On Successive Approximations for Compact-Valued Nonexpansive Mappings

We show that for a given initial point the typical, in the sense of Baire category, nonexpansive compact valued mapping F has the following properties: there is a unique sequence of successive approximations and this sequence converges to a fixed point of F. In the case of separable Banach spaces we show that for the typical mapping there is a residual set of initial points that have a unique trajectory.


Introduction
Fixed point theorems play an important role in various branches of mathematics.While Banach's fixed point theorem is valid for every (strictly) contractive self-mapping of a complete metric space, unfortunately there is no general fixed point theorem for nonexpansive mappings.Let C be a closed, bounded and convex subset of a Banach space X .If X is finite dimensional, Brouwer's fixed point theorem guarantees a fixed point for every continuous self-mapping of C. Since Benyamini and Sternfeld in [4] showed that in infinite dimensional spaces the sphere is contractible, an analogous statement as Brouwer's cannot hold in infinite dimensional Banach spaces, even for Lipschitz-continuous mappings.For uniformly convex X and for every nonexpansive self-mapping of C, i.e. for every mapping f : C → C with f (x) − f (y) ≤ x − y the Browder-Göhde-Kirk fixed point theorem implies the existence of a, possibly non-unique, fixed point.Herein lies the motivation for looking at nonexpansive mappings for fixed point theory.But also in these cases de Blasi and Myjak in [8,9] showed that the typical, in the sense of Baire category, nonexpansive mapping has a unique fixed point which can be approximated by iterating the mapping f .Since by [2] the typical nonexpansive mapping is not a strict contraction, this behaviour cannot be explained by Banach's fixed point theorem.S. Reich and A. Zaslavski showed in [20,21,23] that, even in more general situations, the typical nonexpansive self mapping is contractive in the sense of Rakotch, i.e. it satisfies f (x) − f (y) ≤ φ( x − y ) x − y for some decreasing mapping φ : [0, diam(C)] → [0, 1] with φ(t) < 1 for t > 0.
The mentioned fixed point theorems for single-valued mappings have one thing in common: The fixed point can be attained by iterating the mapping.This tends to be more difficult in the setting of a set-valued mapping, as it is a priori unclear what iterating such a mapping means.One way to achieve this is to take the union of the point images, which leads to lifting a compact-valued mapping to a mapping on the hyperspace of compact sets in C, i.e. lifting which was done by Reich and Zaslavski in [19,22,24].In the mentioned papers the authors have also shown that the typical compact-valued mapping is a contraction in the sense of Rakotch and has a fixed point, i.e. in the example above: there is a set A ∈ K(C) such that A = F(A).Yet there are different approaches, for example the approach of successive approximations: starting with a point x 0 we take a sequence satisfying x n+1 ∈ F(x n ), where we let x n+1 be a minimizer of { x n − y : y ∈ F(x n )}.This was studied already in [3,17], but these papers about successive approximations of set-valued mappings focus on special mappings of the form F = { f , g} with two nonexpansive single valued mappings f , g : C → C. We show in this work that for a given initial point x 0 ∈ C the set of mappings F : C → K(C) with unique trajectory starting in x 0 is residual, improving results on successive approximations of compact-valued mappings.Moreover this sequence of successive approximations will converge to a fixed point of F, this can be seen in [19].
Even though our main interest lies on mappings with compact images, a lot of our results will apply to mappings with closed and bounded point images.Set-valued mappings are of interest in various areas, for example Lipschitzian set-valued mappings are being studied in [5].These set-valued mappings occur in a natural way, for example when computing the subdifferential of a convex mapping at a point.Recently unions of nonexpansive mappings have been considered by M. K. Tam and others; see, for example, [6,26].Also recently, global convergence and acceleration of fixed point iterations for union upper semicontinuous operators is being investigated by J. H. Alcantara and C. Lee; see [1].
The properties of the typical compact subset of certain spaces have been extensively studied.In particular the question of the size of the sets of nearest and farthest points is well understood, see e.g.[7,10,11].J. Myjak and R. Rudnicki, in [16], have shown that the typical compact set is hispid, i.e. the projection is non-unique at many points.Regarding the question how big the set of points is that have a single nearest point in an arbitrary subset of a strictly convex normed space, see [25].In Section 3 we provide an answer to the converse question in Theorem 3.3: Given an arbitrary point, how big is the set of closed and bounded sets that have a unique metric projection for the chosen point?

Preliminaries
Throughout this work let X be a Banach space and C ⊂ X be a non-empty, closed, convex and bounded subset of X .In particular C does not have to be compact.When we talk about C as a metric space we use the metric d(x, y) = x − y for x, y ∈ C. For x ∈ C and r > 0 let B d (x, r ) denote the open ball (and B d (x, r ) the closed ball) with radius r and centre x w.r.t. the metric d.If it is clear which metric is used, we will omit the subscript and use B(x, r ) instead of B d (x, r ).
For x, y ∈ C we set and since we assume that x − y , the distance of x to the set M and the diameter of M respectively.

Metric Hyperspace of Sets
We consider the space of non-empty compact sets and equip it with the Pompeiu-Hausdorff distance H , defined by It is well known that the metric space (K(C), H ) is complete.
For two sets A, B ⊂ X we set Proof For all a ∈ A and b ∈ B we have by the triangle inequality Therefore we obtain inf Since this holds for every b ∈ B, we get 123 Nonexpansive and Contractive Mappings Let M denote the set of all nonexpansive mappings T : C → K(C).In this sense nonexpansive means that Moreover we call mappings in M strict contractions, if there exists L ∈ [0, 1) such that We equip M with the metric of uniform convergence, i.e.
which makes (M, ρ) a complete metric space.A mapping T ∈ M is called contractive in the sense of Rakotch, see [18], if there exists some decreasing function for all x, y ∈ C, i.e. φ only depends on the distance between x and y.

Meagre and residual
We call a subset E of a topological space X meagre, if it is a countable union of nowhere dense sets.A residual set is the complement of a meagre set.
Porosity We call a subset E of a metric space X porous at a point x ∈ E if there exists r 0 > 0 and α > 0 such that for all r ∈ (0, r 0 ) there exists a point y ∈ B(x, r ) such that B(y, αr ) ∩ E = ∅.
We call E porous if it is porous at all of its points and σ -porous if it is a countable union of porous sets.Whether r 0 and α are chosen independent of x or not does not matter for the σ -porosity of E, see [2, p. 93].For the sake of completeness we repeat here the proof displayed there.Take a σ -porous set E = ∞ i=1 E i , where E i are porous for all i.For all j, k ∈ N set and each E j,k i is porous with r 0 and α independent of x.

Metric Projections
For a set M ⊂ X and x ∈ X we call the set the metric projection of x onto M. Naturally this set can be empty, even for closed sets M, or contain more than one point.Since continuous mappings attain their minimum on compact sets, compact sets always have a non-empty projection.

Successive Approximations
An infinite sequence of successive approximations or trajectory {x n } ∞ n=0 is called regular, if Strict Contractions are Dense in M The proof of the following Lemma is the same as in the case of single-valued mappings, which can be found for example in [8].We include it for the sake of self-containedness.

Lemma 2.2 The set of strict contractions is dense in M.
Proof Let F ∈ M, x 0 ∈ C, λ ∈ (0, 1) and ε > 0. We will show that then the mapping is a strict contraction satisfying ρ(F λ,x 0 , F) < ε, i.e. for an arbitrary mapping F ∈ M we can find a strict contraction arbitrarily close to F. By [2, p.119,120] the mapping F is well defined, i.e. it is a compact set for every x ∈ C.
For x, y ∈ C and x ∈ F(x) we obtain The roles of x and y respectively are reversible in above calculation.Therefore we obtain Now choose λ < ε/diam(C) and we obtain that which concludes the proof.

How many sets have a unique projection onto them for a fixed point?
As mentioned in the introduction we show in some sense the converse direction to Theorem 1 in [25].For a fixed x ∈ X we take the complement of sets M in K(X ) which have a unique metric projection P M (x) and show that it is σ -porous.The following example shows that being close to a set with unique projection does not imply that the projection is unique.
Example 3.1 Let x ∈ X and r , ε > 0 be arbitrary and set One can take various variations of the sets presented above, which leads to the question how many sets there actually are with a unique projection onto them.In the following Lemma we use a base idea similar to the idea in Lemma 1 in [25].
Theorem 3.3 Let X be a Banach space and C ⊂ X a closed, bounded and convex set.Fix an arbitrary point x ∈ C. Then there is a set and For every 0 < r < r 0 choose a z as in the statement of Lemma 3.2 with dist(z, M) = r .Then z and N as in (3.1) satisfy the conditions from Lemma 3.2.In particular it follows directly that P N (x) = {z}, since x − z < x − y for all y ∈ N by construction of z, and therefore Summing up for any M ∈ A n (x) there is a set N ∈ K(C), which is r -close to M and fulfils B H (N , αr ) ∩ A n (x) = ∅ with α as set above, hence we have that A n (x) is porous.

Successive Approximations for Set-Valued Mappings
Throughout this chapter let F ∈ M be a strict contraction and {x F n } n∈N ⊂ C be a trajectory w.r.t.F. We will omit the F in the superscript, if it is clear which mapping is meant.
For technical reasons we always assume x 0 / ∈ F(x 0 ).The aim is to prove the following claims.

Claim 1
For n ∈ N, r > 0 and a strict contraction F ∈ M, there is a mapping G n ∈ M such that the trajectory starting from x 0 w.r.t.G n is unique in the first n steps and ρ(G n , F) < r .
We will achieve this by following an inductive procedure that we summarize as the following claim.
Claim 2 Let F, G 0 , . . ., G m−1 be already constructed and x 0 , . . ., x m ∈ C with and σ m > 0 small enough, where the upper bound depends on the previous steps.There is a strict contraction G m and x m+1 with ρ(G m−1 , G m ) < σ m and

Some Auxiliary Lemmas
We will state a few auxiliary lemmas before working on the claims.
The above Lemma is well known and is a particular case of the much more general Lemma 2 in [15].
We now recall bounds on the Lipschitz constant of retractions onto balls in Banach spaces by C. F. Dunkl and K. S. Williams, [13].Similar bounds have been obtained by D. G. de Figueiredo and L. A. Karlovitz in [12].Even though the retraction on the closed unit ball does not need to be nonexpansive, the following mapping R ε,x 0 is.Lemma 4.2 Let X be a Banach space, x 0 ∈ X and ε > 0. Denote by B ε the open ball around 0 with radius ε and by B ε the closed ball around 0 with radius ε.The mapping Proof The mapping r ε is a 2-Lipschitz retraction by Dunkl-Williams [13].Therefore R ε,x 0 is a continuous mapping, since r ε is Lipschitz and so in particular continuous.We show that R ε,x 0 is nonexpansive by applying Lemma 4.1 to B(x 0 , ε) and X \ B(x 0 , ε). 123 The case where x, y ∈ B(x 0 , ε) is obvious, since here r ε is the identity and R ε,x 0 constant with value x 0 .Let x, y ∈ X be any two points not in B(x 0 , ε).
Remark 4. 3 The above lemma can be interpreted in the following way: The mapping R ε,x 0 adds a small vector to x in direction of x 0 , i.e.R ε,x 0 (x) = x + ξ with ξ = −r ε (x − x 0 ) and ξ ≤ ε.In particular for x ∈ B ε (x 0 ) the mapping is constant with value x 0 .
Remark 4.4 A somewhat similar way of perturbing was independently developed by M. Dymond, in [14].
The following Lemma is well-known but for the convenience of the reader we include it with a proof.

Lemma 4.5 Let
Therefore the last inequality stated in the Lemma is true and the first one follows as a special case.

Mappings with Isolated Fixed Points and Without Fixed Points
In this part of the section we show that for a fixed x 0 ∈ C the typical mapping in M has no fixed point in a ball around x 0 .Moreover we show first that every mapping with a fixed point has a mapping close to it with the same fixed point being an isolated point. 123 Then the mapping L x 0 ,z,δ : Proof Note that (F(x 0 ) ∩ B(z, δ/2)) and {z} are compact sets, and therefore their sum is compact too.Hence L x 0 ,z,δ is well defined.In [2, p.119, 120] the authors show for t, s ∈ [0, 1] and A ∈ K(C) with A = {z} that holds.Observe that μ x 0 ,δ is a continuous function that is constant on C ∩ B(x 0 , δ/4) and has Lipschitz constant 4 3δ on C ∩ B(x 0 , δ) \ B(x 0 , δ/4) and therefore is Lipschitz with Lip(μ x 0 ,δ ) = 4 3δ by Lemma 4.1.In (4.1) plugging in x, y ∈ B(x 0 , δ), It is easy to see that μ x 0 ,δ (x) = 1 for x with x − x 0 = δ, and therefore L x 0 ,z,δ (x) = F(x 0 ) ∩ B(z, δ/2).So the mapping L x 0 ,z,δ is continuous, constant on C \ B(x 0 , δ) and C ∩ B(x 0 , δ/4), has Lipschitz constant 2/3 on C ∩ B(x 0 , δ) \ B(x 0 , δ/4).Therefore by applying Lemma 4.1 twice the proof is concluded.
Lemma 4.7 Let F ∈ M and x 0 ∈ C be such that x 0 ∈ F(x 0 ).Then for every ε ∈ (0, diam(C)) and for 0 < r ≤ ε/24 there is a mapping where L x 0 ,x 0 ,δ is as in Lemma 4.6.We use that We will show that Lip(G) ≤ max{2/3, Lip(F)}.Let first x, y ∈ B(x 0 , ε ) ∩ C. Then we have that ) where we applied Lemma 4.5 in the penultimate inequality and Lemma 4.6 in the last inequality.Further for x, y ∈ C \ B(x, ε ) we have since Lip(R ε ,x 0 ) = 1 by Lemma 4.2.Note that for x with x −x 0 = ε we have L x 0 ,x 0 ,δ (x) = F(x 0 ) ∩ B(x 0 , δ/2) and hence G is continuous and by the above observations together with Lemma 4.1 we have Lip(G) ≤ max{2/3, Lip(F)}.Now set r = δ/6 and observe that for all v ∈ B(x 0 , r ) we have where we used the triangle inequality, split up the set F(x 0 ) in a union of two sets, in order to apply Lemma 4.5 and then used that L x 0 ,x 0 ,δ (x) ⊂ B(x 0 , δ/2), therefore H (L x 0 ,x 0 ,δ (x), B(x 0 , δ/2)) ≤ δ, in the penultimate inequality.Proof First we investigate the case where x 0 / ∈ F(x 0 ).Since F(x 0 ) is a compact set we have a positive distance s = dist(x 0 , F(x 0 )) > 0. Observe now that for all v ∈ B(x 0 , s/3) we have and hence the inequality dist(v, F(v)) ≥ s/3 holds.So one can take F = G and r = s/3.

123
Now assume x 0 ∈ F(x 0 ).Set ε = ε/2 and δ = ε/4.We use the mapping R ε ,x 0 (x) = x − r ε (x − x 0 ) from Lemma 4.2.Choose an arbitrary z ∈ C with z − x 0 = δ/4.Define the mapping and therefore this mapping is continuous.By construction we have that dist(x 0 , G(x 0 )) ≥ δ/4 and therefore we can choose r := δ/12 to guarantee that no v ∈ B(x 0 , r ) is a fixed point of G by the same argument as in (4.6).We still have to show that Lip(G) ≤ 1 and ρ(F, G) < ε holds.The first of the claims follows by the same reasoning as in (4.2) and (4.3).For the latter one we reason as in (4.4) and (4.5) again to obtain ρ(F, G) < ε.
Corollary 4.9 Let x 0 ∈ C and define the set of mappings in M that do not have x 0 as fixed point as Then the set M \ A 0 (x 0 ) is nowhere dense.
Proof In view of Proposition 4.8 we are left to show that there is a ball around F ∈ A 0 (x 0 ) fully contained in A 0 (x 0 ).We set s = dist(x 0 , F(x 0 )) and observe that every G ∈ B ρ (F, s/3) has to satisfy H (G(x 0 ), F(x 0 )) < s/3.Therefore x 0 / ∈ G(x 0 ).

The Inductive Construction
Given a strict contraction we will construct a new strict contraction, such that the new mapping has a unique projection from a given point and is still close to the mapping we started from.
In the following we fix n ∈ N and 0 < r < 1.
Then there exists a mapping G ∈ M and a point z ∈ C \ G(z) that satisfy: We will prove that proposition in smaller parts, but want to show the bigger picture first.Inserting F for the mapping G and x 0 for z in Proposition 4.11 will yield the mapping G 0 and x 1 , so we get and a sequence of points x 0 , . . ., x n+1 .Using suitable conditions on the constants, this sequence satisfies the conditions for the inductive construction in Claim 2.

Proof of Proposition 4.11
We divide the proof into a number of lemmas.Let G and z be as in Proposition 4.11.Let y ∈ P G(z) (z) be arbitrary and set Moreover let λ z,σ be as in Lemma 4.10 and We define and note that for all x / ∈ B(z, σ/2) the mapping is constant with g(x) = y.Note that the Lipschitz constant of g is bounded from above by 2 ε σ .To see this, let x 1 , x 2 ∈ C and note that (4.9) Further recall R σ,z from Lemma 4.2 and define G : 123 Lemma 4. 12 The mapping G satisfies Lip( G) ≤ max{Lip(G), 2ε/σ } ≤ 1.In particular since G is a strict contraction, so is G.
Proof First observe that for all x ∈ C with x − z = σ we have since g(x) ⊂ G(z) for all x with distance σ to z. Therefore the mapping G is continuous.Using (4.9), Lemma 4.2, Lemma 4.5 and Lemma 4.10 for where we used that R σ,z (x 1 ) = z = R σ,z (x 2 ).In the case of Now Lemma 4.1 and the choice of ε yield where equality would only be attained when Lip(G) = 1 and therefore this yields the claimed result.
The next lemma shows (v).
Taking x ∈ C ∩ B(z, σ ) and using Lemma 4.5 together with the triangle inequality, we obtain that This yields The way G is constructed yields a very useful property.Not only the projection from z onto G(z) is unique, but every point close to z is projected on z, i.e. (vi) holds: Lemma 4.14 For every x ∈ C with x − z < ε /3 the identity P G(z) ( x) = {z} holds.
Proof Let x ∈ C ∩ B(z, ε /3) and assume there would be a ỹ ∈ G(z) \ {z} such that This concludes the proof of Proposition 4.11.

Proof of Claim 1 and Claim 2
For the purpose of proving Claim 1 stated above, we will need further restrictions on σ and set and δ 0 , ε 0 , ε 0 as in (4.7).Let F, G 0 , x 0 and x 1 be G, G, z and z as in (4.10) or construct G 0 as in Lemma 4.7 if as in Claim 2. We are presented with two possibilities.Either x m / ∈ G m−1 (x m ) or x m is a fixed point of G m−1 and is the first in the sequence with this property.In the first case we set In particular the σ i satisfy 0 < σ i < r /(n + 1).In the case of δ m = 0, that is when ) becomes an isolated point.Then we set all the following G m+ j = G m and x m+ j = x m for j ≥ 1.Further we continue the sequences σ m , ε m , ε m by setting ) 3 and (4.17) The number σ m as in (4.14)only vanishes when δ m vanishes, this is a consequence of the following Lemma.
The first inequality is Lemma 2.1.Applying this inductively yields the desired inequality.Now, assume w.l.o.g.i > j, we observe that and equality only holds when ζ j = ζ j+1 , i.e. when ζ j is already a fixed point.If none of the ζ i is a fixed point we have and (4.10) we construct the mapping G = G m and z = x m+1 .
In the case that x m ∈ G m−1 (x m ) we distinguish two cases: If m is the first index that we reach a fixed point, we use Lemma 4.7 to construct G m such that x m+1 = x m is an isolated point in G m (x m ).If m is not the first index that we reach a fixed point, we set G m = G m−1 .Further set σ m , ε m , ε m as in (4.16), (4.17) and (4.18).We show now that G m for m ≤ n is r -close to F and so one part of Claim 1 holds true for m = n and the choice of σ i as in (4.14) for 0 ≤ i ≤ m.Lemma 4. 16 The mappings G 0 , . . ., G m satisfy ρ(G i , F) < r for 0 ≤ i ≤ m ≤ n.
It is helpful at this point to rephrase Proposition 4.11 (vi) Lemma 4.17 Let 0 ≤ i < m and x ∈ B(x i , ε i /3).Then Further, for j < i, a consequence of above statement is that for x ∈ B(x j , ε i /3) With this preparation we can prove the following proposition, which is the major tool to prove Claim 1 and Claim 2. The proposition and the lemmas before show how to construct the mappings G 0 , . . ., G m , but the following proposition shows that we can indeed continue inductively.
Proposition 4. 18 Let the mappings F, G 0 , . . ., G m and the trajectory x 0 , . . ., x m , x m+1 be constructed as before.Then the following holds:

and for all
x ∈ B(x k , ε j /3).
Assume now that none of the x k for 0 ≤ k ≤ m are a fixed point of G m .Recall that G j and σ j are defined as in (4.10) and (4.14).In particular, because we cannot be in the second case of (4.10).So we always have Also we get therefore by (4.10).Assume now that from some 0 ≤ i < m the sequence becomes constant, i.e.
This yields that (4.21) holds also for x instead of x k .Thus rewriting G j ( x) as in (4.20) and using (4.21) we obtain for x ∈ B(x k , ε j /3).To conclude (iv) we use (4.22) and insert it in Lemma 4.17.
Thus the construction of G 0 , . . ., G n for m = n yields that the first n steps of the trajectory starting at x 0 w.r.t.G n are x 0 , . . ., x n , which shows the final piece of Claim 1. Furthermore Proposition 4.18(iii) yields the last missing part of Claim 2.

Mappings with Non-Regular Trajectories Form a Meagre Subset
Consider, for x 0 ∈ C, the sets where β n n→∞ − −−→ 0 is a decreasing null sequence.We omit F in the superscript, if it is clear to which mapping the trajectory refers.Observe also that k=0 is a regular trajectory w.r.t.F}.
We will now show that the sets M \ A n (x 0 ) are nowhere dense.
Thus this R is suited for the one in the definition of the sets A n and we will show now that G n is in A n (x 0 ).Lemma 4.12 guarantees that G n ∈ M and Lemma 4.17 implies that the trajectory w.r.t.G n and starting with v 0 is unique in the first n steps, i.e.G n ∈ A n (x 0 ). Set First we will show that each point in a trajectory {v 0 , z i } n i=1 starting at v 0 w.r.t.G in the first n steps will be ε n /6 close to points in the unique trajectory {v 0 , we can apply Lemma 3.2 and obtain that the projection of v 0 onto G (v 0 ) has to be contained in a ball around x 1 , i.e.
Thus we have that Take an arbitrary point z 1 ∈ P G (v 0 ) (v 0 ), hence ) creates an isolated point, in order to gain control over the trajectory.Therefore this step is done in any case.
In the construction of the mappings G i we need F to be a strict contraction, so the G i in the construction maintain a Lipschitz constant less than one.This is in particular needed for the trajectories to not repeat unless they are at a fixed point already, see Lemma 4.15.
As a direct consequence of Proposition 5.1 we obtain the following theorem.Theorem 5.5 Let X be a separable Banach space.Then there is a residual set F ⊂ M such that for every F ∈ F there is a residual subset U ⊂ C with the following property: For every x 0 ∈ U the mapping F admits a regular trajectory starting at x 0 .
Proof Since X is separable we can find a countable dense subset D ⊂ C. We use for n ∈ N the sets A n (y i ) from Theorem 5.3, set 123 Then the sets U n are dense in C, as D is dense itself.These sets consist of all points such that the trajectory starting at one of those points behaves well in the first n steps.Further this yields that the set is a residual subset of C. By the observations above for every x 0 ∈ U the mapping F has a regular trajectory starting at x 0 .

Proposition 4 . 8
Let x 0 ∈ C, F ∈ M and diam(C) > ε > 0.Then there are a mapping G ∈ M and a radius r > 0 such that ρ(F, G) < ε and for all v ∈ B(x 0 , r ) we have v / ∈ G(v).

Lemma 4 .
13 G and G satisfy ρ(G, G) < σ.Proof In the case of x ∈ C \ B(z, σ ) we may use Lemma 4.2 to obtain

Lemma 4 . 15
For a strict contraction G, ζ 0 ∈ C, n ∈ N and the trajectory {ζ i } ∞ i=0 starting at ζ 0 w.r.t.G we have that all ζ i are different as long as none of them is a fixed point, i.e. ζ i = ζ j if and only

Theorem 5 . 3 Corollary 5 . 4
The sets A n (x 0 ) contain an open and dense subset of M.Moreover, the setA(x 0 ) = n∈N A n (x 0 )is residual.For every x 0 ∈ C, there is a residual subset F ⊂ M such that all F ∈ F admit a regular trajectory starting with x 0 , which converges to a fixed point of F. Proof For the convergence to a fixed point see Theorems 4.2, 4.3 and 4.4 in[19].
A n and F are residual subsets of M. Fix F ∈ F and let n ∈ N. Then F ∈ A n and hence also F ∈ A n (y) for all y ∈ D. In particular this means that there exists an R n,y , see the definition of the sets A n , such that diamP F(v k ) (v k ) ≤ β n for all v 0 , . . ., v n with v 0 ∈ B(y, R n,y ) and v k+1 ∈ P F(v k ) (v k ).Define for n ≥ 1 the sets U n := y∈D B y, R n,y = x ∈ C | x − y < R n,y for a y ∈ D .
ProofIn the case that none of the x i are fixed points of G m the first part is a direct application of Proposition 4.11(vi) to G i and x i .The second part follows since ε i < ε j , i.e.B(x j , ε i /3) ⊂ B(x j , ε j /3) 123 and again Proposition 4.11(vi).In the case that one x i is a fixed point of G m , and therefore every following x i+1 , x i+2 , ... point too, by construction x i is an isolated point of G m (x i ) by Lemma 4.7.Therefore, since ε m ≤ σ m /2, Lemma 4.7 implies ε m defined in (4.18) is small enough to satisfy above claims.