ X and V = V 0 ∩ X; X = Y implies U 0 ∩ V 0 = ∅.
There are closed subsets A0 and B 0 of Y such that Y = A0 ∪ B 0 , U 0 ∩ A0 6= ∅,
U 0 ∩ B 0 6= ∅, and A0 ∩ B 0 ⊆ V 0 . Let A = A0 ∩ X and B = B 0 ∩ X. Clearly A and
B are closed in X, X = A ∪ B, and A ∩ B ⊆ V . Note that U 0 ∩ A0 = U 0 \ B 0
and U 0 ∩ B 0 = U 0 \ A0 are non-empty open subsets of Y . So X = Y implies that
U ∩ A 6= ∅ and U ∩ B 6= ∅.
The following is implicit in the proof of Lemma 9.8 in [12].
Lemma 2. If U and V are disjoint open subsets of X and W is open in βX such
that W ∩ X = U ∪ V , then W0 := W ∩ clβX U and W1 := W ∩ clβX V are disjoint
open sets in βX and W0 ∪ W1 = W .
Proof. By density of X in βX we have W ⊆ clβX U ∪ clβX V . Thus W0 ∪ W1 = W .
We need to show that W ∩ clβX U ∩ clβX V = ∅. Suppose for a contradiction that
there exists p ∈ W ∩ clβX U ∩ clβX V . By Urysohn’s Lemma there is a mapping
F : βX → [0, 1] such that F (p) = 0 and F [βX \ W ] = 1. Define f : X → [0, 1] by
(
1
if x ∈ U
f (x) =
F (x) if x ∈
/ U.
If T is open in [0, 1] then
(
f
−1
[T ] =
F −1 [T ] ∩ X ∪ U
F −1 [T ] ∩ V
if 1 ∈ T
if 1 ∈
/ T,
so f is continuous. Since p ∈ clβX V and f V = F V , we have βf (p) = F (p) = 0.
On the other hand, p ∈ clβX U implies that βf (p) = 1, a contradiction.
Theorem 3. βX is indecomposable if and only if X is strongly indecomposable.
Proof. Suppose that βX is indecomposable. To show that X is strongly indecomposable, by Lemma 1 it suffices to show that βX is strongly indecomposable. Let U
and V be non-empty disjoint open subsets of βX. Let T be a non-empty open subset
of βX such that T ⊆ V . Since U is not contained in a component of βX \ T , by Theorem 6.1.23 in [5] there are disjoint closed sets A0 and B 0 such that A0 ∪B 0 = βX \T ,
A0 ∩ U 6= ∅, and B 0 ∩ U 6= ∅. Then A := A0 ∪ T and B := B 0 ∪ T are as desired.
Suppose that X is strongly indecomposable. Let K be a proper closed subset
of βX with non-empty interior; we show that K is not connected. Let S and T be
4
D. Lipham
non-empty open subsets of βX such that S ⊆ K and K ∩ clβX T = ∅. Let A and B
be disjoint closed subsets of X such that A ∪ B = X \ T , A ∩ S 6= ∅, and B ∩ S 6= ∅.
Note that U := A \ clβX T and V := B \ clβX T are open in X, U ∩ S 6= ∅, and
V ∩ S 6= ∅. By Lemma 2, W := βX \ clβX T is the union of the two disjoint open
sets W0 := W ∩ clβX U and W1 := W ∩ clβX V . Since K ⊆ W and Wi ∩ K 6= ∅ for
each i < 2, K is not connected.
Corollary 4. If X is compact Hausdorff, then X is indecomposable if and only if
X is strongly indecomposable.
Proof. βX ' X when X is compact Hausdorff.
Corollary 5. βX is an indecomposable continuum if and only if X is strongly
indecomposable and connected.
Proof. βX is a continuum if and only if X is connected.
Widely-connected spaces are usually constructed as dense subsets of indecomposable continua. Gary Gruenhage [6] constructed completely regular and perfectly
normal examples by a different method, assuming Martin’s Axiom and the Continuum Hypothesis, respectively. Their co-infinite subsets are totally disconnected (it’s
worth noting that there is no connected metric space with this property), thus both
examples are strongly indecomposable, and so their Stone-Čech compactifications
are indecomposable.
If W is not strongly indecomposable, then it has a perfect hereditarily disconnected subset which also fails to be strongly indecomposable. In point of fact, let U
and V be open subsets of W for which the definition fails. Let T be a non-empty
open set with T ⊆ V . T is the union of two relatively clopen subsets A and B. Then
Y := W \ A is an open hereditarily disconnected subset of W . Since W is connected,
Y is perfect. And no clopen partition of Y \ B divides U .
Example 1. A perfect hereditarily disconnected set which is not strongly indecomposable.
Let C be the middle-thirds Cantor set in [0, 1]. If Q and P denote the rationals and
irrationals, respectively, then let X = (C 0 × Q) ∪ (C 00 × P), where C 0 is the set of
endpoints of intervals removed during the construction of C, and C 00 = C \ C 0 . The
significant property of X is that for each c ∈ C, X ∩ ({c} × R) is a quasi-component
of X. Clearly X ∩ ({c} × R) is also dense in {c} × R for each c ∈ C. Thus X is
connectible, as defined in [8].
Let X1 = X ∩ (R × [0, 1]), and let X10 = {h−y, xi : hx, yi ∈ X1 } be the copy of
X1 rotated 90◦ about the origin. Then Y := X1 ∪ X10 is hereditarily disconnected,
Quasi-components
5
and X10 is contained in a quasi-component of Y \ (C ∩ (1/2, 1]) × [0, 1]. To prove the
latter, suppose that A is a clopen subset of Y \ (C ∩ (1/2, 1]) × [0, 1] and A ∩ X10 6= ∅.
Since A is open, there exists c ∈ C 0 \ {0, 1} such that A ∩ ((−1, 0) × {c}) 6= ∅.
Since c ∈ Q and X10 is connectible, h0, ci ∈ A. Then X1 ∩ ({0} × (0, 1)) ⊆ A because
X1 ∩ ([0, 1/2) × (0, 1)) is connectible. So X10 ∩ ((−1, 0) × C 0 \ {0, 1}) ⊆ A. Since A
is closed, we have X10 ⊆ A.
Figure 1: Y ⊆ [−1, 0] × C ∪ C × [0, 1]
From now on, assume that X is a separable metric space. The standard metric on
P the Hilbert cube [0, 1]ω is given by ρ(y, y 0 ) = n∈ω y(n) − y 0 (n) · 2−n . ([0, 1]ω )X is
the set of functions from X into [0, 1]ω , endowed with the complete metric %(f, g) =
sup{ρ(f (x), g(x)) : x ∈ X}. Lemmas 6 through 8 will be used to show Questions
1(a) and 1(b) are equivalent for separable metric spaces – Theorem 9.
Lemma 6 (§46 V Theorem 3 in [7]). There is a continuous ϕ : X → 2ω such that
the quasi-components of X coincide with the non-empty point inverses of ϕ.
Lemma 7 (§44 V Corollary 4a and §44 VI Lemma in [7]). If A and B are disjoint
closed subsets of X, then
n
o
X
g ∈ [0, 1]ω : g[A] ∩ g[B] = ∅
is a dense open subset of ([0, 1]ω )X .
Lemma 8 (§44 VI Theorem 2 in [7]). The set of homeomorphic embeddings from
X into [0, 1]ω is a dense Gδ -set in ([0, 1]ω )X .
Theorem 9. βX is indecomposable if and only if X has an indecomposable metric
compactification.
Proof. Suppose that βX is indecomposable. Let {Un : n ∈ ω} be a countable basis
for X. For each n ∈ ω let ϕn : X \ Un → 2ω be given according to Lemma 6.
6
D. Lipham
Let {Ci : i ∈ ω} be the canonical clopen basis for 2ω . For each hn, ii ∈ ω 2 , if
−1 ω
Ahn,ii = ϕ−1
n [Ci ] and Bhn,ii = ϕn [2 \ Ci ] then
n
o
X
g ∈ [0, 1]ω : g[Ahn,ii ] ∩ g[Bhn,ii ] = ∅
is a dense open subset of ([0, 1]ω )X by Lemma 7. By Lemma 8 and the fact that
([0, 1]ω )X is complete, there is a homeomorphic embedding e : X ,→ [0, 1]ω such that
e[Ahn,ii ] ∩ e[Bhn,ii ] = ∅ for each hn, ii ∈ ω 2 .
The metric compactification e[X] is indecomposable. Let K be a proper closed
subset of e[X] with non-empty interior. There exists n ∈ ω such that K ∩ e[Un ] = ∅.
Since X is strongly indecomposable, there exists i ∈ ω such that K∩e[Ahn,ii ] 6= ∅ and
K∩e[Bhn,ii ] 6= ∅. As e[X] = e[Un ]∪e[Ahn,ii ]∪e[Bhn,ii ], we have K ⊆ e[Ahn,ii ]∪e[Bhn,ii ].
Finally, e[Ahn,ii ] ∩ e[Bhn,ii ] = ∅, so K is not connected.
Now suppose that γX is a (metric) indecomposable compactification of X. By
Corollary 4, γX is strongly indecomposable. X also is strongly indecomposable by
Lemma 1, thus βX is indecomposable by Theorem 3.
If X is a connected space and p, q ∈ X, then X is irreducible between p and q
if no proper closed connected subset of X contains both p and q. A continuum is
said to be irreducible if it is irreducible between some pair of its points, i.e., if it has
more than one composant.
Corollary 10. If X is an indecomposable connected space, then βX is indecomposable if and only if X has an irreducible compactification (which is necessarily
indecomposable).
Proof. If βX is indecomposable, then by Theorem 9 X has a metric indecomposable
compactification which necessarily has an uncountable number of composants.
Now suppose that βX is decomposable. Let γX be a compactification of X. If
ι : X → X is the identity map, then the Stone-Čech extension βι : βX → γX
maps proper subcontinua to proper subcontinua and maps onto γX. Thus to prove
γX has only one composant, it suffices to show that βX has only one composant.
To that end, let p, q ∈ βX. Let H and K be proper subcontinua of βX such that
βX = H ∪K. Assume that p ∈ K. If q ∈ K, then p and q are in the same composant.
Otherwise, q ∈ U := βX \ K. Since X is indecomposable and U ∩ X 6= X, U ∩ X is
not connected, whence U is not connected by Lemma 2. Let T be a proper clopen
subset of U with q ∈ T . Then K ∪ T is a proper subcontinuum of βX containing p
and q. Again p and q are in the same composant of βX.
Suppose that γX is irreducible (between two points p and q); we show that γX
is indecomposable. Well, otherwise there is a proper subcontinuum K ⊆ γX with
Quasi-components
7
non-empty interior such that p ∈ K. Then βι−1 [K] is a proper closed subset of βX
with non-empty interior. Thus, if q 0 ∈ βι−1 {q} then there is a proper subcontinuum
L ⊆ βX such that q 0 ∈ L and L ∩ βι−1 [K] 6= ∅ (composants are dense). We know
that βX is indecomposable, so L is nowhere dense. Thus L ∪ βι−1 [K] is a proper
closed subset of βX, implying that βι[L] ∪ K is a proper subcontinuum of γX
containing p and q, a contradiction.
Remark. The main result of [2] is that every separable metric nowhere locally
compact space densely embeds into the Hilbert space `2 ' (0, 1)ω . Widely-connected
Hausdorff spaces have no compact neighborhoods by Theorem 6.2.9 in [5]. Thus,
every widely-connected separable metric space has a compactification equal to the
Hilbert cube [0, 1]ω , a decomposable continuum with exactly one composant.
Observe that a widely-connected space is a connected space that is irreducible
between every two of its points. By Theorem 9 and Corollary 10, Question 1(b) is
equivalent to the following.
Question 1(b)0 . If W is a connected separable metric space that is irreducible between every two of its points, then does W have an irreducible compactification?
3
3.1
Examples in R3
Destroying indecomposability with one limit point
Let K and K0 be as defined in Section 3.2. Wojciech Dębski [3] constructed a closed
set A ( K such that K \ A is connected and A intersects every composant of K.
In [8], we constructed a completely metrizable widely-connected set W by removing
ω-many copies of Dębski’s set from K. Assume that A is one of the copies in K0 \W ,
c := W ∪ A to be on the surface of the unit sphere S2 ⊆ (R3 , d).
and consider W
c → R3 be the mapping x 7→ inf{d(x, y) : y ∈ A} · x; q shrinks A
Let q : W
to the single point at the origin h0, 0, 0i. By the construction of W , A intersects
c ∩ K0 (see Section 6 of [8]), so q[W
c ∩ K0 ] is connected.
every quasi-component of W
c ] is not indecomposable. Meanwhile, compactness of A implies that
Therefore q[W
q W is a homeomorphism. In summary q[W ] is a widely-connected subset of R3
that fails to be indecomposable when a single limit point is added to it.
3.2
A composant-locked widely-connected set
In this section we show that W may not have a compactification that is irreducible
between two points in W . This provides a complete negative answer to Question
1(c) (see the statement of Theorem 14). By Corollary 10, a positive answer to any
8
D. Lipham
part of Question 1(c) would have implied a proof of Bellamy’s conjecture (a positive
answer to Question 1(b)). Our counterexample demonstrates that a different line of
attack is needed.
Let e ∈ 2Z be the point defined by concatenating all finite binary sequences in
S
the positive and negative directions of Z. That is, if {bi : i ∈ ω} enumerates n∈ω 2n ,
then let e Z+ = b0_ b1_ b2_ ..., and define e(n) = e(−n) for n ∈ Z− . Let η : 2Z → 2Z
be the shift map η(x)(n) = x(n + 1); clearly η is a homeomorphism. The backward
and forward orbits of e, E0 = {η n (e) : n ∈ Z− } and E1 = {η n (e) : n ∈ Z+ \ {0}},
are dense in 2Z by construction. Finally, E0 ∩ E1 = ∅. For if n, m ∈ ω such that
η −n (e) = η m (e), then η n+m (e) = e. Density of E1 implies that e is not a periodic
point. So n + m = 0, whence n = m = 0. We have η −n (e) = η m (e) = e ∈ E0 \ E1 .
Figure 2 shows a rectilinear version of the bucket-handle continuum K ⊆ [0, 1]2
minus the diagonal Cantor set ∆ = {hx, xi ∈ K : x ∈ [0, 1]}. K \ ∆ is the union of
two open sets K0 = {hx, yi ∈ K : x < y} and K1 = {hx, yi ∈ K : x > y}.
K0
K0
K1
K3
Figure 2: K \ ∆ = K0 ∪ K1
Let X be any widely-connected subset of K (see [11] and [8] for examples), and
for each i < 2 put Wi = (X ∩Ki )∪∆. Note that each Wi is hereditarily disconnected,
but W0 ∪ W1 = X ∪ ∆ is connected since X is connected and (necessarily) dense in
K. Define W ⊆ 2Z × K by
[
W = (Ei × Wi );
i<2
W is hereditarily disconnected, but it will become connected when certain pairs of
its points are identified. Let p, q ∈ ∆ be such that K is irreducible between p and q.
Those familiar with the Cantor set and the composants of K may choose p = h1, 1i
and q = h 41 , 14 i, for instance. Define a relation on 2Z × K by ∼ = hx, pi, hη(x), qi :
f = W/ ∼ .
x ∈ 2Z , and finally, put W
K2
Quasi-components
9
Remark. Recall that the quotient of a compact metric space is metrizable whenever
Hausdorff. The entire quotient (2Z × K)/ ∼ is easily seen to be Hausdorff, and is
therefore metrizable (and compact). Together with dim((2Z × K)/ ∼) = 1, this
implies via the Menger-Nöbeling Theorem (1.11.4 in [4]) that (2Z × K)/ ∼ embeds
f also embeds into R3 .
into Euclidean 3-space R3 , and so of course W
Lemma 11. Let X be a space. If U ⊆ X is open, A0 is clopen in U , A1 is clopen
in X \ U , and A0 ∩ ∂U = A1 ∩ ∂U , then A0 ∪ A1 is clopen in X.
Proof. Clearly A0 ∪ A1 is closed in X and (A0 ∪ A1 ) \ ∂U is open in X. We show
that A0 ∪ A1 is open in X by showing that it is an X-neighborhood of each of its
points in ∂U . Suppose that a ∈ (A0 ∪ A1 ) ∩ ∂U . Then a ∈ A0 ∩ A1 . There are two
open subsets of X, V0 and V1 , such that a ∈ V0 ∩ U ⊆ A0 and a ∈ V1 ∩ (X \ U ) ⊆ A1 .
Then a ∈ V0 ∩ V1 ⊆ A0 ∪ A1 ; to prove the inclusion, suppose x ∈ V0 ∩ V1 and consider
the two cases x ∈ U and x ∈ X \ U .
Lemma 12. For each i < 2, {0} × Wi is a quasi-component of
[
Y := ({0} × Wi ) ∪
({1/n} × W1−i ).
n≥1
Proof. Fix i < 2. Clearly {0} × Wi contains a quasi-component of Y . Now let A be a
clopen subset of Y such that A∩{0}×Wi 6= ∅; we show that {0}×Wi ⊆ A. For each
a ∈ A ∩ ∆ there is an integer n(a) and an open U (a) ⊆ ∆ such that a ∈ [0, 1/n(a)] ×
U (a) ⊆ A (by [0, 1/n] we mean {0}∪{1/k : k ≥ n}). By compactness of A∩∆, there
S
is a finite F ⊆ A such that A ∩ ∆ ⊆ {[0, 1/n(a)] × U (a) : a ∈ F }. Similarly, there
is a finite E ⊆ Y \ A and a collection of open sets {[0, 1/n(b)] × U (b) : b ∈ E} such
S
that (Y \ A) ∩ ∆ ⊆ {[0, 1/n(b)] × U (b) : b ∈ E}. Let m = max{n(y) : y ∈ F ∪ E}.
Let A0 = A∩({0}×Wi ) and A1 = π[A∩({1/m}×W1−i )], where π : R3 → {0}×R2
is the projection onto the y-z-plane. A0 ∩ ({0} × ∆) = A1 ∩ ({0} × ∆) by the
choice of m. Applying Lemma 11 with U = {0} × (Wi \ ∆), we have A0 ∪ A1 is
clopen in {0} × (W0 ∪ W1 ). Since {0} × (W0 ∪ W1 ) = {0} × (X ∪ ∆) is connected,
A0 ∪ A1 = {0} × (W0 ∪ W1 ). Thus {0} × Wi ⊆ A0 ⊆ A.
f is widely-connected.
Theorem 13. W
Proof. For each i < 2 and e0 ∈ Ei , {e0 }×Wi is a quasi-component of W . This follows
immediately from Lemma 12 since there is a sequence of points (en ) ∈ (E1−i )ω such
that en → e0 as n → ∞. Thus, the relation ∼ links together the quasi-components
f is connected.
of W , so that W
f is widely-connected, we let C be a non-empty connected subset
To prove that W
f with C 6= W
f , and show that |C| = 1.
of W
10
D. Lipham
Let U × V be a non-empty open subset of 2Z × K such that {p, q} ∩ V = ∅ and
C ∩ (U × V ) = ∅. Let x ∈ C. Let x0 ∈ W such that x = x0 / ∼, and let e0 be the first
coordinate of x0 . Since E0 and E1 are dense in 2Z , there exist n, m > 0 such that
{η −n (e0 ), η m (e0 )} ⊆ U . Since η n+m is a homeomorphism and 2Z has a basis of clopen
sets, there is a clopen A ⊆ U such that η −n (e0 ) ∈ A and η n+m [A] ⊆ U . Let L and M
be disjoint closed subsets of K such that K \ V = L ∪ M , p ∈ L, and q ∈ M . Then
[
T := W \ (U × V ) ∩ (A × L) ∪ (η n+m [A] × M ) ∪
(η i [A] × K)
∼
0<i<n+m
f \ (U × V ).
is a clopen subset of W
Note that x ∈ T since e0 ∈ η n [A] and 0 < n < n + m. So C ⊆ T and
x ∈ Q := T ∩
η i (e0 ) : −n ≤ i ≤ m × K / ∼ .
If y ∈ T \ Q then by the construction of T there is a relatively clopen S ⊆ T such
that Q ⊆ S ⊆ T \ {y}. Thus C ⊆ Q, which implies |C| = 1 (see Figure 3).
p
q
Figure 3: Superset of Q if n = m = 2 and e0 = e
f is a compactification of W
f , then there is a composant P of
Theorem 14. If γ W
f such that W
f ⊆ P.
γW
f there is a non-empty
Proof. It suffices to show that for every two points x, y ∈ W
f such that x and y are in the same quasi-component of W
f \ T . This
open T ⊆ W
f is a compactification of W
f and T 0 is open in γ W
f such that
would imply that if γ W
f = T , then x and y are contained in a quasi-component of γ W
f \ T 0 . By
T0 ∩ W
f \ T 0 , a proper
Theorem 6.1.23 in [5], x and y are contained in a component of γ W
f.
subcontinuum of γ W
Quasi-components
11
Let x and y be given. Pick two points x0 , y 0 ∈ W such that x = x0 / ∼ and
y = y 0 / ∼, and let n, m ∈ Z such that η n (e) and η m (e) are the first coordinates of x0
and y 0 , respectively. Assume that n ≤ m. There is a non-empty clopen A ⊆ 2Z such
that A ∩ {η i (e) : n − 1 ≤ i ≤ m + 1} = ∅.
For i ∈ Z, set δ(i) = 0 if i ≤ 0 and δ(i) = 1 if i > 0. Then
[ i
R :=
η (e) × Wδ(i) ∼
n≤i≤m
f \ T where T = (A × K)/ ∼. And {x, y} ⊆ R. Each fiber {η i (e)} ×
is a subset of W
Wδ(i) , n ≤ i ≤ m, is a quasi-component of W \ (A × K) by density of E1−δ(i) in
2Z \ A (Lemma 12). These fibers are linked together by ∼, so that R is contained in
f \ T.
a quasi-component of W
To make further progress on Question 1(a)-(b), one may try to construct a T1
or T2 widely-connected space that is not strongly indecomposable, or prove that
βW indecomposable when W is a (widely-connected) subset of the plane. Now a
question of our own:
Question 2. Let P be an indecomposable metric space which is the union of an
increasing collection of countably many continua.
(a) Is P strongly indecomposable?
(b) Is there a metric continuum with a composant equal to P ?
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