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PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 126, Number 1, January 1998, Pages 85–95
S 0002-9939(98)04462-1
A RIGID SPACE HOMEOMORPHIC TO HILBERT SPACE
NGUYEN TO NHU AND PAUL SISSON
(Communicated by James West)
Abstract. A rigid space is a topological vector space whose endomorphisms
are all simply scalar multiples of the identity map. This is in sharp contrast
to the behavior of operators on `2, and so rigid spaces are, from the viewpoint
of functional analysis, fundamentally different from Hilbert space. Neverthe-
less, we show in this paper that a rigid space can be constructed which is
topologically homeomorphic to Hilbert space. We do this by demonstrating
that the first complete rigid space can be modified slightly to be an AR-space
(absolute retract), and thus by a theorem of Dobrowolski and Torunczyk is
homeomorphic to `2.
Rigid topological vector spaces first appeared in the literature in 1977 with an
example by Waelbroeck, in the paper [11]. This first rigid space, however, was not
complete, and the existence of a complete rigid space was first confirmed by Kalton
and Roberts in [5]. In that paper, the constructed space is not only complete and
rigid, but is also quotient-rigid and a subspace of L0[0,1] (quotient-rigid meaning
that every quotient of the space inherits the rigid character). A rigid space which
serves as the domain space of a non-trivial compact operator was constructed in
[9], illustrating that rigid spaces can have relatively rich topologies. In this paper
we demonstrate that rigid spaces can in fact be topologically homeomorphic to
Hilbert space, thus illustrating the degree to which the two concepts of topological
homeomorphism and topological vector space isomorphism can differ.
To do this, we will first modify slightly the rigid space constructed in [5], and then
employ a characterization of ANR-spaces (absolute neighborhood retract spaces)
due to the first author, which appeared in its original form in [6] and in a refined
form in [7]. It is this second version that we will apply in this paper.
The first section of the paper contains the characterization we will apply, as well
as an explanation of the modifications we must make to the first rigid space. We will
then show in the second and third sections that the rigid space under consideration
is an AR-space.
1. Theorems and a construction
We begin with an explanation of the characterization of ANR spaces to be found
in [7].
Received by the editors January 23, 1996.
1991 Mathematics Subject Classification. Primary 46A16, 54F65; Secondary 46C05, 54G15.
c
1998 American Mathemati cal Society
85
86 NGUYEN TO NHU AND PAUL SISSON
Let {Un}be a sequence of open covers of a metric space X. For a given open
cover Un,let
mesh(Un)=sup{diam(U):U∈U
n
}.
We say that Unis a zero sequence if mesh(Un)→0asn→∞. For a given open
cover U,weletN(U)denotethenerve of U. The nerve of an open cover is the
simplicial complex
{σ:σ=hU1,...,U
ni,U
i
∈U,n∈N}
made up of all σ=hU1,...,U
nifor which Tn
i=1 Ui6=∅.N(U) is endowed with
the Whitehead (or weak) topology (see [1] or [3] for a discussion). Finally, define
U=S∞
i=1 Unand let K(U)=S∞
n=1 N(Un∪U
n+1 ), and for any σ∈K(U)let
n(σ)=max{n∈N:σ∈N(U
n∪U
n+1 )}.
We use the following version of the ANR-characterization theorem; see [6], [7],
[8].
Theorem 1.1. AmetricspaceXwith no isolated points is an ANR if and only if
there is a zero sequence {Un}of open covers of Xand a map g:K(U)→Xsuch
that g|U → Xis a selection (i.e. g(U)∈Ufor every U∈U) and such that for
any sequence of simplices {σk}in K(U)with n(σk)→∞and g(σ0
k)→x0∈Xwe
have g(σk)→x0∈X. Here, σ0
krepresents the collection of vertices {U1,...,U
n}
making up the simplex σk,andg(σ
0
k
)and g(σk)represent the sets of images of,
respectively, the vertices of σkand the convex combinations of the vertices of σk.
The goal in this paper is to show that the rigid space to be constructed has this
property. This then shows that the rigid space is an AR space, and so by the result
of Dobrowolski and Torunczyk [2] the space is homeomorphic to `2(Dobrowolski
and Torunczyk showed that for a complete, separable infinite-dimensional linear
metric space, X∼
=`2⇐⇒ Xis an AR).
The topology of the rigid space will be generated by an F-norm on the space.
Recall that an F-norm is defined as follows.
Definition 1.1. Let Xbe a vector space. A map ||·|| :X→[0,∞)isanF-norm
if
(1) ||x|| =0 ⇐⇒ x=0,
(2) ||x+y|| ≤ ||x|| +||y||,
(3) ||αx|| ≤ ||x|| whenever |α|≤1, and
(4) ||αx|| → 0 whenever |α|→0.
In addition, the construction in [5] often makes use of quasi-norms, similar to
F-norms but with the following characteristics:
(1) ||x|| =0 ⇐⇒ x=0,
(2) ||x+y|| ≤ C(||x|| +||y||), Cindependent of xand y,and
(3) ||αx|| =|α|||x||,αa scalar.
The type of “norm” in use at a given point in the construction will be clear from
the context.
We now proceed with an overview of the construction of the first rigid space,
along with our modifications. For simplicity of terminology we first introduce the
following definition.
A RIGID SPACE HOMEOMORPHIC TO HILBERT SPACE 87
Definition 1.2. We say that a sequence {pn}is a 1-approaching sequence if the
following conditions hold:
1/2=p
0<p
1< ... < pn< ... < 1,
lim
n→∞ pn=1.
Let {pn}be a 1-approaching sequence. We define the space `({pn})by
`({p
n
})={x≡
∞
X
n=0
xnen:kxk≡
∞
X
n=0
|xn|pn<∞},(1)
where endenotes the characteristic function of [n, n +1].
The space `({pn}) is equipped with the F-norm k·k defined in (1).
In [5], the existence of a sequence of finite-dimensional spaces {Vn}∞
n=0 is demon-
strated, each a subspace of Lpnand each with basis {vn,k}l(n)
k=1, with the basis el-
ements possessing certain properties and with 1/2=p
0<p
1<··· <1. For the
purposes of that paper an explicit construction of the basis elements was not nec-
essary, but for our purposes we will find it convenient to be more precise. To that
end, let vn,k be the characteristic function of the kth sub-interval of [0,1] of length
l(n)−1,where[0,1] has been sub-divided into l(n) essentially disjoint sub-intervals
of equal length. Note that with this specification of the basis elements, the spaces
Vnhave the properties of Lemma 3.1 of [5].
In [5], each Vnis translated by the map τn, which takes functions of [0,1] to
functions of [n, n +1] by τ
n
f(x)=f(x−n). Keeping the notation of [5], we will
let Un=τnVn.LetMbe the closed linear span of {en}, and let Ybe the closure
of SUn, where closure in both cases is relative to the F-norm defined on the space
Zof real-valued functions on [0,∞)by
||f|| =
∞
X
n=0 Zn+1
n
|f(x)|pndx.
As in [5], let Z(a, b) be the subspace of Zof functions with support on [a, b].
Lemma 3.2 of [5] proved the following:
Lemma. Suppose f∈Z(0,n)with ||f|| =1. Then there exists a linear operator
A:Z(n, n +1)→Z(0,n)with Aen=fand ||A|| =1.
(In this context, Z(0,n)andZ(n, n + 1) are equipped with quasi-norms and, as
in Banach spaces, ||A|| =sup
||x||≤1||Ax||.)
We will have to modify this lemma slightly, so that in addition to the above,
A:Un→(U0+··· +U
n−1). To do this, we have to assume of course that
f∈(U0+···+U
n−1), but this will be the case in the application of the lemma. As
in the proof of Lemma 3.2 of [5], decompose fso that f=h0+···+h
n−1,where
h
i∈U
i
.Foreachi, decompose hiso that hi=hi
1+···+h
i
l(i), with the support of
hi
jbeing the jth sub-interval of [i, i + 1] (recall that [i, i + 1] has been partitioned
into l(i) sub-intervals of equal length). Let en
jdenote the characteristic function of
the jth sub-interval of [n, n + 1], fo r e ach j=1,...,l(n). For the moment, fix i.By
Lemma 2.2 of [5] there exist linear operators Fi
j, each mapping the pnth–integrable
functions with support on the jth sub-interval of [n, n +1] to the p
i
th–integrable
88 NGUYEN TO NHU AND PAUL SISSON
functions with support on the jth sub-interval of [i, i + 1], for each j=1,...,l(i).
Furthermore,
Fi
jen
j=hi
jand
Fi
j
=
hi
j
.
For j=l(i)+1,...,l(n), let Fi
jbe the zero operator, and let Fi=Fi
1+···+Fi
l(n).
Then Fien=hiand Fi:Un→Ui. Also, due to the partitioning of the intervals,
||Fi||pi=sup
||x||≤1Z|Fi
1x+···+Fi
l(n)x|
p
i=sup
||x||≤1Z|Fi
1x|pi+···+Z|Fi
l(n)x|
p
i
≤
Fi
1
p
i+···+
Fi
l(n)
p
i=
h
i
1
p
i+···+
h
i
l(i)
p
i=||hi||pi.
At this point the remainder of the proof is as in Lemma 3.2 of [5], with the operator
Abeing defined by A=F0+···+F
n−1.
The next step in the construction of the rigid space involves defining an operator
Smapping Zto Z. This begins with the selection of a sequence of elements {γk}∞
k=1,
with γk∈U0+···+U
k−1. By Lemma 3.2 of [5], operators Ak:Z(k, k+1) →Z(0,k)
with ||Ak|| =1andA
k
e
k=γ
kcan be chosen. Note that by our modification to
Lemma 3.2 above, we can assume also that Ak:Uk→(U0+··· +U
k−1). A
map T:Z→Zis then defined by T=P∞
k=1 ckAkEk(each Ekis the projection
map from Zonto Z(k,k +1), and {c
k}is a sequence of reals). Our modification to
Lemma 3.2 implies that, in addition, Tmaps Yinto Y.Let˜
Tdenote the restriction
of Tto the subspace Y. Finally, the map S:Z→Zis defined by S=I−T.We
will want to work with the restriction map ˜
S:Y→Yas well, defined by ˜
S=I−˜
T.
In [5] it is shown that ||T|| ≤ 1/4, and that therefore Sis invertible. Combining
this argument with the fact that we have made T:Y→Y,weobtain
˜
T
≤1/4,
and therefore ˜
S:Y→Yis also invertible.
Kalton and Roberts then showed in [5] that the sequences {l(n)}and {pn}can
be chosen so that the quotient space X=Y/S(M), where M=`({pn}), is a rigid
space.
Remark 1.1. In [5], the sequences {pn}and {l(n)}are constructed inductively. In
the inductive step, pnis chosen sufficiently close to 1 so as to obtain the desired
behavior, and l(n) then depends on pn. In our modification of the rigid space, we
will want to choose pnpossibly closer to 1 in the inductive step, for reasons given
in the proof of Theorem 3.5. Since our choice of pnis, if anything, larger than the
choice of pnin [5], this has no impact on the construction of the rigid space.
2. The AR-property for Y
In this section we prove the following theorem:
Theorem 2.1. Yis an AR.
Proof. We aim to verify the conditions of Theorem 1.1. Let {Un}be a zero sequence
of open covers of Y.LetU=
S
∞
n=1 Un,K(U)=S
∞
n=1 N(Un∪U
n+1), and let
f0:U→Ybe a selection.
We extend f0toamapf:K(U)→Yas follows. For any simplex σ=
hU1,...,U
mi∈K(U), Uj∈U for j=1, ..., m.Sincef
0
(U
j
)∈Y,wehave
f
0
(U
j
)=
∞
X
n=0
l(n)
X
i=1
xn
j,ien
i,j=1, ..., m,
A RIGID SPACE HOMEOMORPHIC TO HILBERT SPACE 89
where en
irepresents the characteristic function of ith sub-interval of [n, n +1].
For every x∈σ, with
x=
m
X
j=1
λjUj,λ
j
≥0,j =1,...,m and
m
X
j=1
λj=1,
we define
f(x)=
∞
X
n=0
l(n)
X
i=1
m
X
j=1
λj|xn
j,i|pnτ(xn
j,i)
1/pn
τn
ien
i,(2)
where τ:R→Ris the sign function:
τ(t)=
1ift>0,
0ift=0,
−1ift<0,
(3)
and
τn
i=τ
m
X
j=1
λj|xn
j,i|pnτ(xn
j,i)
.
Observe that for every U∈U we have
f(U)=
∞
X
n=0
l(n)
X
i=1
|xn
i|pnτ(xn
i)
1/pnτ(xn
i)en
i
=
∞
X
n=0
l(n)
X
i=1
xn
ien
i=f0(U).
Therefore f|U =f0.
Now assume that {σk}is a sequence of simplices in K(U) with n(σk)→∞,such
that f(σ0
k)→x0∈Yas k→∞. We have to show that f(σk)→x0as k→∞.
Since x0∈Y,wehave
x
0=
∞
X
n=0
l(n)
X
i=1
xn
ien
i.
Let σk=hUk
1,...,Uk
m(k)i.Thenwehave
f(U
k
j)=
∞
X
n=0
l(n)
X
i=1
xn
j,i(k)en
i,j=1,...,m(k).
For every xk∈σk, with
xk=
m(k)
X
j=1
λj(k)Uk
j,λ
j
(k)≥0,j =1,...,m(k)and
m(k)
X
j=1
λj(k)=1,
we have
f(xk)=
∞
X
n=0
l(n)
X
i=1
m(k)
X
j=1
λj(k)|xn
j,i(k)|pnτ(xn
j,i (k))
1/pn
τn
i(k)en
i,see (2),
90 NGUYEN TO NHU AND PAUL SISSON
where
τn
i(k)=τ
m(k)
X
j=1
λj(k)|xn
j,i(k)|pnτ(xn
j,i(k))
,see (3).
Since f(σ0
k)→x0,wehave
max {kf(Uk
j)−x0k,j =1, ..., m(k)}→0ask→∞.
Therefore
max
∞
X
n=0
l(n)
X
i=1
l(n)−1|xn
j,i(k)−xn
i|pn,j =1, ..., m(k)
→0(4)
as k→∞.
Now given >0, take k0∈Nsuch that
max
∞
X
n=0
l(n)
X
i=1
l(n)−1|xn
j,i(k)−xn
i|pn,j =1, ..., m(k)
<(5)
whenever k>k
0
.
Take N∈Nso that
∞
X
n=N+1
l(n)
X
i=1
l(n)−1|xn
i|pn<.
Then for k>k
0and j=1, ..., m(k)weget
∞
X
n=N+1
l(n)
X
i=1
l(n)−1|xn
j,i(k)|pn≤
∞
X
n=N+1
l(n)
X
i=1
l(n)−1|xn
j,i(k)−xn
i|pn
+
∞
X
n=N+1
l(n)
X
i=1
l(n)−1|xn
i|pn
<+=2.
(6)
Therefore
∞
X
n=N+1
l(n)
X
i=1
l(n)−1
m(k)
X
j=1
λj(k)|xn
j,i(k)|pn=
m(k)
X
j=1
λj(k)
∞
X
n=N+1
l(n)
X
i=1
l(n)−1|xn
j,i(k)|pn
<
m(k)
X
j=1
λj(k)(2)=2.
(7)
Denote
An
i(k)=
m(k)
X
j=1
λj(k)|xn
j,i(k)|pnτ(xn
j,i(k))
1/pn
τn
i(k)−xn
i
pn
.(8)
We claim that there exists a δn
i>0 such that
An
i(k)<2−nwhenever |xn
j,i(k)−xn
i|<(δn
i)1/pn.(9)
To prove the claim we consider two cases:
A RIGID SPACE HOMEOMORPHIC TO HILBERT SPACE 91
Case 1. xn
i=0. Take δ
n
i=2
−n
.Thenweget
A
n
i(k)=
m(k)
X
j=1
λj(k)|xn
j,i(k)|pnτ(xn
j,i(k))
<
m(k)
X
j=1
λj(k)2−n=2
−n
.
Case 2. xn
i6= 0. We may assume that xn
i>0 (the case xn
i<0 is similar). Take
δn
i<min{xn
i,2−n}. Then the inequality
|xn
j,i(k)−xn
i|<(δn
i)1/pn
implies that
xn
i−(δn
i)1/pn<x
n
j,i(k)<x
n
i+(δ
n
i)
1/pn.
Since Pm(k)
j=1 λj(k) = 1, it follows that
m(k)
X
j=1
λj(k)|xn
j,i(k)|pnτ(xn
j,i(k))
1/pn
τn
i(k)−xn
i
pn
<δ
n
i<2
−n
.
The claim is proved.
From (4) we get
|xn
j,i(k)−xn
i|→0ask→∞
for every n=0,1, ... and i=1, ..., l (n).
For every n=0, ..., N ,takeK(n)∈Nsuch that
max |xn
j,i(k)−xn
i|pn,j =1, ..., m(k)<(δn
i)1/pn
(10)
whenever k≥K(n)andi=1, ..., l(n). Denote
K=max{k
0
,K(0), ..., K(N)}.
Then from (8), (9) and (10) we have
An
i(k)=
m(k)
X
j=1
λj(k)|xn
j,i(k)|pnτ(xn
j,i(k))
1/pn
τn
i(k)−xn
i
pn
<2−n
whenever k>K.
Consequently for k>K,weget
N
X
n=0
l(n)
X
i=1
l(n)−1An
i(k)<
N
X
n=0
l(n)
X
i=1
l(n)−12−n
<
∞
X
n=0
2−n<2
(11)
92 NGUYEN TO NHU AND PAUL SISSON
and
∞
X
n=N+1
l(n)
X
i=1
l(n)−1An
i(k)≤
∞
X
n=N+1
l(n)
X
i=1
l(n)−1
m(k)
X
j=1
λj(k)|xn
j,i(k)|pn
+
∞
X
n=N+1
l(n)
X
i=1
l(n)−1
m(k)
X
j=1
λj(k)|xn
i|pn
<2+=3.
(12)
Therefore from (11) and (12) we obtain
kf(xk)−x0k=
∞
X
n=0
l(n)
X
i=1
l(n)−1
m
X
j=1
λj|xn
j,i(k)|pnτ(xn
j,i)(k)
1/pn
τn
i−xn
i
pn
=
N
X
n=0
l(n)
X
i=1
l(n)−1An
i(k)+
∞
X
n=N+1
l(n)
X
i=1
l(n)−1An
i(k)
<2+3=5
whenever k>K.
Consequently f(σk)→x0as k→∞.Hence Yis an ANR by Theorem 1.1. Since
Yis contractible, Yis an AR.
Remark 2.1. Observe that if l(n) = 1 for every n∈N,thenY=M. Therefore from
Theorem 2.1 we obtain that Mis also an AR. In the next section we shall show
that for a certain choice of 1-approaching sequence {pn},thespaceM=`({p
n
})is
locally convex.
3. Proof of the main result
We will begin with a lemma concerning 1-approaching sequences.
Lemma 3.1. There exists a 1-approaching sequence {pn}such that for any n∈N
and for every x1,...,x
n≥0,withP
n
i=0 xi≤1,
xp0
0+... +xpn
n+(1−x
0−... −xn)pn+1 <3.(13)
Proof. We will use the fact that the function
f(x0, ..., xn)=x
p
0
0+... +xpn
n+1−x
0−... −xn,
where x1,...,x
n≥0andP
n
i=0 xi≤1, attains the maximum
fmax =p
p0
1−p0
0+... +pn
pn
1−pn+1−p
1
1−p
0
0−... −pn
1
1−pn
at xi=p
1
1−pi
i,i=0, ..., n. This may be easily confirmed by the use of Gundelfinger’s
Rule (see, for instance, p. 219 of [10]).
Now, to prove the lemma, first choose p0=1/2. Assume that pihas been chosen
up to n, with
(1 + 2−i)−1≤pi<1,i=0, ..., n,(14)
such that condition (13) holds. Consider the function
f(x0, ..., xn)=x
p
0
0+... +xpn
n+1−x
0−... −xn,
A RIGID SPACE HOMEOMORPHIC TO HILBERT SPACE 93
where xi≥0, i=0, ..., n,andP
n
i=0 xi≤1. By the above fact and by (14) we have
f(x0, ..., xn)≤p
p0
1−p0
0+... +pn
pn
1−pn+1−p
1
1−p
0
0−... −pn
1
1−pn
=1+1
4+
n
X
i=1
p
1
1−pi
i(p−1
i−1)
<5
4+
n
X
i=1
(p−1
i−1) ≤5
4+
n
X
i=1
2−i<9
4
for any xi≥0, i=0, ..., n, with Pn
i=0 xi≤1.
Therefore we can choose pn+1, with
(1+2
−n−1
)
−1≤p
n+1 <1
such that
xp0
0+... +xpn
n+(1−x
0−... −xn)pn+1 <3
for any xi≥0, i=0, ..., n, with Pn
i=0 xi≤1.
We will use Lemma 3.1 to show that the space M=`({pn}) can be assumed to
be locally convex.
Theorem 3.2. There exists a 1-approaching sequence {pn}such that `({pn})is a
locally convex space.
Proof. Let x0, ..., xk∈`({pn}), xi=P∞
n=0 xi
nen, with
kxik=
∞
X
n=0
|xi
n|pn<1 for every i=0, ..., k .
Let αi≥0,i=0, ..., k, with Pk
i=0 αi= 1. We will show that kPk
i=0 αixik<3.
Observe that
k
X
i=1
αixi=
∞
X
n=0
(
k
X
i=1
αixi
n)en=
∞
X
n=0
λnen,
where λn=Pk
i=1 αixi
n.Thenweget
∞
X
n=0
|λn|=
∞
X
n=0
|
k
X
i=1
αixi
n|≤
k
X
i=1
αi
∞
X
n=0
|xi
n|
≤
k
X
i=1
αi
∞
X
n=0
|xi
n|pn<
k
X
i=1
αi=1.
Therefore from Lemma 3.1 we get
k
k
X
i=1
αixik=
∞
X
n=0
|λn|pn<3.
This uniform bound on convex combinations of elements from within the unit
ball implies the existence of a local base at 0 of convex sets, and so `({pn}) is locally
convex.
94 NGUYEN TO NHU AND PAUL SISSON
Remark 3.1. Observe that by the proof of Theorem 3.2 we get the following stronger
result: there exists a 1-approaching sequence {p0
n}such that for any 1-approaching
sequence {pn}, with pn≥p0
n, the resulting space M=`({pn}) is locally convex.
Remark 3.2. It is natural to ask whether `({pn}) is locally convex for any 1-
approaching sequence {pn}. The answer to this question is no, as we shall see
by the following example.
Example. Let {tn}be any 1-approaching sequence. Take a sequence {m(n)}of
natural numbers such that m(n)1−tn>nfor every n∈N. Now let {pk}be any
1-approaching sequence for which pm(n)=tn. Then the resulting space `({pk})is
not locally convex.
Proof. We hav e
k
m(n)
X
i=1
1
m(n)eik=
m(n)
X
i=1
|1
m(n)|pi>
m(n)
X
i=1
|1
m(n)|tn
=m(n)1−tn>n→∞
as n→∞. Therefore `({pk}) is not locally convex.
Let Xbe a linear metric space, let Mdenote a closed linear subspace of X,let
E=X/M denote the quotient space and let π:X→Edenote the quotient map.
We say that a m a p g:E→Xis a selection if g(x)∈π−1(x) for every x∈E.The
proof of our result uses the following theorem of Michael, see [1], Proposition 7-1,
p.87.
Theorem 3.3. Let Mbe a locally convex closed linear subspace of a complete linear
metric space X. Then there exists a continuous selection g:E→X.
From Theorem 3.3 we get
Theorem 3.4. Let Mbe a locally convex closed linear subspace of a complete linear
metric space X.IfXis an AR,thenX/M is an AR.
Proof. Let f0:A→X/M be a continuous map from a closed subset Aof a
metric space Zinto X/M.SinceMis locally convex, by Theorem 3.3 there exists
a selection g:X/M →X.SinceXis an AR, there exists a continuous map
h:Z→Xsuch that h|A=g◦f0.Thenf=π◦h,whereπ:X→X/M denotes
the quotient map, is an extension of f0.
Consequently, X/M is an AR, and the theorem is proved.
Now we are able to prove our main result in this paper.
Theorem 3.5. There exists a 1-approaching sequence {pn}such that the result-
ing rigid space X=Y/S(M)constructed in Section 1 is an AR, and therefore is
homeomorphic to `2.
Proof. First observe that by Remark 1.1 we can assume that the 1-approaching
sequence {pn}used in the construction of the rigid space is such that the space M
is locally convex, by first identifying a sequence {˜pn}which will satisfy Lemma 3.1
and then specifying that in the inductive step in the creation of the rigid space, pn
is chosen to be at least as large as ˜pn.
Now, since Sis an isomorphism, S(M) is locally convex. By Theorem 2.1 Yis
an AR,andsoY/S(M)isanAR by Theorem 3.4.
A RIGID SPACE HOMEOMORPHIC TO HILBERT SPACE 95
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Institute of Mathematics at Hanoi, P.O. Box 631, Bo Ho, Hanoi, Vietnam
Current address: Department of Mathematics, New Mexico State University, Las Cruces, New
Mexico 88003
E-mail address:nnguyen@emmy.nmsu.edu
Department of Mathematics, Louisiana State University - Shreveport, One Univer-
sity Place, Shreveport, Louisiana 71115
E-mail address:psisson@pilot.lsus.edu