MATRIZ ESTRATEGICA Y PROGRAMATICA SECTOR SALUD
2.3 SECTOR AGUA POTABLE Y SANEAMIENTO BÁSICO.
LetM =S2\S{c:c∈ Dn, c∩e=∅}. If ann-cellc does not meete, then it
contains neitherunorv. Hence by (5.12) we have
S2\M ⊂(S2\Wn(u))∩(S2\Wn(v)) =S2\Wn(e), and soWn(e)
⊂M.
Conversely, let x ∈ M be arbitrary, and c be the unique n-cell c such that x ∈ int(c). Then c∩e 6= ∅ and therefore u ∈ c or v ∈ c. It follows that x ∈ Wn(u)
∪Wn(v) =Wn(e). We conclude that M
⊂Wn(e), and soM =Wn(e) as
claimed.
(ii) By Lemma 5.3 (i) an n-tile X meets e if and only if X contains u or v. Hence by (i) and Lemma 5.28 (ii) we have
Wn(e) =Wn(u)∪Wn(v) =[{X ∈Xn:X
∩e6=∅}
as desired.
For the second claim suppose thatcis ann-cell andX ann-tile withc∩e=∅, X∩e6=∅, andc⊂∂X. Thenc⊂S2
\Wn(e) andc must be ann-edge or consist
of ann-vertex. Moreover,c⊂X ⊂Wn(e). It follows thatc⊂∂Wn(e).
Conversely, let x be a point in ∂Wn(e). Then by (i) the point x is also a
boundary point ofWn(u) orWn(v), sayx
∈∂Wn(u).
By Lemma 5.28 (ii) there exist ann-cellc′ and ann-tileX withx
∈c′,u
∈X, u /∈ c′, and c′
⊂ ∂X. If x is an n-vertex, we let c = {x}. Then c is an n-cell and we havec∩e=∅, becauseWn(e) is an open neighborhood ofe andc lies in
∂Wn(e) ⊂S2
\Wn(e). Moreover,X
∩e6=∅ and c ⊂c′
⊂∂X. So c is ann-cell with the desired properties containing x.
Ifxis not a vertex we putc=c′. Again ifc∩e=c′∩e=∅, thencis ann-cell
with the desired properties containing x.
The other case, wherec∩e6=∅, leads to a contradiction. Indeed, then we have v ∈ c. Moreover, since xis not a vertex, it follows that x∈ int(c). Note that c then is necessarily ann-edge. It follows that x∈int(c)⊂Wn(v)
⊂Wn(e) which
is impossible, becausex∈∂Wn(e) ⊂S2
\Wn(e).
(iii) Ifc is an n-cell and c∩e=∅, thenc ⊂S2
\Wn(e). If c
∩e6=∅, thenc containsuorv, and soc⊂Wn(u)∪Wn(v) =Wn(e).
5.7. Joining opposite sides In this section f: S2
→S2 will again be a Thurston map, and
C ⊂ S2 be a
Jordan curve with post(f)⊂ C. In addition, we assume that # post(f) ≥3. We fix a base metric d on S2 that induces the given topology and consider the cell
decompositionsDn =Dn(f,C) as discussed in Section 5.3.
We will define a constantδ0>0 such that any connected set of diameter< δ0
(with respect to the base metricd) is contained in a single 0-flower (as introduced in Section 5.6). However, there is a slight difference between the cases # post(f) = 3 and # post(f)≥4. In order to treat these two cases simultaneously, the following definition is useful.
Definition 5.32 (Joining opposite sides). A set K⊂S2 joins opposite sides
ofC if # post(f)≥4 andK meets two disjoint 0-edges, or if # post(f) = 3 andK meets all three 0-edges.
The case # post(f) = 2 is excluded here, and so the concept of “joining opposite sides” (as well as the constantδ0below) remains undefined for such Thurston maps
f.
We will mostly use Definition 5.32 for connected sets K (when the phrase “joining” really makes sense), but it is convenient to allow arbitrary sets here.
We now define
δ0=δ0(f,C) = inf{diam(K) :K⊂S2 is a set
(5.14)
joining opposite sides ofC}. Thenδ0>0. Indeed, if # post(f) = 4, then
δ0= min{dist(e, e′) :eande′ are disjoint 0-edges}>0.
If # post(f) = 3 and we hadδ0= 0, then it would follow from a simple limiting
argument that the three 0-edges had a common point. This is absurd.
Lemma 5.33. A connected setK ⊂S2 joins opposite sides ofC if and only if
K is not contained in a single 0-flower.
Proof. IfKis contained in a 0-flowerW0(p), wherep
∈ C is a 0-vertex, then K meets at most two 0-edges, namely the ones that have the common endpointp. SoKdoes not join opposite sides of C.
Conversely, supposeKdoes not join opposite sides ofC. We have to show that K is contained in some 0-flower. Note thatK cannot meet three distinct 0-edges.
IfKdoes not meet any 0-edge, thenK does not meetC and is hence contained in the interior of one of the two 0-tiles. This implies that K is actually contained in every 0-flower.
IfK meets only one 0-edgee, thenK is contained in the 0-flowersW0(u) and W0(v), where uandv are the endpoints ofe.
If K meets two edges, then these edges share a common endpoint v ∈ V0 = post(f). This is always true if # post(f) = 3 and follows from the fact thatKdoes not join opposite sides of C if # post(f)≥ 4. Moreover, K cannot meet a third
0-edge which implies thatK⊂W0(v).
By the previous lemma every connected setK ⊂S2 satisfying diam(K)< δ 0
is contained in a 0-flower.
Lemma 5.34. Let n∈N0, andδ0>0 be as in (5.14).
(i) IfK⊂S2is a connected set withdiam(K)< δ0, then every connected set
K′⊂f−n(K)is contained in somen-flower.
(ii) Ifγ: [0,1]→S2 is a path such thatdiam(γ)< δ0, then each liftγeofγ by
fn has an image that is contained in some n-flower.
Here by definition aliftofγbyfn is any patheγ: [0,1]→S2withγ=fn◦eγ. Proof. (i) The set K is contained in some 0-flower W0(p), p
∈ V0, by Lemma 5.33 and the definition of δ0. So if K′ is a connected subset of f−n(K),
thenK′ is contained in a component off−n(W0(p)), and hence in ann-flower by
Lemma 5.29 (ii).
(ii) The reasoning is exactly the same as in (i). The (image of the) path γ is contained in some 0-flower; by Lemma 5.29 (ii) this implies that any lifteγof γby
5.7. JOINING OPPOSITE SIDES 141
We will often have to estimate how many tiles are needed to connect certain points. If we have a condition that is formulated “at the top level”, i.e., for con- necting points inC, then the mapfn can be used to translate this ton-tiles.
Lemma 5.35. Let n∈N0, and K ⊂S2 be a connected set. If there exist two disjoint n-cellsσandτ with K∩σ6=∅ andK∩τ 6=∅, thenfn(K)joins opposite sides ofC.
Proof. It suffices to show that K is not contained in any n-flower, because thenfn(K) is not contained in any 0-flower (Lemma 5.29 (iii)) and so fn(K) joins
opposite sides ofC (Lemma 5.33). We consider several cases.
Case 1: One of the cells is an n-vertex, say σ ={v}, where v ∈ Vn. Then v∈K; so the onlyn-flower thatKcould possibly be contained in isWn(v), because
no othern-flower contains then-vertexv. But sinceσandτ are disjoint, we have v /∈τ, and soτ ⊂S2\Wn(v). Hence K∩(S2\Wn(v))6=∅, and so Wn(v) does
not containK.
Case 2: Suppose one of the cells is an n-edge, say σ = e∈ En. Thene has two endpointsu, v ∈Vn. The onlyn-flowers that meet eareWn(u) and Wn(v);
so these n-flowers are the only ones that could possibly contain K. But the set Wn(e) = Wn(u)
∪Wn(v) does not containK, because K meets the set τ which
lies in the complement ofWn(e).
Case 3: One of the cells is ann-tile, sayσ∈Xn. ThenKmeets∂X. Since∂X consists ofn-edges, the set K meets ann-edge disjoint fromτ. So we are reduced
to Case 2.
Forn ∈N0 we denote by Dn(f,C) the minimal number ofn-tiles required to
form a connected set joining opposite sides ofC; more precisely,
Dn(f,C) = minN ∈N: there existX1, . . . , XN ∈Xn such that
(5.15)
K=
N [ j=1
Xj is connected and joins opposite sides ofC .
We simply write Dn forDn(f,C) iff and C are clear from the context (as in this
section).
From Lemma 5.35 we can immediately derive the following consequence. Lemma5.36. Letn, k∈N0. Every set of(n+k)-tiles whose union is connected and meets two disjointn-cells contains at least Dk elements.
Proof. SupposeKis a union of (n+k)-tiles with the stated properties. Then the images of these tiles under fn are k-tiles and fn(K) joins opposite sides ofC
by Lemma 5.35. Hence there exist at leastDk distinctk-tiles in the union forming
fn(K) and hence at leastD
k distinct (n+k)-tiles inK.
The following two lemmas give some motivation why we introduced flowers. Namely, the number of (n−1)-tiles or the number of (n+ 1)-tiles required to cover some n-tile X may not be bounded by a constant independent of X and n. Similarly, in general there will be no universal bound on the number ofn-tiles defined with respect to a different Jordan curveCeneeded to coverX. Both issues are resolved by considering flowers instead of tiles. Note that in both lemmas we allow # post(f) = 2 for our given Thurston mapf.
Lemma 5.37. There existsM ∈Nwith the following properties:
(i) Each n-tile,n∈N, can be covered byM (n−1)-flowers.
(ii) Each n-tile,n∈N0, can be covered byM (n+ 1)-flowers.
For easier formulation of this lemma and the subsequent proof, we assume for simplicity that a cover byat mostM elements contains preciselyM elements. This can always be achieved by repetition of elements in the cover.
Proof. We first consider the special case when # post(f) = 2. Then there are exactly two n-vertices, and hence exactly two n-flowers for each n∈ N0 (see
Lemma 5.18). These two n-flowers coverS2 (see Lemma 5.29 (iv)). Thus both
statements are true withM = 2 in this case.
Assume now that # post(f)≥3. It suffices to consider the statements (i) and (ii) separately and find a corresponding numberM for each of them.
(i) Letδ0>0 be as in (5.14). Then there existsM ∈Nsuch that each of the
finitely many 1-tilesX is a union ofM connected setsU ⊂X with diam(U)< δ0.
IfY is an arbitraryn-tile,n≥1, thenZ=fn−1(Y) is a 1-tile andfn−1
|Y a homeo- morphism ofY ontoZ. HenceY is a union ofM sets of the form (fn−1|Y)−1(U),
whereU ⊂Zis connected and diam(U)< δ0. Each set (fn−1|Y)−1(U) is connected
and so by Lemma 5.34 (i) it lies in an (n−1)-flower. HenceY can be covered by M (n−1)-flowers.
(ii) There existsM ∈Nsuch that each of the two 0-tilesX can be covered by M connected setsU ⊂X with diam(f(U))< δ0. IfY is an arbitraryn-tile, then
Z =fn(Y) is a 0-tile. By the same reasoning as above, the setY is a union of M sets of the form (fn
|Y)−1(U), whereU
⊂Z is connected and diam(f(U))< δ0.
Then U′ = (fn
|Y)−1(U) is connected, and fn+1(U′) = f(U) which implies
diam(fn+1(U′)) < δ
0. Hence by Lemma 5.34 (i) the set U′ is contained in some
(n+ 1)-flower. SinceM of the setsU′ coverY, it follows thatY can be covered by
M (n+ 1)-flowers.
Lemma5.38. LetCandCebe two Jordan curves inS2that both containpost(f). Then there exists a numberM such that eachn-tile for(f,Ce),n∈N0, can be covered by M n-flowers for(f,C).
Proof. The argument is very similar to the proof of Lemma 5.37. Again the case # post(f) = 2 is trivial; so we may assume # post(f)≥3.
Letδ0=δ0(f,C)>0 be the number as defined in (5.14). There exists a number
M such that each of the two 0-tiles X for (f,Ce) is a union of M connected sets U ⊂X with diam(U)< δ0. IfY is an arbitraryn-tile for (f,Ce), thenZ=fn(Y) is
a 0-tile for (f,Ce) andfn
|Y is a homeomorphism ofY ontoZ. HenceY is a union ofM sets of the form (fn|Y)−1(U), where U ⊂Z is connected and diam(U)< δ
0.
Each set (fn|Y)−1(U) is connected and so by Lemma 5.34 (i) it lies in ann-flower
CHAPTER 6