Monovex Sets^{†}^{†}thanks: The first author also uses the spelling “Buhovski” for his family name. The work of L. Buhovsky was partially supported by the ISF grant 1380/13, by the Alon Fellowship, and by the Raymond and Beverly Sackler Career Development Chair. The work of E. Solan was partially supported by ISF grant 323/13.
Abstract
A set in a finite dimensional Euclidean space is monovex if for every two points there is a continuous path within the set that connects and and is monotone (nonincreasing or nondecreasing) in each coordinate. We prove that every open monovex set as well as every closed monovex set is contractible, and provide an example of a nonopen and nonclosed monovex set that is not contractible. Our proofs reveal additional properties of monovex sets.
Keywords: Monovex sets, contractible sets.
1 Introduction
A set in a finite dimensional Euclidean space is monovex if for every two points there is a continuous path within the set that connects and and is monotone (nonincreasing or nondecreasing) in each coordinate. In particular, whether or not a set is monovex depends on the choice of basis for the space.
Monovex sets arise in the study of stochastic games (Solan, 2016), where an extension of the Kakutani’s fixedpoint theorem (Kakutani, 1941) for setvalued functions with a closed graph and nonempty monovex values is needed.^{1}^{1}1Kakutani’s fixedpoint theorem states that any setvalued function from a convex and compact subset of to itself with a closed graph and nonempty convex values has a fixed point. By EilenbergMontgomery fixedpoint theorem (Eilenberg and Montgomery, 1946) any setvalued functions from a convex compact subset of to itself with a closed graph and nonempty contractible values has a fixed point. Consequently, our goal is to study contractibility of monovex sets.
In this paper we prove that every open monovex set, as well as every closed monovex set, is contractible. We also provide an example of a nonopen and nonclosed monovex set that is not contractible.
2 Definition and Main Results
The concept that this paper studies is monovex subsets of a finite dimensional Euclidean space.
Definition 2.1
A set is monovex if for every there is a continuous path that satisfies the following properties:

and .

is a monotone function (nondecreasing or nonincreasing) for every .
A path that satisfies Condition (M2) is called monotone.
The image of a monovex set under a diagonal affine transformation is monovex, yet a rotation of a monovex set need not be monovex. Every convex set is in particular monovex. If is a monovex set, then so is the projection of onto any “coordinate subspace”, that is, a subspace spanned by a collection of elements of the standard basis of . Every monovex subset of is convex, yet there are monovex subsets of that are not convex (see Figure 1).
Figure 1: A monovex set (Part A) and a nonmonovex set (Part B) in the plane.
As the following example shows, monovex sets may be complex objects. In particular, they need not be CWcomplexes.
Example 2.2
Let be the following set (see Figure 2):
It is evident that the set is monovex, yet it is not a CWcomplex.
Figure 2: The monovex set in Example 2.2.
The Minkowski sum of two convex sets is a convex set. This property is not shared by monovex sets. In fact, as the following example shows, the Minkowski sum of a monovex set and a convex set need not be a monovex set. In Lemma 3.4 below we will prove that the Minkowski sum of a monovex set in and an dimensional box whose faces are parallel to the axes is a monovex set.
Example 2.3
Let be the union of the two line segments and , which is monovex. Let be the line segment . The intersection of the set and the line is the two points and . Indeed, all points satisfy , while the only points that satisfy are and . Hence a point is on the line if and only if or .
Since the intersection of and the line contains two points, there is no monotone path that connects these points and lies in , and therefore the set is not monovex.
The Minkowski sum of the sets in Example 2.3 is contractible. As the following example shows, the Minkowski sum of a monovex set and a convex set can be homotopy equivalent to the circle .
Example 2.4
Let be the union of the three line segments , , and , which is monovex. Let , which is convex. Denote by the triangle in whose vertices are , , and . The Minkowski sum of and is , which is homotopy equivalent to the circle .
As mentioned in the introduction, our goal is to study whether monovexity implies contractibility. It is a little technical but not difficult to show that every monovex subset of is contractible. As the following example shows, not every threedimensional monovex set is contractible.
Example 2.5
Let be the set of all points that have at least one negative coordinate and at least one nonnegative coordinate. The reader can verify that the set is monovex. The set is disjoint of the line , and it contains the loop that is depicted in Figure 3 and is not contractible in . In particular, the set is not contractible. In fact, one can show that the set is homotopy equivalent to the circle .
Figure 3: The path in Example 2.5 (the dark curve).
We next observe that every open monovex set is contractible.
Theorem 2.6
Every open monovex subset of is contractible.
Proof. The proof is by induction on . For , an open monovex set is an open interval, hence contractible. Assume now that is an open monovex subset of with . Let be the projection of onto its first coordinates, and let be the setvalued function whose graph is ; that is
Note that is an open interval for every . The set is open and monovex, and by the induction hypothesis it is contractible. The setvalued function satisfies the conditions of Michael’s selection theorem (Michael, 1956, Theorem 3.1”’),^{2}^{2}2Michael’s selection theorem implies in particular that for every subset and every setvalued function with an open graph and nonempty convex values there exists a continuous function such that for every . hence there is a continuous function such that for every . This implies that is contractible; indeed first contract to , and then contract to a point.
The main result of the paper is the following.
Theorem 2.7
Every closed monovex subset of is contractible.
We provide two proofs to Theorem 2.7, each one uses different properties of monovex sets, which may have their own interest. The first proof, provided in Section 3.1, relies on the property that one can assign, in a continuous way, to every pair of points in a monovex set a (not necessarily monotone) path that connects these points and lies in the set. The second proof, provided in Section 3.2, relies on the stronger property that the complement of a monovex set can be continuously projected onto the set. In the first proof we will provide a direct argument that shows the existence of a continuous map from pairs of points in the monovex set to paths that connect the points and lie in the set.
Several open problems regarding contractibility of monovex sets still remain. We prove that every closed monovex set is contractible. We do not know whether for every such set there is a Lipschitz continuous contraction. Another issue that remains open is whether our results hold for infinite dimensional spaces.
3 Proof of Theorem 2.7
Throughout the paper we use the maximum metric in , that is, for every . The distance between a point and a set is , and the distance between two sets is the Hausdorff distance .
For every and every we denote by the open ball around with radius , and by the closed ball around with radius . We denote by the vector in .
A (closed) box in is a set of the form , where for each . A box is dimensional if the number of indices such that is . The set of vertices of a box is denoted . The smallest box that contains a set is called the bhull of and denoted .
A blattice is a set of the form , where . Denote by the set of dimensional elementary boxes having vertices in the lattice, that is, the collection of all sets such that for each either or for some , and moreover the second condition happens for exactly values of . Denote the set of fulldimensional elementary boxes with vertices in the lattice .
3.1 First Proof
The following lemma states that any function from an dimensional grid to a monovex set can be extended to a continuous function from the dimensional space to with the property that the image under of any elementary dimensional box whose vertices are points in the grid is a subset of the bhull of the image under of the vertices of the box.
Proposition 3.1
Let be a closed monovex set and let be a blattice. Let be a (finite or infinite) union of boxes in , and let be the set of all vertices of these boxes. Let . Then can be extended to a continuous function that satisfies the following property:

For every , , and every box that is a subset of , the image is a subset of the bhull of .
Proof. Assume w.l.o.g. that . To prove the result, we will define the function iteratively on the sets , , and show that this definition can be extended to a continuous function over .
For every natural number and every integer , , define a function as follows. Let , let be an index in which is attained, and let be an index in which is attained. Choose a continuous monotone curve connecting and (if , the curve is constant). By continuity there exists such that . Set
We now extend the function from to , for . Suppose then that is given, and set . Every is the center of a unique dimensional box that is contained in (where ). Define
where are the vertices of . Note that this definition verifies Property (P).
The function is locally uniformly continuous,^{3}^{3}3A function is locally uniformly continuous if it is uniformly continuous on every bounded subset. and in fact, locally Hölder. Indeed, for and a box denote . Let be the maximum of the corresponding quantity over all subboxes of that belong to . If then , while if then . Since infinitely often with step as increases, and since satisfies Property (P), it follows that is indeed locally Hölder continuous.
This implies that can be extended to a continuous function . The set is dense in , hence can be extended to a continuous function from to that satisfies Property (P). The extended function is locally Hölder continuous as well.
We would like to prove that there is a continuous function that satisfies and for every . In the next lemma we prove an approximate version of this result. We will use it in Proposition 3.3 below to prove the stronger version of the claim.
Lemma 3.2
Let be a closed monovex set. For every there exists a continuous function such that for every we have:

and .

for every .
Proof. Fix . Consider the lattice , and denote by the union of all boxes that satisfy . Denote , , and .
Let be any function that satisfies the following property: for every we have and . Such a function exists since every is a vertex of a box whose sidelength is with .
By Proposition 3.1, the function can be extended to a continuous function that satisfies Property (P). In particular, for every two boxes lying in , we have:

is contained in the bhull of ;

is contained in the bhull of .
Moreover, for every , every , every , and every we have , , , and . By the triangle inequality it follows that and . We conclude that given and , for every and we have and . Therefore, since and , we have and .
In addition, since satisfies Property (P), the image is contained in the bhull of , and hence we also conclude that for every .
To summarize, for every and every we have

,

, and

.
To end the proof of the lemma, define to be the restriction of to .
Proposition 3.3
There exists a continuous function such that and for every .
We note that Proposition 3.3 implies Theorem 2.7. Indeed, choose an arbitrary . The function defined by for every and is a homotopy between and .
Proof of Proposition 3.3. Let be a sequence of positive reals such that . We define the function in steps by a Cantor set construction. Define , , and for every let be the collection of all closed intervals where and , for some , (see Figure 4).
Figure 4: The sets for .
For , in step we define on . For set
For , consider an interval and set and . If , the points and are 0 and 1, which lie in . If , one of these points is an endpoint of an interval in and the other is an endpoint of an interval in . In both cases and were already defined. Set
where satisfies the statement of Lemma 3.2. The procedure described above defines on . The set is dense on , hence is defined in a dense subset of . Since , the function is in fact locally uniformly continuous, hence it can be extended to a continuous function , as desired.
3.2 Second Proof
We first argue that if is a monovex set and is an open box whose faces are parallel to the axes, then is monovex. We note that the proof is valid also when the box is closed.
Lemma 3.4
If the set is monovex and is an open box whose faces are parallel to the axes, then the set is monovex.
Proof. Let . Then and , where and . Assume w.l.o.g. that for every , and let be a continuous monotone path that connects to . Let . There are a diagonal matrix and a vector such that and for every coordinate . Define for every coordinate a continuous function as follows:

If then .

If then is any continuous monotone function that satisfies and .
Since is monotone for every coordinate and since and , we have for every .
The path satisfies the following properties, which imply that is a continuous monotone path in from to .

and .

for every .

For every coordinate we have . Since is a diagonal matrix, the function is monotone.

For every coordinate we have , and therefore in this case is monotone as well.
Since and are arbitrary, the result follows.
We will use the following extension of Michael’s selection theorem to monovexvalued functions.
Lemma 3.5
Let and let be a setvalued function with open graph and nonempty monovex values. Then has a continuous selection: there is a continuous function that satisfies for every .
Proof. We prove the result by induction on . If then the values of are convex, hence by Michael’s selection theorem has a continuous selection .
Assume now that . Let be the projection of to its first coordinate:
Let be the setvalued function defined by
The setvalued functions and have open graphs and monovex values, hence by the induction hypothesis applied to both of them there are continuous selections of and of . The function defined by is a continuous selection of .
Definition 3.6
Let be an open set, let be a continuous function, and let be a setvalued function. The neighborhood of is the set
The following result states that every setvalued function with a relatively closed graph and compact monovex values can be approximated by a setvalued function with an open graph and monovex values.
Lemma 3.7
Let be an open set, let be a continuous function, and let be a setvalued function with a relatively closed graph and compact monovex values. There exists a setvalued function with an open graph and monovex values satisfying .
Proof.
Step 1: Definitions.
Define a function by
This function is uppersemicontinuous function: for every sequence that converges to a limit we have . Given , let be the collection of elementary dimensional boxes in the lattice . Let be the union of all boxes in that have nonempty intersection with :
The set contains , it is a uniion of closed boxes, hence closed, and it approximates : for every .
Step 2: The set is monovex for every .
Let and let . By the definition of , there are and two boxes such that and . Since is monovex, there is a continuous monotone path that connects to within . Assume w.l.o.g. that for every .
We now define a path .

If there is such that , set .

Otherwise there is such that . We let be the projection of to the line segment :
The reader can verify that is contained in . However, need not be and need not be . Indeed, for every for which Condition (B2) holds we have and . Define then two points by
(1)  
(2) 
A monotone path in that connects and is the concatenation of (a) a monotone path that connects and , (b) the path , and (c) a monotone path that connects to .
Step 3: For every there is such that and for every .
Since the function is uppersemicontinuous and its image is discrete, for every there is such that , for every . We turn to prove the analogous property for . If the property does not hold, then for every there exists such that . That is, there is . Since , the point belongs to some box of the lattice , and in particular there is a point . Since (i) has compact values, (ii) the image of is discrete, and (iii) is locally bounded from below, it follows that the number of boxes that satisfy these properties is finite, hence by taking a subsequence we can assume that (a) for every and (b) the sequence converges to some point . In particular, . Since the graph of is relatively closed, . In particular, , and hence , which implies that for every , a contradiction.
Step 4: Definition of the setvalued function .
For every define a set by
Thus, if the two points and are close, when the distance is measured by . Note that for every , and therefore has nonempty values. Define
We will prove that the setvalued function satisfies the desired conditions.
Note that is a union of open sets, and hence it is open. In addition, since , we have , hence .
Step 5: If then either (a) and , or (b) and .
Let and assume w.l.o.g. that . Since we deduce that . In particular
which implies that . By Step 3 this implies that and .
Step 6: The set is monovex for every .
Let . Then there are such that and . By Step 5 we can assume w.l.o.g. that . By Step 2 and Lemma 3.4 the set is monovex.
Step 7: The graph of is an open subset of the neighborhood of .
By the definition of ,
It follows that is a union of open sets, hence open. Moreover, is a subset of the