AMERICAN MATHEMATICAL SOCIETY Volume 359, Number 5, May 2007, Pages 2059–2074 S 0002-9947(06)04026-8
Article electronically published on December 15, 2006
GENERATING CONTINUOUS MAPPINGS WITH LIPSCHITZ MAPPINGS
J. CICHO ´N, J. D. MITCHELL, AND M. MORAYNE
Abstract. IfX is a metric space, thenCX and LX denote the semigroups
of continuous and Lipschitz mappings, respectively, from X to itself. The relative rank ofCXmoduloLXis the least cardinality of any setU\LXwhere UgeneratesCX. For a large class of separable metric spacesX we prove that the relative rank of CX modulo LX is uncountable. When X is the Baire spaceNN, this rank isℵ1. A large part of the paper emerged from discussions about the necessity of the assumptions imposed on the class of spaces from the aforementioned results.
1. Introduction and preliminaries
Given a mathematical structureX and a subsetY ofX, it is natural to ask how the ‘size’ of Y compares to that of X. If X and Y are both infinite, there is no single obvious way of answering this question. In fact, a large part of mathematics is devoted to answering this type of question. For instance, in analysis and topology, two standard notions of size are measure and category. In this paper we consider an algebraic notion of size. Let X be a semigroup and let Y be a subset of X. Therelative rank ofX moduloY, denoted by rank(X :Y), is defined as the least cardinality of a subsetZ ofX such thatY ∪Z generatesX. We may also refer to the relative rank ofY in X.
Relative ranks have been intensely investigated for semigroups of different types of mappings. For example, a classical result in the field, proved by Sierpi´nski, states that every countable set of mappings from an infinite setXto itself can be generated using two such mappings; see [1], [7] or [12]. The semigroup of all mappings from
X to X is denoted by TX. As an immediate consequence of Sierpi´nski’s result,
it follows that any subset of TX has relative rank 0,1,2 or uncountably infinite.
An important subset ofTX is, of course, the symmetric group on X, denoted by
SX. Another early result says that rank(TX : SX) = 2; see [6] and [9]. In [5],
to mention another interesting result, it has been shown that wheneverG ⊆ SX
then rank(SX :G) is 0, 1 or uncountable. Slight modifications of Banach’s proof
of Sierpi´nski’s result can be used to prove analogous results for binary relations, partial mappings, and partial bijections of infinite sets; see [7].
Received by the editors January 28, 2005.
2000Mathematics Subject Classification. Primary 54H15, 20M20.
Key words and phrases. Relative ranks, functions spaces, continuous mappings, Lipschitz map-pings, Baire space.
c
In this paper, we investigate the relative ranks of continuous mappings modulo Lipschitz mappings on certain metric spaces. For a metric space X, let CX
de-note the semigroup of continuous mappings from X to X and letLX denote the
semigroup of Lipschitz mappings from X to X. We show that rank(CX :LX) is
uncountable for two large classes of metric spaces (Theorems 2.1 and 3.2). The fact that we considered these function spaces and the particular metric spaces on which they act, was, in part, motivated by connections to relative ranks of endomorphisms of ordered sets and graphs. In fact, the first of our results concerning rank(CX:LX)
may be applied to endomorphisms of graphs and ordered sets (Corollaries 2.4 and 2.5).
Furthermore, in Sections 4 and 5 we give examples that demonstrate that none of the assumptions made about the metric spaces, that are the subject of Theorems 2.1 and 3.2, can be removed. Related to this discussion is Theorem 6.1, in which we determine the exact rank of CX moduloLX, whereX is a space that plays an
important role in descriptive set theory, the Baire space.
It is known that for many natural metric spaces the family of Lipschitz mappings is a meager subset of the space of all continuous functions with the topology of uniform convergence. Thus from a topological point of view LX is ‘small’ in CX.
Our main results say that the same is true, from an algebraic point of view, for some classes of metric space.
In the remainder of this section, we will give all the definitions and notation necessary for the rest of the paper.
As usual, for a semigroup S and U ⊆ S, denote by U the subsemigroup generated byU. For brevity, ifU, V ⊆S, we writeU, V instead of, the formally correct,U∪V .
A metric space X is concentric if it is unbounded and it is the union of an increasing sequence of compact balls. Such spaces are common, for example every euclidean space Rn is concentric. The induced topology of a concentric metric
space is noncompact, locally compact and second countable. On the other hand, for a topological space (X,U) which is noncompact, locally compact and second countable there exists a metricdsuch that (X, d) is concentric; see [14].
Letσ be a sequence in a metric spaceX and let Y be the set of elements that occur inσ, and its limitl, if it exists. Thenσis said to beextendible ifY is infinite and every continuous map from Y to Y that fixesl, if it exists, can be extended to an element ofCX. A retract is a subset Rof a metric space that occurs as the
image of a continuous mapping that acts as the identity on R. Throughout N<N
denotes the set of all finite sequences of natural numbers. An element of N<N is
denoted byi, and |i|denotes the length of the sequence.
2. Concentric spaces
In this section we prove that the relative rank ofCX moduloLX is uncountable
ifX is a concentric space satisfying a particular property. This is accomplished in the next theorem. At the end of the section, we discuss various corollaries to the main result of this section.
Proof. It suffices to prove that for every countable subset of CX there exists a
continuous mapping that is not generated by the union of the Lipschitz mappings and this subset. We construct such a mappingγrecursively.
To this end, let µ1, µ2, . . . ∈ CX be arbitrary, let ρ = (x0, x1, x2, . . .) be an extendible sequence with no convergent subsequence, and let B(p,1) ⊆ B(p,2)
⊆ · · · be compact balls that comprise X for some p∈ X. Since all the balls we consider in this proof are centered onp, for brevity we writeB(n) instead ofB(p, n). The elements of LX, µ1, µ2, . . .are finite compositions of the form
Φm+1µim· · ·Φ2µi1Φ1,
for somei1, i2, . . . , im∈Nand Φ1,Φ2, . . . ,Φm+1∈LX. We representLX, µ1, µ2, . . . as a countable union of families of such compositions. These families are deter-mined, roughly speaking, according to the elements of {µ1, µ2, . . .} that appear, the Lipschitz constants involved, and the balls that contain images of x0. The mapping γ is initially defined on the sequence ρ. Thus there are countably many steps in the definition, at each of which we ensure thatγdoes not belong to one of the given families. This process exhausts every possibility.
Precisely speaking, we consider the set
(1) Σ ={[i, n] :i∈N<N, n∈N}
and for [i, n]∈Σ, withi= (i1, i2, . . . , ik) andn∈N, the family of compositions
(2)
F[i,n] ={Φk+1µik· · ·Φ2µi1Φ1: Φj+1µijΦjµij−1· · ·Φ2µi1Φ1(x0)∈B(n),
Φj+1∈ LX with constantnfor each 0≤j≤k}.
Obviously,
[i,n]∈Σ
F[i,n]= LX, µ1, µ2, . . ..
Since Σ is countable we may enumerate its elements asσ1, σ2, . . .. Step 0 in the definition ofγ is made by settingγ(x0) =x0.
At step r > 0 we define γ on xr in such a way that when the definition is
complete no extension ofγto an element ofCX is an element ofFσr. Assume that
γwas defined onx0, x1, . . . , xr−1in the previous steps such that any element ofCX
agreeing with γ on these points does not lie in Fσi for i < r. Also assume that
σr= [i, n] withi= (i1, . . . , ik) andn∈N.
This step is completed by proving that every element of Fσr mapsxr into the
same ballB(m) for somem. In fact, it suffices to prove that if for each 0≤j≤k−1 and Φj+1 as in (2) we have Φj+1µij· · ·Φ2µi1Φ1(xr) ∈ B(mj+1), for some mj+1, then every Φj+2µij+1· · ·Φ2µi1Φ1(xr)∈B(mj+2), for somemj+2. Thus within step
rwe perform a finite induction.
For the base case, Φ1(x0)∈B(n) andd(Φ1(x0),Φ1(xr))≤nd(x0, xr). Therefore d(Φ1(xr), p)≤n+nd(x0, xr),and so Φ1(xr)∈B(m1), wherem1=n+nd(x0, xr).
Note that the choice ofm1does not depend on Φ1, but on its Lipschitz constant. The inductive hypothesis states that for every Φj+1µij· · ·Φ2µi1Φ1∈ F[(i1,...,ij),n]
we have
u= Φj+1µij· · ·Φ2µi1Φ1(xr)∈B(mj+1) for somemj+1. By the definition ofF[i,n]
If M is the maximum of mj+1 and n, then µij+1(u), µij+1(v) ∈ µij+1(B(M)). Since the continuous image of a compact set is compact, there exists M such that µij+1(B(M))⊆B(M). Thusd(µij+1(u), µij+1(v))≤2M. But, again by the definition ofF[i,n], Φj+2µij+1(v)∈B(n). Therefore
d(p,Φj+2µij+1(u))≤n+ 2nM.
We deduce that Φj+2µij+1· · ·Φ2µi1Φ1(xr)∈ B(mj+2), where mj+2 =n+ 2nM. Defineγ(Xr)∈B(mk+1).
When the recursion is complete, the extension of γ to an element of CX is not
contained inFσ for anyσ∈Σ. This implies thatγ∈ LX, µ1, µ2, . . ..
Next, we give three corollaries to the last result that demonstrate its potential applications. We start with a special case.
Corollary 2.2. Let X = N, Z or R with the usual euclidean metric. Then rank(CXn:LXn)is uncountable for any n∈N.
The next easy example, in conjunction with Corollary 2.2, demonstrates that al-though two different metrics on a space may induce the same topology, the conclu-sion of Theorem 2.1 can be true or false. In other words, Theorem 2.1 is dependent upon the metric used.
Example 2.3. LetX =N with metric ρ(x, y) = 1 when x=y, and ρ(x, y) = 0 when x=y. Then the topology induced byρis the same as that induced by the usual euclidean metric. However, rank(CX:LX) = 0.
Next, we apply Theorem 2.1 in the context of endomorphisms of graphs. Recall that EndX(Γ) is the monoid of endomorphisms (i.e. mappings that preserve edges)
of a graph Γ with vertex setX and TX is the monoid of all mappings from X to X.
Corollary 2.4. LetΓbe a countably infinite connected graph with the vertex setX
with every vertex having finite degree. Then rank(TX : EndX(Γ))is uncountable.
Proof. Define a metricdonX byd(x, y) =nif and only if the minimum length of a path fromx to y isn. Of course, d(x, y) = 0 if and only if x= y. Our metric satisfies the following conditions:
(i) for everyk∈Nthere existx, y∈X such thatd(x, y)> k;
(ii) for everyx∈X and everyk∈Nthe set{y∈X : d(x, y)≤k} is finite.
Thus (X, d) is a concentric space. Moreover, every mapping from X to X is con-tinuous, and every endomorphismαof Γ satisfiesd(α(x), α(y))≤d(x, y). In other words,CX=TX and EndX(Γ)⊆ LX. Hence by Theorem 2.1, rank(TX : EndX(Γ))
is uncountable.
Our next application of Theorem 2.1 concerns partially ordered sets. Let X
be a partially ordered set and let OX denote the monoid of all order preserving
mappings fromX toX. Recall, that forx, y∈X apath from xtoy of lengthnis a sequence
(x=p0, p1, . . . , pn−1, pn=y)
such that for each i either pi+1 ≤ pi or pi+1 ≥ pi. A partially ordered set is
connected if there is a path between every pair of elements. For x∈X, the sets
Corollary 2.5. LetX be a connected partially ordered set with|x∨|and|x∧|finite for everyx∈X. Thenrank(TX:OX)is uncountable.
Proof. Forx, y∈X defined(x, y) to be the minimum lengthnof a path fromxto
y.
The remainder of the proof follows by the same argument as used in the proof
of the previous corollary.
In fact, the converse of Corollary 2.5 is true; see the main theorem of [8].
3. Countable unions of compact sets
A topological space is calledKσif it is a countable union of compact sets.
Exam-ples of such spaces are second countable locally compact spaces and all countable spaces. In particular, Qn is Kσ, n∈ N. In this section we prove an analogue of
Theorem 2.1 for a class ofKσ spaces.
We require the following routine lemma.
Lemma 3.1. Let(X, d1),(Y, d2)be metric spaces, letµbe a continuous map fromX
toY and letKbe a compact subset ofX. Then for each >0there existsδ >0such that for eachx∈K and any y∈X withd1(x, y)< δ we haved2(µ(x), µ(y))< .
Next, we give the main result of this section.
Theorem 3.2. Let X be a Kσ metric space containing an extendible convergent
sequence. Thenrank(CX :LX) is uncountable.
Proof. We follow an argument similar, but not identical, to the proof of Theorem 2.1. As before we construct a continuous mappingγrecursively that is not generated byLX and a fixed countable set of mappings inCX.
Let (x1, x2, . . .) be an extendible convergent sequence in X with limit x, let
K1 ⊆ K2 ⊆ . . . be compact sets that comprise X, and let µ1, µ2, . . . ∈ CX. Of
course, we may assume that x∈ {x1, x2, . . .}, that xi =xj when i= j and that d(xi, x)≤1/i. The set Σ is defined in (1). For [i, n]∈Σ withi= (i1, i2, . . . , ik), n∈N, the family of compositionsG[i,n] is defined by
G[i,n]={Φk+1µik· · ·Φ2µi1Φ1: Φj+1µij· · ·Φ2µi1Φ1(x)∈Kn,
Φj+1∈ LXwith constantn, 0≤j≤k}.
As before, enumerate the elements of Σ asσ1, σ2, . . .. Step 0 in the definition ofγ is to setγ(x) =x.
At step r > 0 we will define γ(xpr) = xr for a point xpr in such a way that
when the recursion terminates γ∈ Gσr. Moreover, this definition will ensure that
γ|{x,x1,x2,...}is continuous when the recursion is complete. Assume thatγhas been defined on pointsxp1, xp2, . . . , xpr−1 with imagesx1, x2, . . . , xr−1, respectively, and thatσr= [i, n],i= (i1, . . . , ik) andn∈N.
This step of the proof is completed by finding xpr, via a finite induction, such
that every composition Φk+1µik· · ·Φ2µi1Φ1 ∈ Gσr takesxpr to a point closer tox
thanxr.
For the base case, by Lemma 3.1, if δk+1=d(x, xr)/2, then there existsδk >0
such that for u ∈ Kn and v ∈ X with d(u, v) < δk we have d(µik(u), µik(v)) <
The inductive hypothesis is as follows. For 0 ≤ j ≤ k−2 we have chosen a sequence of positive numbersδk+1, δk, . . . , δk−j such that if 0≤t≤j,u∈Kn and v∈X withd(u, v)< δk−t, then
d(µik−t(u), µik−t(v))< δk−t+1/n.
For the inductive step, by Lemma 3.1, there existsδk−j−1>0 such that foru∈
Kn andv∈X withd(u, v)< δk−j−1 we haved(µik−j−1(u), µik−j−1(v))< δk−j/n. When this finite induction is finished, we have a sequenceδk+1, δk, . . . , δ1 with the above properties. Letprbe chosen large enough to ensure thatd(x, xpr)≤δ1/n
and thatxpr ∈ {/ xp1, . . . , xpr−1}. It follows that
d(Φk+1µik· · ·Φ2µi1Φ1(x),Φk+1µik· · ·Φ2µi1Φ1(xpr)) ≤δk+1=d(x, xr)/2< d(x, xr).
So, if we defineγ(xpr) =xr, then no element ofCXthat agrees withγon the points
xp1, . . . , xpr is an element ofGσr.
When the recursion is finished and γ is already defined on every element of
{xp1, xp2, . . .} ∪ {x}, we extend γ to the whole set {x1, x2, . . .} ∪ {x} by defining
γ(xs) = x for all s /∈ {p1, p2, . . .}. As the sequence xn is convergent to x the
function γ, as defined so far, on the elements x, x1, x2. . ., is continuous. Thusγ can be extended to an element of CX, and, by construction, this element is not a
member of anyGσr.
In the next section we give examples that demonstrate the necessity of every assumption in Theorem 3.2 and Corollary 3.3.
Corollary 3.3. Let X be a locally compact separable metric space containing an extendible convergent sequence. Thenrank(CX :LX)is uncountable.
Proof. Every locally compact separable metric space is aKσ metric space.
Corollary 3.4. Let X be a 0-dimensional Kσ metric space that contains a
con-vergent sequence with an infinite number of distinct terms. Then rank(CX:LX)is
uncountable.
Proof. Assume that (x1, x2, . . .) is a convergent sequence with limitx, thatxi=xj
ifi=j andx∈ {x1, x2, . . .}. It is enough to show that this sequence is extendible. As X is 0-dimensional and metric, we can take pairwise disjoint clopen neigh-bourhoodsHi ofxi,i∈N, none of which containsxand such that diam(Hi)→0,
with i → ∞. Let µ be any continuous mapping from{xn : n ∈ N} ∪ {x} to
{xn : n∈N} ∪ {x}. Defineµ∗:X →X by
µ∗(u) =
µ(xi), u∈Hi, µ(x), otherwise.
Obviously,µ∗∈ CX.
As every countable metric space is a 0-dimensional Kσ space, we have the
fol-lowing corollary.
4. Counterexamples
In this section we give two examples. Namely, an example that the extendible sequence assumption in Theorem 3.2 may not be abandoned, and a further example that Corollary 3.3 does not hold when X is locally compact but not separable. Another example that demonstrates that Theorem 3.2 does not hold whenX has an extendible sequence but is notKσis given in the next section. Our first example
comes from the literature.
Example 4.1. In [2] it is proven that there exist metric continua (i.e. compact and connected metric spaces) with no continuous mappings other than the constant mappings and the identity. It follows that every continuous mapping on such a continuaCis Lipschitz, and so rank(CC:LC) = 0. Such a space is clearly Kσ but
contains no extendible convergent sequences. Continua with this property can even be planar; see [11].
Our next example demonstrates that the separability assumption in Corollary 3.3 cannot, in general, be removed.
Example 4.2. LetX be a direct sum of continuum copies of the unit interval [0,1] with the usual euclidean metric. In other words,X is the product space
X = [0,1]×c,
where c denotes the cardinal number continuum with the discrete topology. We prove that there exists a mapping Ψ such that LX,Ψ=CX. Define a metricd
onX by
d((s, α),(t, β)) =
|s−t|, α=β,
1, α=β.
The metricdis consistent with the topology ofX. Clearly,Xis locally compact but not separable. Note that a continuous mappingµ:X →X maps each component [0,1]× {α} into some other component [0,1]× {β}. Thus componentwise µ is a continuous function from [0,1] into [0,1]. There are continuum many members of
C[0,1]. LetC[0,1]={µα : α <c},and letτ:c×c→cbe a bijection.
We define the mapping Ψ∈ CX by
Ψ(t, τ(α, β)) = (µα(t), β).
Letµ∈ CX be arbitrary. By what we said above, there are two mappingsξ:c→c
andζ:c→csuch that
µ(t, α) = (µξ(α)(t), ζ(α))
for each t∈[0,1] andα∈c. Let Φ :X→X be defined as
Φ(t, α) = (t, τ(ξ(α), ζ(α))).
It is easy to see that Φ∈ LX. Now we have
µ(t, α) = (µξ(α)(t), ζ(α)) = Ψ(t, τ(ξ(α), ζ(α))) = Ψ◦Φ(t, α).
5. A further counterexample
In this section we give a further example. This example takes the form of a theo-rem and shows that Theotheo-rem 3.2 does not necessarily hold whenXis a (separable) metric space that has an extendible sequence but is notKσ. The space we consider
is closely related to a classical space that plays a crucial role in descriptive set the-ory, namely, the Baire spaceNN=N. Recall, that forx= (xi)i∈N, y= (yi)i∈N∈ N
the distanced(x, y) = 1/n,wherenis the least number wherexn=yn. For clarity,
we may write thenth element x
n of a sequence xas x(n). Of course,N contains
an extendible convergent sequence but is notKσ.
The space that we wish to consider is justN with a single point removed, the sequence (1,1, . . .); we denote this space by N∗. The new spaceN∗ is separable and contains an extendible convergent sequence but is not Kσ (and hence not
locally compact). The reader may ask why we considerN∗ rather thanN. This question is answered by the main result in the next section, where we prove that rank(CN : LN) = ℵ1. However, removing a single point from N we obtain the following result, which is the main theorem of this section.
Theorem 5.1. The relative rank of CN∗ modulo LN∗ is1.
Before giving the proof of this theorem, we require some notation and additional lemmas. We now define a metricρon the product spaceNN. First, defineκ:N2→ Nby
κ(i, j) = (i+j−1)(i+j−2)
2 +j.
This function corresponds to enumeratingN2as follows:
(1,1), (2,1), (1,2), (3,1), (2,2), (1,3), . . . .
Thusκis a bijection. Note that
(3) ifp+q < s+t, thenκ(p, q)< κ(s, t),
andκ(p, q)≥p, q. Letmandn be the coordinate functions ofκ−1, i.e. functions such thatκ−1(i) = (m(i), n(i)). Letx= (x1, x2, . . .) and defineϕ:N → NNby
(4) ϕ(x) = (y(1), y(2), . . .) where yq(p)=xκ(p,q),
(i.e. xk =xκ(m(k),n(k))=y (m(k))
n(k) ). Now, foru= (u
(1), u(2), . . .), v= (v(1), v(2), . . .)
∈ NN, define
(5) ρ(u, v) =d(ϕ−1(u), ϕ−1(v)) =d((unm(1)(1), umn(2)(2), . . .),(vmn(1)(1), vmn(2)(2), . . .)) = 1
i,
wherei= min{j∈N : un(m(j()j))=vn(m(j()j))}.
In other words, the metric ρ is the metric d transported from the space N to the spaceNN. The definition of the metricρis contrived to makeϕand its inverse
ϕ−1 Lipschitz homeomorphisms with constant 1. From now on, unless specified otherwise, we always treatNN and its subsets as metric spaces equipped with the metricρ. It is known that the product topology onNNis generated by the metric
(6) ρ¯(x, y) =
∞
i=1 1 2id(x
where x = (x(1), x(2), . . .) and y = (y(1), y(2), . . .). As the topology on (NN, ρ) is given as the image byϕof the topology onN, we infer that the topologies onNN generated by ¯ρandρare identical.
Lemma 5.2. Let A ⊆ N and let ζi : A → N be Lipschitz mappings with
con-stant 1, i ∈ N. Then the mapping φ : AN → NN defined by φ(x(1), x(2), . . .) = (ζ1(x(1)), ζ2(x(2)), . . .) is Lipschitz with constant1.
Proof. Let x = (x(1), x(2), . . .), y = (y(1), y(2), . . .) ∈ AN such that ρ(x, y) = 1/p, for some p. From (5), we have that p = min{i : xn(m(i()i)) = yn(m(i()i))}. Thus for
any j < p, x(nm(j()j)) = yn(m(j()j)). Moreover, if r < n(j), then r+m(j) < n(j) +
m(j), and so κ(m(j), r) < κ(m(j), n(j)) = j by (3). It follows that x(rm(j)) = yr(m(j)), and so d(x(m(j)), y(m(j)))<1/n(j). Sinceζm(j) is Lipschitz with constant 1,d(ζm(j)(x(m(j))), ζm(j)(y(m(j))))≤d(x(m(j)), y(m(j)))<1/n(j).
So, ifζm(j)(x(m(j))) =u(m(j))andζm(j)(y(m(j))) =v(m(j)), thenu (m(j))
n(j) =v (m(j))
n(j) for allj < p. Butd(φ(x), φ(y)) = 1/q whereq= min{i:un(m(i()i))=vn(m(i()i))}, and so
d(φ(x), φ(y))<1/p=d(x, y).
Forφ ∈N<N denote by [φ] the clopen set of all elements in N that agree with
φ on the first |φ| components. For φ ∈ N<N and θ ∈ N<N or θ ∈ N denote by
φ∧θ orφ∧θ the element ofN<NorN, respectively, formed by concatenating the sequencesφ and θ or θ. For φ ∈N<N denote byφ[i], φ concatenated with itself i
times; ifi= 0, then suppose thatφ[i] is empty.
Lemma 5.3. There is a homeomorphism η from (N∗)N to N∗ that is Lipschitz with constant1.
Proof. We defineη by defining a sequence of homeomorphismsφ: (N∗)N → NN,
ψ:NN→ N andθ:N → N∗, each of which is Lipschitz with constant 1. Before making the definition ofφ, defineζ:N∗→ N by
(7) ζ(x) =
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x ifx= (1, a2, . . .), a2>1, (1)[k]∧(2a
k+1−2, ak+2, . . .) ifx= (1)[k]∧(ak+1, ak+2, . . .),
ak+1>1, k >1, (1)[a2+1]∧(2a
3+ 1, a4, . . .) ifx= (2, a2, a3, . . .), (a1−1, a2, . . .) ifx= (a1, a2, . . .), a1>2.
It may be verified that this definition ofζ gives a homeomorphism that is Lipschitz with constant 1 from N∗ to N. Note that ζ−1 is not Lipschitz. The mapping
φ: (N∗)N→ NN defined by applyingζ componentwise is a homeomorphism that is Lipschitz with constant 1 by Lemma 5.2.
The mapping ψ is defined by ψ =ϕ−1 : NN → N, where ϕ is defined by (4), and is a Lipschitz homeomorphism with constant 1.
Penultimately, defineθ:N → N∗ by
θ(x) =
(i+ 1, a2, a3, . . .) ifx= (2i−1, a2, a3, . . .), i≥1,
(1)[i]∧(a2+ 1, a3, . . .) ifx= (2i, a2, a3, . . .), i≥1.
Ultimately, the composition η =θ◦ψ◦φ: (N∗)N → N∗ is a homeomorphism
that is Lipschitz with constant 1.
A subsetA of a topological spaceX isdiscrete if for each pointxinA there is an open setV in X such thatV ∩A={x}. We will call a continuous mappingµ
fromA toB discrete ifµ(A) is discrete.
Lemma 5.4. Let S be a dense subset ofN∗ and let ν :N∗→ N be a continuous mapping. Then there exist continuous discrete ν1, ν2, . . .:N∗→S such that (8)
for alli, j≥nandx∈ N∗, d(νi(x), νj(x))≤1/n andνi→ν pointwise as i→ ∞.
Proof. Leti∈Nbe arbitrary. The clopen sets [γ], forγ∈N<Nand|γ|=i, partition
N. Thus the sets ν−1[γ] partitionN∗. Defineν
i(x) =sγ for allx∈ν−1[γ], where sγ ∈[γ]∩S is fixed for eachγ ∈ N<N. Thus each νi, being constant on each of
the setsν−1[γ], is continuous. Moreover, the sequenceν
1, ν2, . . . has the required
property.
Lemma 5.5. Let S be a countable dense subset of N∗. Then there exists a con-tinuous mapping Ψ : N∗ → N∗ such that for every continuous discrete mapping
µ:N∗→S there is a Lipschitz mapping Φµ:N∗→ N∗ with constant1 such that µ= Ψ◦Φµ.
Proof. Take any sequenceσ= (s1, s2, . . .) of elements ofSsuch that every element of S occurs infinitely many times in σ. We define the function Ψ in two stages. First, for everyn∈Nandx∈[(1)[n]∧(2)] define Ψ by
Ψ(x) =sn.
Note that the setn∈N[(1)[n]∧(2)], on which Ψ was defined at the first stage, is a clopen set inN∗. It follows that the definition of Ψ can be completed so that Ψ is continuous onN∗. This can be done, for instance, by setting Ψ to be constant on the remaining part ofN∗.
Letµ:N∗→Sbe any discrete continuous mapping. We partition the preimage of each s ∈ µ(N∗) into open sets [σs,k] for natural numbers k and σs,k ∈ N<N.
Now, we may define Φµ onx= (x1, x2, . . .)∈[σs,k] by
Φµ(x) = (1)[N(σs,k)]∧(2, xm+1, xm+2, . . .),
where m =|σs,k| and N(σs,k) is the least number greater than m such that s= sN(σs,k).
In order to prove that Φµis a Lipschitz mapping with constant 1, letx, y∈σs,k, x=y, for somes∈µ(N∗) and k∈N. Again letm=|σk,s|and letd(x, y) = 1/p,
wherep=m+ min{i:xm+i=ym+i}. Then
d(Φµ(x),Φµ(y)) = 1/(N(σs,k) + 1 +p−m).
SinceN(σs,k)≥mwe have d(Φµ(x),Φµ(y))≤d(x, y).
The other case is thatx∈[σs,k] andy∈[σt,l] with [σs,k]= [σt,l] (in which case
[σs,k]∩[σt,l] =∅). Now, ifM = min{|σs,k|,|σt,l|}, thend(x, y)≥1/M. The images
Φµ(x) and Φµ(y) agree on at least the first P = min{N(|σs,k|), N(|σt,l|)} terms.
Therefored(Φµ(x),Φµ(y))≤1/P. ButP ≥M, and sod(Φµ(x),Φµ(y))≤d(x, y).
Then there existk, ssuch that x∈[σk,s]⊆µ−1(s). Ifm=|σk,s|, then
Ψ◦Φµ(x) = Ψ((1)[N(σk,s)]∧(2, xm+1, xm+2, . . .)) =sN(σk,s)=s=µ(x),
and the proof is complete.
Lemma 5.6. The mapping τ :N → NN defined by τ(x) = (x, x, . . .) is Lipschitz with constant1.
Proof. Letx, y∈ N,x=y, and let ϕ:N → NNbe the mapping defined by (4). Then ϕ−1(τ(x)) = ϕ−1(x, x, . . .) = (x
n(1), xn(2), . . .) and, likewise, ϕ−1(τ(y)) = (yn(1), yn(2), . . .). Hence
ρ(τ(x), τ(y)) =d((xn(1), xn(2), . . .),(yn(1), yn(2), . . .)) = 1
i,
where i= min{j : xn(j) =yn(j)}. In particular, xn(i) =yn(i), and so d(x, y)≥ 1/n(i). Buti=κ(m(i), n(i))≥n(i), and sod(x, y)≥1/i=ρ(τ(x), τ(y)).
We may now commence with the proof of the main theorem of this section.
Proof of Theorem 5.1. Letζ:N∗→ N andη: (N∗)N→ N∗be homeomorphisms, such as those defined in (7) and Lemma 5.3, and let Ψ :N∗→ N∗be the mapping defined in Lemma 5.5 for a fixed countable dense subsetS ofN∗ (and hence N).
LetZ denote the set of all elements (z(i))
i∈Nof (N∗)Nsatisfying
d(z(i), z(j))≤1/kwheneveri, j≥k.
The set Z is closed in (N∗)N(this is easy to see when one considers NN with the metric ¯ρdefined by (6)). Therefore, by [4], Z is a retract of (N∗)N. Denote byσ
the retraction from (N∗)Nto Z. Define Ψ: (N∗)N→(N∗)Nby
Ψ(x(1), x(2), . . .) = (Ψ(x(1)),Ψ(x(2)), . . .).
Of course, Ψ is continuous since Ψ is. Ify∈ N∗, thenη−1(y)∈(N∗)N. Therefore
σ◦Ψ◦η−1(y) is an element (z(1), z(2), . . .) ofZ. The mappingλ:Z → N defined by
λ(z(1), z(2), . . .) = lim
i→∞z
(i)∈ N,
is well defined and continuous. Define Λ∈ CN∗ by
Λ(y) =ζ−1◦λ◦σ◦Ψ◦η−1(y).
It is this mapping Λ that will, together with the Lipschitz mappings, allow us to generate every continuous mapping fromN∗ to itself.
In order to prove this, letµ∈ CN∗ be arbitrary. We will writeµas a composition of a single element of LN∗ and Λ. For the time being consider the continuous mapping ν = ζ◦µ : N∗ → N. By Lemma 5.4, there exist continuous discrete
ν1, ν2, . . . : N∗ → S that satisfy condition (8). Thus, for x ∈ N∗ the sequence (ν1(x), ν2(x), . . .) is an element ofZ. By Lemma 5.5, each of the mappingsνi can
be written as a composition Ψ◦Φνi for some Φνi∈ LN∗ with constant 1. Now, by
Lemmas 5.6 and 5.2, the mapping αthat takes x∈ N∗ to (Φν1(x),Φν2(x), . . .)∈ (N∗)N is continuous and Lipschitz. Therefore the mapping Φ :N∗ → N∗ defined by
The proof is concluded by observing that forx∈ N∗ we have
Λ◦Φ(x) =ζ−1◦λ◦σ◦Ψ◦η−1◦η◦α(x) =ζ−1◦λ◦σ◦Ψ(Φν1(x),Φν2(x), . . .) =ζ−1◦λ◦σ(Ψ◦Φν1(x),Ψ◦Φν2(x), . . .) =ζ
−1◦λ◦σ(ν1(x), ν2(x), . . .)
=ζ−1◦λ(ν1(x), ν2(x), . . .) =ζ−1( lim
i→∞νi(x)) =ζ
−1◦ν(x)
=ζ−1◦ζ◦µ(x) =µ(x).
In other words,µ∈ LN∗,Λ, and so LN∗,Λ=CN∗.
6. The Baire space
It is well known that N is homeomorphic to N2 and that such a homeomor-phism may be chosen to be Lipschitz in both directions. For example, the map-pingσ(x1, x2, x3, x4, . . .) = ((x1, x3, . . .),(x2, x4, . . .)) is such a Lipschitz homeomor-phism. The spaceN is homeomorphic to the irrationals with the usual euclidean topology [10, Section 3, IX].
We now state the main result of this section.
Theorem 6.1. The relative rank of CN moduloLN isℵ1.
The proof of this theorem is divided into two parts, the first being a proof that rank(CN :LN)≥ ℵ1 and another that rank(CN :LN)≤ ℵ1.
Now, we begin a sequence of results leading to the first of these proofs. Let
A ⊆ TN and X ⊆ N × N. A mapping Ψ : X → N is called universal for A if
A ⊆ {Ψ(x,·) : x∈ N }.
The following proposition is well known, however, for completeness a proof is included.
Proposition 6.2. There is no continuous mapping from N × N to N that is universal forCN.
Proof. Assume that there is such a mapping, call it Γ. Then the mappingγ(x) = Γ(x, x) is continuous. Now, for allξ∈ CN there existsx∈ N such thatγ(x) =ξ(x). But if γ1, γ2, . . . are the coordinate functions of γ, then the continuous mapping
ξ:N → N defined by ξ(x) = (γ1(x) + 1, γ2(x), . . .)∈ CN shares no common value
withγ, a contradiction.
However, there are continuous mappings from some proper subsets ofN × N to
N that are universal forCN. We use this fact later; see Lemma 6.5.
Lemma 6.3. There exists a continuous mapping Λ : N × N → N universal for
LN.
Proof. Letτk :Nk−1 →N be a bijection. For a sequence x= (x1, x2, . . .)∈ N let
x(i) =xi and letL
(k)
N denote the class of Lipschitz functions with constantk. We
define a continuous function Φk : NN× N → N that will eventually allow us to
define a universal function forL(Nk):
Of course, Φk is continuous. For φ∈ L
(k)
N we definec(φi):N→N:
c(φi)(τk(i+1)(a1, a2, . . . , ak(i+1)−1)) = (φ(a1, a2, . . . , ak(i+1)−1,1,1, . . .))(i),
for (a1, . . . , ak(i+1)−1)∈Nk(i+1)−1.
We now show thatφ(x) = Φk((c
(i)
φ )i∈N, x) for eachx∈ N. Fixx∈ N andn∈N.
Thend(x,(x1, x2, . . . , xk(n+1)−1,1,1, . . .))≤1/k(n+ 1), and so
d(φ(x), φ(x1, x2, . . . , xk(n+1)−1,1,1, . . .))≤1/(n+ 1).
Thus
(φ(x))(n) = (φ(x1, x2, . . . , xk(n+1)−1,1,1, . . .))(n)
= c(φn)(τk(n+1)(x1, x2, . . . , xk(n+1)−1))
= (Φk((c
(i)
φ )i∈N, x))(n).
Next, define Φ : (N× NN)× N → N by
Φ((n,(c(i))i∈N), x) = Φn((c(i))i∈N, x).
Note that the spacesN× NNand N are homeomorphic. Letθ:N →N× NN be such a homeomorphism. Finally, define
Λ(y, x) = Φ(θ(y), x),
which, by construction, is universal forLX.
We use Lemma 6.3 to prove the next lemma.
Lemma 6.4. Let µ1, µ2, . . . ∈ CN be arbitrary and leti = (i1, i2, . . . , in)∈ N<N.
Then there exists a continuous mappingΨiuniversal for the family of compositions
Ai={Φn+1µinΦn· · ·Φ2µi1Φ1 : Φ1,Φ2, . . . ,Φn+1∈ LN}.
Proof. Let Λ be universal for LN, as in Lemma 6.3, and let τ : N → Nn+1 be any homeomorphism with coordinate functionsτ1, . . . , τn+1. For eachx∈ N define
Λx:N → N by Λx(y) = Λ(x, y). The mapping we seek is defined by
Ψi(x, y) = Λτn+1(x)µin· · ·Λτ2(x)µi1Λτ1(x)(y).
It is routine to verify that this mapping is, indeed, continuous and universal for
Ai.
Proof of Theorem 6.1. Part I. In this part of the proof we will show that
rank(CN :LN)≥ ℵ1,
i.e. rank(CN :LN) is uncountable.
Let µ1, µ2, . . . ∈ CN be arbitrary and leti = (i1, i2, . . . , in). Enumerate the
families Ai, from Lemma 6.4, as A1,A2, . . . with universal continuous mappings Ψ1,Ψ2, . . .. The function Ψ :N × N → N defined by
Ψ((n1, n2, . . .),(m1, m2, . . .)) = Ψn1((n2, n3, . . .),(m1, m2, . . .)) is continuous and universal for∞i=1Ai. By Proposition 6.2, it follows that
LN, µ1, µ2, . . .=
∞
i=1
Ai=CN.
Before giving the second part of the proof of Theorem 6.1 we state one further auxiliary lemma.
For a topological space X, we say that a function f :X →[0,1] is lower semi-continuous or upper semicontinuous if the sets f−1(y,1], f−1[0, y), respectively, are open in X for each y ∈ [0,1]. There exists a lower semicontinuous function
L:N2→[0,1] universal for all lower semicontinuous mappings from N2 to [0,1]; see, for example, [3]. Equivalently, there exists an upper semicontinuous function
U : N2 → [0,1] universal for all such functions from N2 to [0,1]. We will con-struct lower and upper semicontinuous functionsL and U, respectively, that are independent and universal for all lower and upper semicontinuous functions, re-spectively. In other words, for any lower semicontinuousl :N →[0,1] and upper semicontinuous u : N → [0,1] there exists x ∈ N such that l(·) = L(x,·) and
u(·) =U(x,·).
Letφ= (φ1, φ2) :N → N2be a homeomorphism and defineθ1(x, y) = (φ1(x), y) and θ2(x, y) = (φ2(x), y). Of course, θ1 and θ2 are continuous. Now, define L :
N2→[0,1] by
L(x, y) =L(φ1(x), y) =L◦θ1(x, y)
and defineU:N2→[0,1] by
U(x, y) =U(φ2(x), y) =U◦θ2(x, y).
It is evident that the functionsL andU are lower and upper semicontinuous and universal for all lower and upper semicontinuous functions, respectively, fromN to [0,1].
Lemma 6.5. There exists a continuousΓ :U→ N, whereUis a Borel set inN2, that is universal forCN.
Proof. The mappingsL andU are defined above. Let
U={(x, y) : L(x, y) =U(x, y)} ∩L−1([0,1]\Q).
As mentioned above, there exist homeomorphisms from N to the irrationals and thus also to [0,1]\Q; letη : [0,1]\Q→ N be one such homeomorphism. Define Γ :U→ N by
Γ =η◦L|U.
We will prove that Γ is continuous and universal forCN.
First,L coincides withU onU. ThusL is lower and upper semicontinuous on
Uand so is continuous onU. Hence Γ is continuous onU.
Second, letµ∈ CN be arbitrary, and defineµ:N →[0,1]\Qbyµ(y) =η−1◦ µ(y). Note that µ is continuous and thus both upper and lower semicontinuous. By the construction ofL andU there existsx∈ N such thatL(x,·) =µ(·) and
U(x,·) =µ(·). Now,{x} × N ⊆U, sinceµ has domainN and by the definition ofU. But then
Γ(x,·) =η◦L(x,·) =η◦µ(·) =µ(·),
We may now conclude the proof of Theorem 6.1.
Proof of Theorem 6.1. Part II. In this part of the proof we will show that
rank(CN :LN)≤ ℵ1,
i.e. that we may findℵ1 mappings such that together withLN they generateCN. Let V = {x ∈ N : {x} × N ⊆ U} with the set U defined in the proof of Lemma 6.5. The setV is coanalytic. Therefore by [10, Section 39, II, Cor. 3], V
is the union ofℵ1 nonempty Borel setsBα, α <ℵ1. Moreover, by [10, Section 37,
I, Th. 1] for eachα <ℵ1 there exists a continuous surjective mappingµα from N
ontoBα.
Let σ : N → N2 be a Lipschitz homeomorphism, the existence of which is discussed at the beginning of this section. Define Φα:N → N by
Φα(σ−1(x, z)) = Γ(µα(x), z),
with Γ defined as in Lemma 6.5.
Letτ∈ CN be arbitrary. We will prove thatτ is the composition of a Lipschitz mapping and Φα for someα. Since Γ is universal forCN there exists y∈ N such
that τ(·) = Γ(y,·). In fact,{y} × N ⊆U, and soy∈V. In particular,y ∈Bα for
some α <ℵ1. Define Λ : N → N such that Λ(z) =σ−1(x, z), where µα(x) =y.
Sincez→(x, z) is Lipschitz andσ−1is Lipschitz, so is Λ. To conclude, for anyz∈ N we have
τ(z) = Γ(y, z) = Γ(µα(x), z) = Φα(σ−1(x, z)) = Φα◦Λ(z).
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Institute of Mathematics, Wroclaw University of Technology, Wybrze ˙ze Wyspia´ n-skiego 27, 50-370 Wroclaw, Poland
E-mail address:[email protected]
Mathematics Institute, University of St Andrews, North Haugh, St Andrews, Fife, KY16 9SS, Scotland
E-mail address:[email protected]
Institute of Mathematics and Computer Science, Wroclaw University of Technol-ogy, Wybrze ˙ze Wyspia´nskiego 27, 50-370 Wroclaw, Poland
