CHARACTERIZATIONS OF INTERVALS VIA CONTINUOUS SELECTIONS
Tsugunori NOGURA and Dmitri SHAKHMATOV
1
Department of Mathematics, Faculty of Science
Ehime University, Matsuyama 790, Japan
E-mail: [email protected] (Nogura)
E-mail: [email protected] (Shakhmatov)
Version of July 19, 1995
Abstract. We prove that: (i) a pathwise connected, Hausdorff space which has a continuous
selection is homeomorphic to one of the following four spaces: singleton, [0, 1), [0, 1] or the
long line L, (ii) a locally connected (Hausdorff) space which has a continuous selection must be
orderable, and (iii) an infinite connected, Hausdorff space has exactly two continuous selections
if and only if it is compact and orderable. We use these results to give various characterizations
of intervals via continuous selections. For instance, (iv) a topological space X is homeomorphic
to [0, 1] if (and only if) X is infinite, separable, connected, Hausdorff space and has exactly
two continuous selections, and (v) a topological space X is homeomorphic to [0, 1) if (and
only if) one of the following equivalent conditions holds: (a) X is infinite, Hausdorff, separable,
pathwise connected and has exactly one continuous selection; (b) X is infinite, separable, locally
connected and has exactly one continuous selection; (c) X is infinite, metric, locally connected
and has exactly one continuous selection. Three examples are exhibited which demonstrate the
necessity of various assumptions in our results.
MSC Classification: Primary 54C65; Secondary 54B20, 54C60, 54D05, 54D30, 54F05, 54F15.
Key words and phrases: Set-valued map, selection, continuous selection, hyperspace, Vietoris topology, ordered space, linear order, interval, closed interval, connected space, locally
connected space, compact space, pathwise connected space, arcwise connected space.
For a topological space X we use F (X) to denote the space of all closed, nonempty subsets
of X endowed with the Vietoris topology [10, 6]. Recall that the basis for the Vietoris topology
consists of sets
hV ; U1 , . . . , Un i = {F ∈ F (X) : F ⊆ V and F ∩ Ui 6= ∅ for all i ≤ n},
with n an arbitrary natural number and V, U1 , . . . , Un arbitrary open subsets of X. A map
f : F (X) → X is called a selection if f (F ) ∈ F for every F ∈ F (X). A continuous selection is
1
The second author would like to acknowledge partial financial support under Grant no. 06740071 from the
Ministry of Education, Science and Culture of Japan.
1
a selection f : F (X) → X which is continuous map. For a linear order < on X and a, b ∈ X we
define (−∞, a) = {x ∈ X : x < a}, (a, +∞) = {x ∈ X : a < x}, (a, b) = {x ∈ X : a < x < b},
[a, b) = {x ∈ X : a ≤ x < b}, (a, b] = {x ∈ X : a < x ≤ b} and [a, b] = {x ∈ X : a ≤ x ≤ b}.
Recall that a topological space X is orderable if there exists a linear order < on X which
generates its topology in a sense that the family
{(−∞, a) : a ∈ X} ∪ {(a, +∞) : a ∈ X} ∪ {(a, b) : a, b ∈ X, a < b}
is a base for the topology of X. In what follows [0, 1] denotes the closed unit interval in its
standard topology, and [0, 1) its subspace. The main goal of this note is to give different
characterizations of [0, 1) and [0, 1] in terms of continuous selections.
In every compact orderable space (X, <) maps F 7→ min F and F 7→ max F for F ∈ F (X)
are continuous selections. It turns out that this condition characterizes compact orderable
spaces among connected Hausdorff spaces:
Theorem 1 An infinite connected, Hausdorff space has exactly two continuous selections if
and only if it is compact and orderable.
Since a separable infinite connected compact linearly ordered space is homeomorphic to
[0, 1] (see [1]), from the last theorem we get the following characterization of [0, 1] in terms of
continuous selections:
Corollary 2 A topological space X is homeomorphic to [0, 1] if (and only if ) X is infinite,
separable, connected, Hausdorff space and has exactly two continuous selections.
We say that a space X is pathwise connected if for every pair x, y of different points of X
there exists a homeomorphic copy of the closed unit interval in X with endpoints x and y.
All pathwise connected spaces are connected, but the converse does not hold. Our next result
completely describes pathwise connected, Hausdorff spaces which have continuous selections:
Theorem 3 A pathwise connected, Hausdorff space which has a continuous selection is homeomorphic to one of the following four spaces: singleton, [0, 1), [0, 1] or the long line L.
Recall that the long line is an ordered space L = {(α, x) : α < ω1 , x ∈ [0, 1)} with the linear
ordering (α, x) < (β, y) iff either α < β, or α = β and x < y, where ω1 is the first uncountable
cardinal.
Let us agree to call a space X locally connected if X is Hausdorff and for every point x of
X and any open neighbourhood U of x there exist an open set V and a connected set C with
x ∈ V ⊆ C ⊆ U. With this definition we have the following two theorems:
Theorem 4 A locally connected space which has a continuous selection must be orderable.
Theorem 5 A locally connected space X has exactly one continuous selection if and only if X
is connected and its topology can be generated by a linear order < such that:
(i) every closed set F has a minimal element with respect to <, and
(ii) either X is a singleton, or X has no maximal element with respect to <.
2
Our last theorem gives three different characterizations of [0, 1) in terms of continuous selections:
Theorem 6 A topological space X is homeomorphic to [0, 1) if (and only if ) one of the following
equivalent conditions holds:
(i) X is infinite, Hausdorff, separable, pathwise connected and has exactly one continuous
selection.
(ii) X is infinite, separable, locally connected and has exactly one continuous selection.
(iii) X is infinite, metric, locally connected and has exactly one continuous selection.
Now we exhibit three examples serving to show the necessity of various conditions in our
results.
Example 7 The disjoint union [0, 1) ⊕ [0, 1) of two copies of [0, 1) is a locally connected space
which is neither compact nor homeomorphic to [0, 1]. It is proved in [8] that [0, 1) ⊕ [0, 1)
has exactly two continuous selections. This example demonstrates that “connected” cannot be
omitted or replaced by “locally connected” either in Theorem 1 or in Corollary 2.
Example 8 Consider a subspace of the plane X = {(0, 0)} ∪ {(x, sin(1/x) : 0 < x ≤ 1}. Then
X is a connected, complete separable metric space which is neither compact, nor orderable (and
thus not homeomorphic to [0, 1)). We will prove later in Lemma 15 that X has exactly one
continuous selection. Therefore the existence of X shows that: (a) “locally connected” cannot
be replaced by “connected” in Theorem 4, (b) “pathwise connected” cannot be weakened to
“connected” in Theorem 6(i), and (c) “locally connected” cannot be replaced by “connected”
in Theorem 6(iii).
Example 9 The long line L is (pathwise) connected, locally connected and has exactly one
continuous selection (by Theorem 5 above). Therefore “separable” cannot be omitted from
Theorem 6(i) and “metric” is essential in Theorem 6(iii).
Kuratowski, Nadler and Young [4] proved that a locally compact separable metric space
which has a continuous selection is homeomorphic to a subset of the real line R. Example 8
demonstrates that “locally compact metric” cannot be weakened to “complete metric” in their
result.
We turn now to the proofs of our results. Our first lemma organizes some scattered results
from [6] in a way necessary for future applications:
Lemma 10 Suppose that (X, T ) is a connected Hausdorff space which has a continuous selection. Then:
(i) there exists a linear order < on X such that the topology T< induced by < on X is weaker
than T , i.e. T< ⊆ T ,
(ii) (X, T ) has at most two continuous selections, which (if exist) are given by formulae
f (F ) = min F and g(F ) = max F for F ∈ F (X, T ) (here min and max is taken with respect to
<).
3
Proof: Let f : F (X, T ) → (X, T ) be a continuous selection. For x, y ∈ X with x 6= y define
x < y iff f ({x, y}) = x. It was proved in [6, Lemma 7.2] that < is a linear order satisfying
the conclusion of item (i) of our lemma. ¿From [6, Lemma 7.3.2] it follows that f (F ) = min F
for all F ∈ F (X, T ) [in particular, min F exists for every T -closed set F ]. Suppose now that
g : F (X, T ) → (X, T ) is another continuous selection different from f . Then from [6, Lemma
7.2.4] we conclude that g({x, y}) = max(x, y) for all x, y ∈ X with x 6= y. Let ≺ be the reverse
order of <, i.e. x ≺ y iff y < x. Form the above discussion it follows that x ≺ y iff x 6= y
and g({x, y}) = x. Applying [6, Lemma 7.3.2] once again we obtain that g(F ) is a ≺-minimal
element of F for every T -closed set F . Since ≺ is the reverse order of <, we conclude that
g(F ) = max F for all F ∈ F (X, T ). Finally, there is no more than two selections on (X, T ) by
[6, Proposition 7.8.2].
Lemma 11 Suppose that (X, T ) is a connected, locally connected space which has a continuous
selection. Then T = T< , where T< is as in Lemma 10.
Proof: Since (X, T ) is locally connected, it is Hausdorff. Let < be the linear order from Lemma
10(i). We are going to show that T< = T . The inclusion T< ⊆ T was already established in
Lemma 10(i). To prove the reverse inclusion pick arbitrarily V ∈ T and x ∈ V . Since
X is locally connected, we can find T -open sets U, W and a T -connected set C such that
x ∈ U ⊆ C ⊆ W ⊆ V . Since C is T -connected and T< ⊆ T , C is also T< -connected. Note that
C 6= {x}, since otherwise {x} = U ∈ T , which contradicts connectedness of (X, T ). Now we
consider three cases.
Case 1 . x is a <-minimal element of X. ¿From T< -connectedness of C and x ∈ C 6= {x} it
follows that there exists some c ∈ X with [x, c) ⊆ C ⊆ V . Since x is a <-minimal element of
X, one has [x, c) = (−∞, c) ∈ T< and we are done.
Case 2 . x is a <-maximal element of X. This case is similar to Case 1.
Case 3 . x is neither <-minimal nor <-maximal element of X. Since T< ⊆ T by Lemma 10,
both sets F − = {y ∈ X \ W : y < x} and F + = {y ∈ X \ W : y > x} are T -closed. By picking
if necessary a smaller W we will also assume, w.l.o.g., that F + 6= ∅ and F − 6= ∅.
Claim 1.
x 6= sup F − .
Proof Assume that x = sup F − . Since F − = {y ∈ X \ W : y < x} and C ⊆ W , it follows from
T< -connectedness of C that C ⊆ {y ∈ X : x ≥ y}. In particular, x ∈ U ⊆ C ⊆ {y ∈ X : x ≥ y}.
Since U ∈ T and (x, +∞) ∈ T< ⊆ T , we conclude that [x, +∞) ∈ T . Since (−∞, x) ∈ T< ⊆ T
and (X, T ) is connected, it follows that (−∞, x) = ∅, i.e. that x is a <-minimal element of X,
a contradiction with our assumption.
Claim 2.
x 6= inf F − .
Proof Assuming the contrary and arguing similar to the proof of previous claim we would
arrive to a conclusion that x is <-maximal element of X, again giving a contradiction with our
assumption.
¿From two claims above we obtain that x 6= sup F − and x 6= inf F + . Therefore there exist
a, b ∈ X such that x ∈ (a, b) ⊆ U ⊆ V , and the proof is complete.
4
Proof of Theorem 1: If (X, <) is a connected compact Hausdorff ordered space, then the
maps F 7→ min F and F 7→ max F for F ∈ F (X) are continuous selections, which are obviously
different if X has more than one element. ¿From Lemma 10 it follows that these are the only
selections on X. Suppose now that X is an infinite connected Hausdorff space which has exactly
two continuous selections. Let < be the linear order < on X given in Lemma 10(i).
Claim 3.
Every T -closed subset F of X has both min F ∈ F and max F ∈ F .
Proof Since X has exactly two continuous selections, these selections are given by formulae
from Lemma 10(ii), which implies that both min F and max F exist and belong to F .
Claim 4.
T< is compact.
Proof It suffices to show that every A ⊆ X has inf A and sup A [3]. In its turn, it is enough to
check that every T< -closed set F ⊆ X has inf F = max F and sup F = max F . Since T< ⊆ T ,
F is also T -closed, and result follows from Claim 3.
Claim 5.
T< = T .
Proof We need to show that every T -closed set F is also T< -closed. Assume that x is in the
T< -closure of F . Then x is in the T< -closure of either F − = {y ∈ F : y ≤ x} or F + = {y ∈ F :
y ≥ x}, which means that either x = sup F − or x = inf F + . Since T< ⊆ T , both F − and F +
are T -closed, and by Claim 3, sup F − = max F − ∈ F − ⊆ F and inf F + = min F + ∈ F + ⊆ F .
In any case one has x ∈ F . Thus F is T< -closed as required.
The proof of Theorem 1 is now complete.
Lemma 12 Suppose that (X, ≺) is a connected, locally connected, (linearly) ordered space
which has more than one point.
(i) If X has neither ≺-minimal nor ≺-maximal element, then (X, ≺) does not have a continuous selection.
(ii) The total number of continuous selections on (X, ≺) is equal to the number of ≺-minimal
and ≺-maximal elements of X.
Proof: Let < be the linear order from Lemma 10(i). ¿From Lemma 11 it follows that both
< and ≺ generate the same connected topology. Applying [1, Theorem II], we conclude that
ordering < and ≺ must be either identical (i.e. x < y iff x ≺ y) or inverse to each other (i.e.
x < y iff y ≺ x). Now the conclusion of our lemma easily follows from Lemma 10.
Corollary 13 (i) (0, 1) does not have continuous selections [2], and
(ii) [0, 1) has exactly one continuous selection.
Proof of Theorem 3: Assume that (X, T ) is a Hausdorff pathwise connected space which has
a continuous selection. Let < be the linear order from Lemma 10(i). Let us say that a subset
C of X is convex if [a, b] ⊆ C whenever a, b ∈ C and a < b. (In what follows all intervals and
convexity are understood with respect to <.)
Claim 6.
If a, b ∈ X and a < b, then [a, b] is homeomorphic to [0, 1].
5
Proof Let f : [0, 1] → (X, T ) be a continuous 1-1 map such that f (0) = a and f (1) = b,
which exists because (X, T ) is pathwise connected. Since T< ⊆ T by Lemma 10(i), f ([0, 1]) is
a T< -connected subset of X. Since every connected subset of a linearly ordered space is convex,
[a, b] ⊆ f ([0, 1]). The reverse inclusion f ([0, 1]) ⊆ [a, b] follows from the “mean value theorem”
and the fact that f is 1-1. Thus f is a 1-1 continuous map with f ([0, 1]) = [a, b], which implies
the conclusion of our claim.
Let L∗ be the space obtained by gluing together the minimal elements of two disjoint copies of
the long line L.
Claim 7.
(X, T ) does not contain a convex subset homeomorphic to L∗ .
Proof Suppose the contrary and let Z be a convex subset of (X, T ) homeomorphic to L∗ . As a
convex subset of a linearly ordered space, Z must coincide with one of the sets in the following
list: X, (−∞, c), (c, −∞), (−∞, c], [c, ∞), (c, d), (c, d], [c, d) or [c, d] (for some c, d ∈ X with
c < d). Proper intervals (c, d), (c, d], [c, d) or [c, d] are ruled out completely by Claim 6. The
case Z = X also cannot happen, because L∗ is a connected, locally connected ordered space
without both minimal and maximal element, so L∗ has no continuous selection by Lemma 12(ii),
while (X, T ) does have a continuous selection. We conclude that there must exist some c ∈ X
such that Z coincides with one of the rays (−∞, c), (c, ∞), (−∞, c] or [c, ∞). Without loss of
generality we will assume that Z is either (−∞, c) or (−∞, c]. Since Z 6= X, there exists b ∈ X
with c < b. Pick some a < c and consider Y = Z ∩ [a, b] with the subspace topology induced
from (X, T ). By Claim 6 [a, b] is homeomorphic to [0, 1], so Y is a separable metric space. Now
note that for every separable metric space Y ⊆ L∗ , the compliment L∗ \ Y of Y in L∗ is not
connected. On the other hand, Z \ Y = (−∞, a) is pathwise connected, which contradicts the
fact that Z is homeomorphic to L∗ .
By transfinite induction on α < ω1 we try to pick two sequences aα ∈ X and bα ∈ X such
that aβ < aα < bα < bβ . If our induction goes on up to stage ω1 , then one can easily see (with
S
the help of Claim 6) that Z = {[aβ , bβ ] : β < ω1 } would be a convex subset of X homeomorphic
to L∗ , which contradicts Claim 7. Thus our induction must stop at some countable α < ω1
S
because it becomes impossible to choose next two points. Since Y = {[aβ , bβ ] : β < α} is a
convex subset of X, one of the following two possibilities must hold:
(i) Y = X, or
(ii) there exists some c ∈ X such that Y coincides with one of the four rays (−∞, c), (c, ∞),
(−∞, c] or [c, ∞).
Assume first that (i) holds, i.e. Y = X. ¿From Claim 6 it follows that Y is homeomorphic
to one of the following spaces: (0, 1), [0, 1) and [0, 1]. The first case is impossible because X
has a continuous selection (Corollary 13(i)). Thus if (i) holds, X is homeomorphic to either
[0, 1) or [0, 1].
Suppose now that (ii) holds, and assume, w.l.o.g., that Y = [c, ∞). Now we use transfinite
induction on α < ω1 again to try to pick a sequence cα ∈ X such that c0 = c and cβ ≤ cα for
all α < β. If our induction stops at some α < ω1 because it becomes impossible to choose next
S
point, then the same argument as above shows that X = {(−∞, cβ ] : β < α} is homeomorphic
6
to either [0, 1) or [0, 1]. Suppose now that the induction goes on up to stage ω1 . Then it is easy
S
to see from Claim 6 that Z = {[cβ , ∞) : β < ω1 } will be homeomorphic to the long line L.
If Z = X, our proof is complete, so it remains only to show that the case X \ Z 6= ∅ cannot
happen. Indeed, if it does, then there must exist some d ∈ X such that Z = (d, ∞). Pick
arbitrarily e ∈ X with d < e. By our construction, there exists some α < ω1 such that cβ < e
for all β ≥ α. Since each of the sets (cβ+1 , cβ ) is open in X, the family {((cβ+1 , cβ ) : β > α} is
an uncountable family of pairwise disjoint nonempty open subsets of [d, e], which contradicts
the fact that [d, e] is homeomorphic to [0, 1] by Claim 6. The proof is now complete.
Proof of Theorem 4: Let X be a locally connected space. Then all connected components of
X are both open and closed in X. Let C be one of such components of X. Being closed in X,
C must have a continuous selection. Since C is connected and locally connected, the topology
of C is generated by some linear order <C (Lemma 11). Since each C is both open and closed
in X, in a straightforward way one can “glue” from <C ’s a linear order < which will generate
a topology of X.
Lemma 14 A space which has exactly one continuous selection must be connected.
Proof: Assume that Y has a continuous selection and Y = Y0 ⊕ Y1 where Y0 and Y1 are
nonempty. Since both Y0 and Y1 are closed in Y , there exist continuous selection f0 : F (Y0 ) →
Y0 and f1 : F (Y1 ) → Y1 . Now for every i ∈ {0, 1} we can define a continuous selection
gi : F (Y ) → Y as follows: gi (F ) = fj (F ) if j ∈ {0, 1} and F ⊆ Yj , and gi (F ) = fi (F ∩ Yi) if
F ∩ Yj 6= ∅ for both j ∈ {0, 1}. Since g0 (Y ) ∈ Y0 and g1 (Y ) ∈ Y1 , these selections are different.
Proof of Theorem 5: If (X, <) is a connected, locally connected space satisfying items (i) and
(ii) of Theorem 5, then X has exactly one continuous selection by Lemmas 11 and 12, which
implies the “if” part of our theorem. To prove the “only if” assume that X is a locally connected
space which has exactly one continuous selection. Lemma 14 imples that X is connected. The
existence of linear order on X satisfying items (i) and (ii) of our theorem follows now from
Lemmas 10, 11 and 12.
Proof of Theorem 6: The “only if” part is obvious in view of Lemma 12, so we need only
to prove the “if” part. The long line L is not separable, and [0, 1] has two selections (Corollary
2), so the “if” part of item (i) follows from Theorem 3.
(ii) Assume that X is a space satisfying item (ii) of our theorem. Then the topology of X is
generated by the linear order < which satisfies the conclusion of Theorem 5. Now observe that
an infinite connected separable ordered space (X, <) satisfying the conclusion of Theorem 5 is
homeomorphic to [0, 1).
(iii) Let X be an infinite, metric, locally connected and has exactly one continuous selection.
First note that X must be connected by Lemma 14. Also X is complete by theorem of van Mill,
Pelant and Pol [9]. Now we conclude that X is pathwise connected by the classical theorem
of Menger [5] and Moore [7] which says that a connected, locally connected, complete metric
space is pathwise connected. Since L is not metric, and [0, 1] has two continuous selections
(Corollary 2), from Theorem 3 it follows that X is homeomorphic to [0, 1).
Lemma 15 The space X from Example 8 has exactly one continuous selection.
7
Proof: For notational convenience only we will assume that X = {(0, 0)}∪{(x, sin(1/x) : −1 ≤
x < 0}. Let φ : [−1, 0] → X be the 1-1 map defined by φ(x) = (x, sin(1/x)) for x ∈ [−1, 0) and
φ(0) = (0, 0). Then φ “copies” natural linear order on [−1, 0] to X, and we will use symbol
<′ for denoting this copy. Now observe that F 7→ min F is a continuous selection on X, so
the order <′ coincides with the order < from Lemma 10 (see the proof of Lemma 10 for the
definition of the order <). Since X is connected, applying this lemma once again we conclude
that there exist at most two different selections on X, and the only other possible candidate
for the second one is the map F 7→ max F . So to finish the proof of our lemma, it suffices to
notice that
!
)
(
(4k + 1)π
2
:k∈N ⊆X
, sin
F =
(4k + 1)π
2
is a closed subset of X such that max F does not exist.
Remark 16 Let F2 (X) be the subspace of F (X) consisting of subsets with no more than 2
points. Let us call a (continuous) 2-selection any (continuous) map f : F2 (X) → X such that
f (F ) ∈ F for all F ∈ F2 (X). Clearly a restriction of a continuous selection to F2 (X) is a
continuous 2-selection, but (0, 1) is an example of the space which has continuous 2-selections
(both F 7→ min F and F 7→ max F will do), but does not have a continuous selection (Corollary
13(ii)). We note that Theorem 4 remains valid under a weaker assumption of the existence of
a continuous 2-selection. Indeed, the proof of Theorem 4 uses only Lemma 11, which in its
turn uses Lemma 10(i) in its proof, and it remains only to note that Lemma 10(i) remains
valid under the weaker assumption of the existence of a continuous 2-selection, see [6]. All
other results in our paper use the full strength of the existence of continuous selection and fail
under the weaker assumption of the existence of continuous 2-selection, as (0, 1) and Example
8 demonstrate. We leave this to the reader as an exercise.
References
[1] S. Eilenberg, Ordered topological spaces, Amer. J. Math. 63 (1941), 39-45.
[2] R. Engelking, R.W. Heath and E. Michael, Topological well-ordering and continuous selections, Inventationes Math. 6 (1968), 150-158.
[3] A. Haar, and D. König, Uber einfach geordnete Mengen, J. für die reine und angew. Math.
139 (1910), 16-28.
[4] K. Kuratowski, S.B. Nadler, Jr., and G.S. Young, Continuous selections on locally compact
separable metric spaces, Bull. Acad. Polon. Sci. sér. math. astron. et phys. 18, no. 1 (1970)
5-11.
[5] K. Menger, Über die Dimension von Punktmengen III. Zur Begründung einer axiomatischen
Theorie der Dimension, Monatsch. für Math. und Phys. 36 (1929), 193-218.
[6] E. Michael, Topologies on spaces of subsets, Trans. Amer. Math. Soc. 71 (1951), 152-182.
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[7] R.L. Moore, Foundations of point set theory, New York, 1932.
[8] T. Nogura and D. Shakhmatov, Spaces which have finitely many continuous selections,
preprint.
[9] J. van Mill, J. Pelant and R. Pol, Selections that characterize topological completeness,
Preprint no. WS-428 (November 1994), Faculteit der Wiskunde en Informatika, Vrije Universiteit, Amsterdam.
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