Compact subspace of products of linearly ordered spaces and co-Namioka spaces

0.1 MB·

by  Volodymyr Mykhaylyuk

COMPACT SUBSPACE OF PRODUCTS OF LINEARLY
ORDERED SPACES AND CO-NAMIOKA SPACES
6
1
V.V.MYKHAYLYUK
0
2

n Abstract. ItisshownthatforanyBairespaceX,linearlyorderedcompact
a spaces Y1,...,Yn, compact space Y ⊆Y1×···×Yn such that forevery par-
J allelepiped W ⊆ Y1×···×Yn the set Y ∩W is connected, and separately
5 continuous mapping f :X×Y →R thereexists adense inX Gδ-setA⊆X
2 suchthatf isjointlycontinuous ateverypointofA×Y.

]
N
G
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h 1. Introduction
t
a
Investigationofjoint continuitypoints setofseparatelycontinuousfunctionwas
m
started by R. Baire in his classical work [1] where he considered functions of two
[
real variables. The Namioka’s result [7] become the impulse to the intensification

1 of these investigations. These result leads to appearance of the following notions
v which were introduced in [5].
9 Let X, Y be topological spaces. We say that a separately continuous function
8 f :X×Y →Rhas the Namioka propertyifthereexistsadenseinX G -setA⊆X
4 δ
6 such that f is jointly continuous at every point of set A×Y.
0 A compactspace Y is calleda co-Namioka space iffor everyBairespaceX each
. separately continuous napping f :X ×Y →R has the Namioka property.
1
0 Verygeneralresultsinthedirectionofstudyofco-Namiokaspacepropertieswere
6 obtained in [2, 3]. It was obtained in [2, 3] that the class of compact co-Namioka
1 spacesisclosedoverproductsandcontainsallValdiviacompacts. Moreover,itwas
:
v shownin[3]thatthelinearlyorderedcompact[0,1]×{0,1}withthelexicographical
i orderisco-Namiokaanditwasreprovedin[2](resultfrom[6])thateverycompletely
X
orderedcompactisco-Namioka. Theseresultsweregeneralizedin[9]. Itwasshown
r
a in [9] that every linearly ordered compact is co-Namioka.
However,an example given by M.. Talagrandin [8] of a compact space which is
notco-Namiokaindicatesthataclosedsubspaceofco-Namiokacompactcanbenot
co-Namioka. Therefore the following question naturally arises: is every compact
subspaceY oftheproductY ×···×Y offinitefamilyoflinearlyorderedcompacts
1 n
Y a co-Namioka?
k
Inthis paperweusinganapproachfrom[9]weshowthatundersomeadditional
assumptions on Y the formulated above question has the positive answer.
2000 Mathematics Subject Classification. Primary 54C08, 54F05; Secondary 54D30, 54C30,
54C05.
Key words and phrases. separately continuous functions, Namioka property, linearly ordered
compact.
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2 V.V.MYKHAYLYUK
2. Continuous mappings on compact subspaces of product linearly
ordered spaces
For a mapping f :X ×Y →Z and a point (x,y)∈X ×Y we put
fx(y)=f (x)=f(x,y).
y
For a linearly ordered space X and points a,b ∈ X with a ≤ b by [a,b], [a,b),
(a,b] and (a,b) we denote the corresponding intervals.
Let X be a topological space. For a mapping f : X → R and a set A ⊆ X by
ω (A) we denote the oscillation
f
sup |f(x)−f(y)|
x,y∈A
of the function f on the set A. Moreover,for a point x ∈X by ω (x ) we denote
0 f 0
the oscillation
inf ω (U)
f
U∈U
of the function f at the point x , where U is a system of all neighborhoods of x
0 0
in X.
Proposition 2.1. Let X ,...,X be a linearly ordered spaces, X ⊆X ×···×X
1 n 1 n
be a compact space, ε > 0 and f : X → R be a continuous mapping. Then there
existsaninteger m∈N suchthat for every collection W ,...,W ofparallelepipeds
1 m
W =[a(k),b(k)]×···×[a(k),b(k)]
k 1 1 n n
such that (a(k),b(k))∩(a(j),b(j)) = ∅ for every i ∈ {1,...n} and distinct j,k ∈
i i i i
{1,...,m}, there exists k ∈{1,...m} such that ω (W ∩X)≤ε.
0 f k0
Proof. WithoutlossofthegeneralitywecanproposethatX ,...,X arecompacts.
1 n
Let g : X × ···× X → R be a continuous extension of the mapping f. We
1 n
choose finite covers U ,...U of spaces X ,...,X by open intervals such that
1 n 1 n
ω (U ×···×U ) ≤ ε for every U ∈ U ,...U ∈ U . For every i ∈ {1,...,n}
g 1 n 1 1 n n
we denote by A the set of all ending points of intervals U ∈ U and put m =
i i
|A |+|A |+···+|A |+1. Show that m is the required.
1 2 n
Let W ,...,W be a collection of parallelepipeds
1 m
W =[a(k),b(k)]×···×[a(k),b(k)]
k 1 1 n n
such that (a(k),b(k))∩(a(j),b(j)) = ∅ for every i ∈ {1,...n} and distinct j,k ∈
i i i i
{1,...,m}. Suppose that ω (W ∩X) > ε for every k ∈ {1,...m}. Then W 6⊆
f k k
U ×···×U foreveryk ∈{1,...m}andeveryU ∈U ,...U ∈U . Thereforefor
1 n 1 1 n n
every k ∈ {1,...m} there exist i ∈ {1,...n} and a ∈ A such that a ∈ (a(k),b(k)).
i i i
Since m > |A |+|A |+···+|A |, there exist i ∈ {1,...n}, a ∈ A and distinct
1 2 n i
j,k ∈{1,...,m} such that a∈(a(k),b(k))∩(a(j),b(j)). But it is impossible. (cid:3)
i i i i
3. Main result
Theorem 3.1. Let Y ,...,Y be linearly ordered spaces, Y ⊆ Y ×···×Y be a
1 n 1 n
compact subspace such that for every (possibly empty) parallelepiped
W =[a ,b ]×···×[a ,b ]⊆Y ×···×Y
1 1 n n 1 n
the set Y ∩W is connected. Then Y is co-Namioka.
COMPACT SUBSPACE OF PRODUCTS OF LINEARLY ORDERED SPACES AND CO-NAMIOKA SPACES3
Proof. Let X be a Baire space and f : X ×Y → R be a separately continuous
function. We prove that the mapping f has the Namioka property.
We fix ε>0 and show that the open set
G ={x∈X :ω (x,y)<ε for every y ∈Y}
ε f
is dense in X.
Let U is an open in X nonempty set. Show that there exists an open nonempty
setU ⊆U∩G . For every x∈U we denote by M(x) the set ofall integersm∈N
0 ε
for which there exists collection W ,...,W of parallelepipeds
1 m
W =[a(k),b(k)]×···×[a(k),b(k)]
k 1 1 n n
such that (a(k),b(k))∩(a(j),b(j)) = ∅ for every i ∈ {1,...n} and distinct j,k ∈
i i i i
{1,...,m} and ωfx(Wk ∩ Y) > 6(nε+1) for every k ∈ {1,...,m}. According to
Proposition 2.1 all sets M(x) are upper bounded. For every x∈U we put ϕ(x) =
maxM(x), if M(x)6=∅, and ϕ(x)=0, if M(x)=∅.
It follows from the continuity with respect to the first variable of the function f
that for every parallelepiped W ⊆ Y ×···×Y the set {x ∈ X : ω (W ∩Y) >
1 n fx
ε } is open in X. Therefore for every nonnegative m ∈ Z the set {x ∈ U :
6(n+1)
ϕ(x) > m} is open in U, i.e. the function ϕ : U → Z is lower semicontinuous on
the Baire space U. According to [4] the function ϕ is pointwise discontinuous, i.e.
ϕiscontinuousateverypointfromsomedenseinU sets. ThereexistanopeninU
nonempty set U and nonegative real m∈Z such that ϕ(x)=m for every x∈U .
1 1
If m = 0, then |f(x,a)−f(x,b)| ≤ ε < ε for every x ∈ U and a,b ∈
6(n+1) 3 1
Y. Then taking a point y ∈ Y and an open nonempty set U ⊆ U such that
0 0 1
ω (U )< ε, we obtain that ω (U ×Y)<ε. In particular, U ⊆G .
fy0 0 3 f 0 0 ε
Nowweconsiderthecaseofm∈N. Takeapointx ∈U andchooseacollection
0 1
W ,...,W of parallelepipeds
1 m
W =[a(k),b(k)]×···×[a(k),b(k)]
k 1 1 n n
such that (a(k),b(k))∩(a(j),b(j)) = ∅ for every i ∈ {1,...n} and distinct j,k ∈
i i i i
{1,...,m} and ωfx0(Wk ∩Y) > 6(nε+1) for every k ∈ {1,...,m}. Using the con-
tinuity of f with respect to the first variable we choose an open neighborhood
U ⊆U of x in U such that ω (W ∩Y)> ε for every k ∈{1,...,m} and
0 1 0 fx k 6(n+1)
x∈U .
0
Further without loss of the generality we can propose that Y ,...,Y are com-
1 n
pacts, i.e. Y =[a ,b ],...Y =[a ,b ].n For every i∈{1,...n} we put
1 1 1 n n n
A ={a(k),b(k) :1≤k ≤m}∪{a ,b }
i i i i i
and denote by V the set of all nonempty intervals [a,b] ⊆ Y such that a,b ∈ A .
i i i
Moreover,we put
W =V ×···×V .
1 n
Show that ω (W ∩Y)≤ ε for all x∈U W ∈W.
fx 6 0
Suppose that ωfx(W ∩ Y) > 6ε for some x ∈ U0 and W = [c1,d1] × ··· ×
[c ,d ] ∈W. We choose z =(z(0),...,z(0)),z = (z(n+1),...,z(n+1))∈W ∩Y
n n 0 1 n n+1 1 n
such that |f(x,z )−f(x,z )| = q > ε. For certainty we propose that z(0) ≤
0 n+1 6 1
z(n+1),...,z(0) ≤ z(n+1). Using the continuity of fx and the fact that for every
1 n n
parallelepiped V ⊆ Y ×···×Y the set V ∩Y is connected, we choose points
1 n
4 V.V.MYKHAYLYUK
z = (z(1),...,z(1)),...,z = (z(n),...,z(n)) ∈ W ∩Y such that z(0) ≤ z(1) ≤
1 1 n n 1 n i i
zi(2) ≤···≤zi(n) ≤zi(n+1) foreveryi∈{1,...,n}and|f(x,zk−1)−f(x,zk)|= n+q1
for every k ∈ {1,...,n+1}. Now for every i ∈ {1,...,n} and k ∈ {1,...,n+1}
we put c(k) = z(k−1), d(k) =z(k) and V = [c(k),d(k)]×···×[c(k),d(k)]. Note that
i i i i k 1 1 n n
(c(k),d(k))∩(c(j),d(j))=∅ for every i∈{1,...n} and distinct j,k ∈{1,...,n+1}
i i i i
and ωfx(Vk∩Y)≥ n+q1 > 6(nε+1) for every k ∈{1,...,n+1}.
Since for every i∈{1,...n} the set {k ≤m :(a(k),b(k))∩(c ,d )6=∅} contains
i i i i
at most one element, for the set
N ={k≤m:(a(k),b(k))∩(c ,d )=∅ ∀i=1,...n}
i i i i
wehave|N|≥m−n. ThereforethesystemP ={W :k ∈N}∪{V :1≤j ≤n+1}
k j
contains at least m+1 parallelepipeds. Besides, ωfx(P ∩Y) > 6(nε+1) for every
P ∈P and (t ,u )∩(v ,w )=∅ for every i∈{1,...,n}, where
i i i i
[t ,u ]×···×[t ,u ],[v ,w ]×···×[v ,w ]
1 1 n n 1 1 n n
are distinct parallelepipeds with P. But this contradicts to ϕ(x) = m. Thus,
ω (W ∩Y)≤ ε for all x∈U and W ∈W.
fx 6 0
Fix y ∈Y and x∈U . Put W ={W ∈W : y ∈W} and V = S (W ∩Y).
0 y y
W∈Wy
Clearly that V is a neighborhood of y in Y. Since ω (W ∩Y) ≤ ε for every
y fx 6
W ∈W andy ∈ T (W ∩Y), ω (V )≤ ε. Now using the continuity of f with
y fx y 3
W∈Wy
respect to the first variable at (x,y) we choose a neighborhood U˜ of x in X such
thatω (U˜)< ε. Thenω (U˜×V )<ε,in particular, ω (x,y)<ε for everyy ∈Y
fy 3 f y f
and x∈U . Thus, U ⊆G .
0 0 ε
(cid:3)
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Departmentof Mathematics,Chernivtsi National University, str. Kotsjubyn’skogo
2,Chernivtsi,58012Ukraine
E-mail address: [email protected]