Download OBTAINING POINTS OF JOINT θ-CONTINUITY OF

ANALELE ŞTIINŢIFICE ALE UNIVERSITĂŢII ”AL.I.CUZA” IAŞI
Tomul XLIII, s.I.a, Matematică, 1997, f2
OBTAINING POINTS OF JOINT θ-CONTINUITY
OF MULTIPLICATION IN A SEMIGROUP
WITH SEPARATELY θ-CONTINUOUS MULTIPLICATION
BY
S.GANGULY and T.BANDYOPADHYAY
0. Introduction Semigroups S provided with topology under which
translations x → s · x, x → x · s, s ∈ S are θ-continuous but not necessarily the multiplication are called semi-θ-topological semigroup (abbreviated
to semi-θ-mob) in contrast to semigroups with θ-continuous multiplication
which we call θ-topological semigroup (abbreviated to θ-mob). In this paper,
we deal with the question of how to find and locate points where the multiplication of a semi-θ-mob is jointly θ-continuous generalizing the famous
’Ellis–Lawson Theorem’ in the process.
1. Preliminaries.
All spaces are supposed to be Hausdorff. A
point x in a space (X, T ) is in the θ-closure of A ⊂ X [x ∈ θ-clX A] iff
clX U ∩ A 6= ∅ for any U ∈ T, x ∈ U . A is θ-closed in X iff A = θ-clX A.
Complement of θ-closed set is θ-open. (X, T ) is called H-closed iff every
open cover of X has a finite subcollection whose closures in X cover X. A
subset A of (X, T ) is an H-set in X iff every open (in X) cover of A admits
a finite subcollection whose closures in X cover A. A space X is Urysohn
iff for distinct x, y ∈ X, there exists nbd. Ux , Uy of x, y respectively with
clX Ux ∩ clX Uy = ∅.
Let (X, T1 ), (Y, T2 ) and (Z, T3 ) be spaces and f : (X, T1 ) × (Y, T2 ) →
(Z, T3 ) be a function. f is jointly θ-continuous at (x0 , y0 ) ∈ X × Y iff for
every open nbd. W of f (x0 , y0 ), there exist open nbds. U and V of x0
and y0 respectively such that f (clX U × clY V ) ⊂ clZ W . f is right (left) θcontinuous at (x0 , y0 ) ∈ X × Y iff there exists open nbd. V (U respectively)
A.M.S. Subject Classification 22A99
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such that f ({x0 } × clY V ) ⊂ clZ W [f (clX U × {y0 } ⊂ clZ W respectively]. f is
separately θ-continuous iff it is both left and right θ-continuous. Semigroup
with right (left) θ-continuous multiplication is called a right (left)-θ-mob
and that with separately θ-continuous one is called a semi-θ-mob.
A quasi-uniformity [12] for a set X is a non-void family U of subsets of X ×X
such that
(i) ∆ = {(x, x) | x ∈ X} ⊂ U, ∀U ∈ U
(ii) ∀U ∈ U, ∃V ∈ U such that V ◦ V = {(x, z) | ∃y ∈ X, (x, y), (y, z) ∈
V}⊂U
(iii) ∀U, V ∈ U, U ∩ V ∈ U
(iv) U ∈ U, U ⊂ V ⊂ X × X ⇒ V ∈ U
(X, U) is a quasi-uniform space. A quasi-uniformity U for X satisfying
the condition
(v) U ∈ U ⇒ U −1 = {(y, x) | (x, y) ∈ U } ∈ U
is called a unformity for X and (X, U) is a uniform space.
If (X, U) is a uniform space, then the topology TU of the uniformity U is
the family of all subsets O of X such that for each x in O there is U ∈ U
such that U [x] = {y ∈ X | (x, y) ∈ U } ⊂ O. A topological space (X, T ) is
uniformizable iff there exists a unformity U on X such that T = TU .
Let F be a family of functions of a space (X, T ) into a space Y and
let V be the quasi-uniformity attached with Y [every topological space is
quasi-uniformizable, [12]]. F is equi-θ-continuous at x ∈ X iff for all V ∈ V,
there exists U ∈ T such that f (clX U ) ⊂ clY V [f (x)], f ∈ F .
Let X be a set and S be a semigroup. Then a map π: X × S → X, (x, s) →
π(x, s) = x·s is a right (left) action of S on X iff (x·s)·t = x·(st) [s·(t·x) =
(st) · x respectively ], ∀s, t ∈ S, ∀x ∈ X.
A function f : (X, T ) → (Y, T1 ) between topological spaces is θ-continuous at x ∈ X iff for each open nbd. V of f (x), there exists open nbd. U
of x such that f (clX U ) ⊂ clY V . f , is a θ-homeomorphism iff f is bijective
and both f and f −1 are θ-continuous. A space (X, T ) is θ-metrizable iff
there exists a metric d on X and a θ-homeomorphism f : (X, T ) → (X, d).
A space (X, T ) is locally θ-H-closed [8] iff for each x ∈ X, there exists θopen nbd. of x in (X, T ) whose closure is an H-set. A space (X, T ) is almost
regular iff for all x ∈ X and for each open nbd. V of x, there exists an open
nbd. U of x such that x ∈ U ⊂ clX U ⊂ intX clX V .
The collection of all θ-open subsets in a space (X, T ) forms a topology Tθ
on (X, T ). (X, Tθ ) is the θ-open topology corresponding to (X, T ).
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2. Joint θ-continuity theorem.
Lemma 2.1. [1] A topological space X is uniformizable iff it satisfies
the following :
Given any point x0 ∈ X and any nbd. V of x0 , there exists a continuous
real valued function on X which takes its values in [0,1], is equal to 0 at x0
and is equal to 1 on X − V .
Let (X, T ) be a space and Tθ be the θ-open topology corresponding to T on X. Following Urysohn’s lemma, existence of such a function
f : (X, Tθ ) → [0, 1] can be guarenteed if (X, T ) be H-closed and Urysohn.
Thus,
Theorem 2.2. Let (X, T ) be an H-closed Urysohn space. Then
(X, Tθ ) is uniformizable.
Lemma 2.3.
[12] Let (X, T ) be a space. For each G ∈ T , let
SG = (G × G) ∪ ((X − G) × X). Then the family G = {SG | G ∈ T } is
a subbase for a quasi-uniformity U with T = TU . Thus, every topological
space is quasi-uniformizable.
Lemma 2.4. [8] A space X is almost regular iff for all x ∈ X and
for each open nbd. U of x, there exists θ-open nbd. V of x in X such that
V ⊂ clX U .
Theorem 2.5. Let X be a space, Y be H-closed Urysohn and (Z, T )
be an almost regular space. Let U be the quasi-uniformity attached to Z
[Lemma 2.3]. Let π: X × Y → Z be a right θ-continuous map. Then for
any x0 ∈ X,
(i) the maps fY : X → Z, x → x · y, y ∈ Y are equi-θ-continuous at x0
implies
(ii) π is jointly θ-continuous at every point (x0 , y) where y ∈ Y is arbitrary. Further (ii) ⇒ (i) holds if Z is H-closed Urysohn.
Proof. (i) ⇒ (ii) Let W ∈ U and consider the nbd. W [x0 · y] of x0 · y
[by Lemma 2.3, {W [x0 · y] | W ∈ U} coincides with the nbd. system at x0 · y
in Z ]. Since Z is almost regular, by Lemma 2.4, there exists a θ-open set
T in Z such that x0 · y ∈ T ⊂ clZ W [x0 · y]. Further we can find open sets
N and M in Z such that x0 · y ∈ N ⊂ clZ N ⊂ intZ clZ M ⊂ clZ M ⊂ T (by
definition of θ-open sets and almost regularity). Let P = intZ clZ M . Since
P is a regular open set in the almost regular space Z, it is θ-open in Z [8].
By Lemma 2.3, it is clear that
T, if m ∈ T
ST [m] = {y | (m, y) ∈ ST =
Z, if m 6∈ T .
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Thus, ∀z ∈ Z, ST [z] is a θ-open nbd. of z in Z. Obviously, x0 ·y ∈ ST [x0 ·y] ⊂
clZ W [x0 · y], SN [x0 · y] = N and SP [x0 · y] = P . Since{fy | y ∈ Y } are equiθ-continuous, there exists a nbd. U of x0 such that u · y ∈ clZ SP [x0 · y], ∀u ∈
clX U, ∀y ∈ Y . Again π is right θ-continuous and so there is a nbd. V of
y such that ∀v ∈ clY V, x0 · v ∈ clZ SN [x0 · y] = clZ N ⊂ P ⊂ ST [x0 · y].
Thus (x0 · v, x0 · y) ∈ ST . Since for v ∈ clY V, x0 · v ∈ P, ∀u ∈ clX U, u · v ∈
clZ SP [x0 · v] = clZ P ⊂ T = ST [x0 · v] (since x0 · v ∈ clZ P ⊂ T ). Thus
(u · v, x0 · v) ∈ ST . Hence (u · v, x0 · y) ∈ ST ◦ ST = ST for all u ∈ clX U
and for all v ∈ clY V . Since U generates T , this proves π to be jointly
θ-continuous.
(ii) ⇒ (i) Let us now suppose Z to be H-closed Urysohn and let
V be the uniformity attached to the θ-open topology Tθ corresponding to
(Z, T ) [Theorem 2.2]. Let W1 ∈ U be arbitrary and choose y ∈ Y . Since
clZ W1 [x0 · y] is a closed nbd. of x0 · y in the almost regular space Z, there
exists, by Lemma 2.4, a θ-open nbd. O of x0 ·y with x0 ·y ∈ O ⊂ clZ W1 [x0 ·y].
By Theorem 2.2 , there exists W ∈ V such that W [x0 · y] = O. Choose
symmetric W2 ∈ V [that is, W2−1 = {(x, y) | (y, x) ∈ W2 } = W2 ] with
W2 ◦ W2 ⊂ W . Since W2 [x0 · y] contains a θ-open of nbd. x0 · y and
since π is jointly θ-continuous at (x0 · y), y ∈ Y , by definition of θ-open
set, we can find open nbds. UY of x0 and VY of y such that ∀u ∈ clX UY
and ∀v ∈ clY VY , u · v ∈ W2 [x0 · y], that is, (u · v, x0 · y) ∈ W2 . Thus
(x0 · v, x0 · y) ∈ W2 and (u · v, x0 · y) ∈ W2 and hence, by symmetry of
W2 , (u · v, x0 · y) ∈ W2 ◦ W2 ⊂ W, ∀u ∈ clX UY and ∀v ∈ clY VY . Since
Y is H-closed and {Vy | y ∈ Y } forms an open cover of Y , there exists
n
n
1
1
{Vy1 , . . . , Vyn } such that Y = ∪clY Vyi . Then U0 = ∩Uyi is a nbd. of x0 and
for all u ∈ clX U0 , we have (u · y, x0 · y) ∈ W . Thus (clX U0 ) · y ⊂ W [x0 · y] =
O ⊂ clZ W1 [x0 · y]. Hence the proof.
Lemma 2.6. [1] Let X be a a space, (Y, dY ) and (Z, dZ ) be two
metric spaces. Let f : X × Y → Z be a map such that y → f (x0 , y) is
continuous on X for each y0 ∈ Y . For each real ε > 0 and each x ∈ X,
let g(x, ε) denote the least upper bound of the numbers δ > 0 such that
dY (y, y 0 ) < δ implies dZ (f (x, y), f (x, y 0 )) ≤ ε. Then the function x → g(x, ε)
is upper semicontinuous.
Theorem 2.7.
A θ-metrizable space (X, T ) is almost regular.
Proof.
Let f : (X, T ) → (X, d) be the corresponding θ-homeomorphism. Let x ∈ U ∈ T . Since f −1 is θ-continuous, corresponding to
the open nbd. U of x = f −1 f (x) in (X, T ), there exists an open nbd. V of
f (x) in (X, d) such that f −1 (cl(X,d) V ) ⊂ cl(X,T ) U . Since (X, d) is regular,
V is θ-open in (X, d) as well and thus f −1 (V ) is θ-open in (X, T ). Hence
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f −1 (V ) ⊂ f −1 (cl(X,d) V ) ⊂ cl(X,T ) U proving (X, T ) to be almost regular by
Lemma 2.4.
Imitating and generalizing the proof of uniform continuity of a continuous function over a compact metric space, we have
Theorem 2.8. Let (Y, T ) and (Z, T1 ) be θ-metrizable spaces with
f : (Y, T ) → (Y, d) and g: (Z, T1 ) → (Z, d1 ) be the corresponding θ-homeomorphisms. Let Y1 be an H-set in (Y, T ) and f1 : Y1 → (f (Y1 ), d) be the
restriction of f . Let h: (Y1 , T ) → (Z, T1 ) be θ-continuous :
f1−1
(f (Y1 ), d)
−→–
–←− (Y1 , T )
f1
g
h
−→– (Z, T1 )
−→–
–←− (Z, d1 ).
g −1
Then ∀ε > 0, ∃λ > 0 (depending only on ε) such that
ghf1−1 Bf1 (Y1 ) (f1 (y), λ) ⊂ B(gh(y), ε).
Definition. Let X be a space. A(⊂ X) is nowehere θ-dense iff θclX A has no θ-interior point. Union of a countabe family of nowhere θ-dense
sets is of θ-1st-category.
Theorem 2.9. [1] Let X be a locally θ-H-closed Urysohn space and
let {fα }α be a family of lower semicontinuous real valued functions on X
such that for every x ∈ X, the upper envelope supfα (x) is finite. Then every
α
non-empty θ-open set in X contains a nonempty θ-open set on which the
family {fα }α is uniformly bounded above.
Theorem 2.10. Let X be a locally θ-H-closed Urysohn space, Y
and Z are θ-metrizable spaces and let Y1 be an H-set in Y . Suppose that
π: X × Y → Z, (x, y) → x · y is a map with the properties :
(a) the map Y1 → Z, y → x · y is θ-continuous, ∀x ∈ X
(b) the map X → Z, x → x · y is θ-continuous for every y in a θ-dense
subset D of Y1 .
Then there is a θ-residual (that is, complement of a θ-1st-category subset)
subset R1 of X such that for all x0 ∈ R1 ,
(i) the maps X → Z, x → x · y, y ∈ Y1 are equi-θ-continuous at x0 with
respect to the quasi-uniformity attached to Z,
(ii) π is jointly θ-continuous at every (x0 , y) where y ∈ Y1 is arbitrary.
Proof. Let fY : Y → (Y, dY ) and fZ : Z → (Z, dZ ) be the θ-homeomorphisms involved. For x ∈ X and ε > 0, let
g(x, ε) = sup{δ ∈ R | y, y 0 ∈ Y1 , dY (fY (y), fY (y 0 )) < δ
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⇒ dZ (fZ (x·y), fZ (x·y 0 )) ≤ ε},
where R stands for the space of real numbers. Existence of δ is guarenteed
by the θ-continuity and hence continuity [with range space being regular,
continuity and θ-continuity coincides] of the map (Y1 , dY ) → Y1 → Z →
(Z, dZ ).
1.
We first prove that for fixed ε > 0, the function X →
R, x → g(x, ε), is upper semicontinuous. To this end, choose ε > 0 and a
θ-convergent net {xλ }λ in X with θ-lim xλ = x, g(xλ , ε) ≥ δ. If y, y 0 ∈ D
λ
and dY (fY (Y ), fY (y 0 )) < δ, then dZ (fZ (xλ · y), fZ (xλ · y 0 )) ≤ ε, ∀λ and
thus
dZ (fZ (x · y), fZ (x, y 0 )) = dZ (fZ θ-lim(xλ · y), fZ θ-lim(xλ · y 0 )) =
λ
λ
(by θ-continuity of the map x → x · y)
= dZ (θ-lim fZ (xλ · y), θ-lim fZ (xλ · y 0 )) (since fZ is θ−continuous)
λ
λ
= θ-lim dZ (fZ (xλ · y), fZ (xλ · y 0 )) (since dZ is continuous
λ
and thus θ-continuous) ≤ ε.
Let (y, y 0 ) ∈ Y1 × Y1 be arbitrary satisfying dY (fY (y), fY (y 0 )) < δ.
0
Since D is θ-dense in Y1 , there exists nets {yλ }λ and {yλ }λ in D with
0
θ-lim yλ = y and θ-lim yλ = y 0 (θ-limits in Y1 ). Since f |Y1 : Y1 → (Y, dY )
λ
λ
is θ-continuous and since (Y, dY ) is regular, f |Y1 is continuous and hence
f1 : Y1 → (f (Y1 ), dY ) is continuous. Thus, fY (y) = f1 (y) = lim f1 (yλ ) =
0
lim fY (yλ ) and fY (y 0 ) = lim fY (yλ ) [note that (f (Y1 ), dY ) being regular,
there is no difference 0 between ordinary limits and θ-limits of a net. Thus
{fY (yλ )}λ and {fY (yλ )}λ are convergent to f1 (y) and f1 (y 0 ) in (f (Y1 ), dY )
and hence in (f (Y ), dY ) . Thus, we can choose λ such that
0
dY (fY (yλ ), fY (y1 )) + dY (fY (yλ ), fY (y 0 )) < δ − dY (fY (y), fY (y 0 )).
0
Then dY (fY (yλ ), fY (yλ )) < δ, ∀λ and consequently
0
dZ (fZ (x · y), fZ (x · y 0 )) = lim dZ (fZ (x · yλ ), fZ (x · yλ )) ≤ ε.
Thus, x → g(x, ε) is upper semicontinuous by Lemma 2.6.
2.
For every x ∈ X and ε < 0, by Theorem 2.8, g(x, ε) > 0.
Therefore, using Theorem 2.9 (which can also be stated in the form that the
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set of points in the nbd. of which the family {fα }α is uniformly bounded
above is a θ-dense θ-open set), to every ε > 0, we can find a θ-dense θ-open
subset Oε of X such that every x0 in Oε has a θ-open nbd. U (x0 , ε) where the
1
function x → g(x,ε)
is bounded, that is, there is a δ(x0 , ε) such that g(x, ε) >
δ(x0 , ε), ∀x ∈ U (x0 , ε). Thus, if x0 ∈ Oε , then dY (fY (y), fY (y 0 )) < δ(x0 , ε)
implies dZ (fZ (x · y), fZ (x · y 0 )) ≤ ε, ∀x ∈ U (x0 , ε). Define R1 = ∩{O1/n |
n = 1, 2, . . .}. The sets O1/n are θ-open and θ-dense in X, so R1 is θ-dense
in X [8] and hence is θ-residual.
3.
Suppose now that x0 ∈ R1 and choose ε > 0. Then, by the
definition of R1 , there is a θ-open nbd. U of x0 in X and δ > 0 such that
dZ (fZ (x · y), fZ (x · y 0 )) ≤ ε, ∀x ∈ U and for all (y, y 0 ) ∈ Y1 × Y1 , with
dY (fY (y), fY (y 0 )) < δ. Since Y1 is an H-set in Y and fY is θ-continuous,
fY (Y1 ) is an H-set [15] in the regular space (Y, dY ) and hence is compact [15].
Further D being θ-dense in Y1 and fY (Y1 ) being regular , fY (D) is dense in
fY (Y1 ). Thus there exists a finite subset F of D with dY (fY (y), fY (F )) <
δ, ∀y ∈ Y1 . Choose a nbd. V of x0 such that clX V ⊂ U (since U is θ-open)
and dZ (fZ (x0 · w), fZ (x · w)) < ε, ∀w ∈ F, ∀x ∈ clX V .
Let y ∈ Y1 be arbitrary. Then for a suitable w ∈ F , we have
dY (fY (y), fY (w)) < δ and therefore
dZ (fZ (x0 , y), fZ (x·y)) ≤ dZ (fZ (x0 ·y), fZ (x0 ·w))+dZ (fz (x0 ·w), fZ (x·w))+
+dZ (fZ (x · w), fZ (x · y)) < ε + ε + ε = 3ε, ∀x ∈ clX V.
Thus, fZ [(clX V ) · y] ⊂ B(fZ (x0 · y), 3ε). Let U be the quasiuniformity
attached to Z and U1 ∈ U be arbitrary. Since Z is almost regular [Theorem
2.7], there exists a θ-open nbd. O of x0 · y such that O ⊂ clZ U1 [x0 · y] [8].
Since fZ is a θ-homeomorphism, fZ (O) is θ-open in (Z, dZ ). Thus there
exists ε > 0 such that B(fZ (x0 · y), 3ε) ⊂ fZ (O). Hence fZ [(clX V ) · y] ⊂
B(fZ (x0 · y), 3ε) ⊂ fZ (O) implying (clX V ) · y ⊂ O ⊂ clZ U1 [x0 · y] (since fZ
is bijective) for an arbitrary U1 ∈ U. This establishes (i).
Obviously, if y0 ∈ Y1 , then x ∈ clX V and dY (fY (y0 ), fY (y)) < δ
implies
dZ (fZ (x0 · y0 ), fZ (x · y)) < dZ (fZ (x0 · y0 ), fZ (x · y0 ))+
+dZ (fZ (x · y0 ), fZ (x · y)) < 3ε + ε = 4ε.
Thus, fZ [clX V ) · fY−1 B(fY (y0 ), δ)] ⊂ B(fZ (x0 · y0 ), 4ε). Since B(fY (y0 ), δ)
is open and thus θ-open in the regular space (Y, dY ), fY−1 B(fY (y0 ), δ) is
θ-open in Y (inverse image of θ-open set under θ-continuous function is θopen). Thus there exists, by definition of θ-open set , an open nbd. W of y0
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in Y with clY W ⊂ fY−1 B(fY (y0 ), δ). Also for any open nbd. O of x0 · y0 in
Z, there exists θ-open nbd. O1 of x0 ·y0 with O1 ⊂clZ O [by almost regularity
of Z]. Further, fZ−1 being θ-continuous, fZ (O1 ) = (fZ−1 )−1 (O1 ) is θ-open in
(Z, dZ ) and thus there exists ε > 0 such that B(fZ (x0 · y0 ), 4ε) ⊂ fZ (O1 ).
Thus, finally, fZ [(clX V ) · (clY W )] ⊂ B(fZ (x0 · y0 ), 4ε) ⊂ fZ (O1 ) implying
clX V ·clY W ⊂ O1 ⊂clZ O, proving the joint θ-continuity at (x0 , y0 ) where
y0 ∈ Y1 is arbitrary.
Corollary. Let S be a θ-metrizable right θ-mob , X be a locally
θ-H-closed subset of S and S1 be an H-set in S. Further let D = {d ∈ S1 |
the map X → S, x → x · d is θ-continuous} be θ-dense in S1 . Then X
contains a subset RX which is θ-residual in X such that
(i) the map X × S1 → X, (x, s) → x · s is jointly θ-continuous at every
point (x0 , s), s0 ∈ RX , s ∈ S1 and
(ii) the maps X → S, x → x · s, s ∈ S are equi-θ-continuous at every
x0 ∈ RX .
Proof. S, being almost regular Hausdorff, s Urysohn. Rest clear.
Lemma 2.11. (Transport argument) If a right θ-mob has an identity
1 and if g ∈ H(1) [H(1) being the maximal subgroup of S containing 1], then
the multiplication µ of S is jointly θ-continuous at (x0 , y0 ) iff it is jointly
θ-continuous at (gx0 , y0 ).
Proof. Obvious.
Lemma 2.12. Let S be a right θ-mob, X be an H-closed Urysohn
space, X1 be an H-set in X and π: S × X1 → X, (s, x) → s · x be a right
θ-continuous action. Assume that (s0 , x0 ) is a point of S × X1 with the
following property C :
To every point z 6= s0 · x0 , there is an element s ∈ S with s · z 6= ss0 · x0 and
such that the restriction of π to sS × X1 is jointly θ-continuous at (ss0 , x0 ).
Then π (unrestricted) is jontly θ-continuous at (s0 , x0 ).
Proof. Let {sλ }λ and {xλ }λ be nets in S and X1 respectively with
θ-lim xλ = x0 (in X1 ) and θ-lim sλ = s0 . Suppose that the net {sλ xλ }
λ
λ
does not θ-converge in X to s0 · x0 . Since X is H-closed, taking subnets
if necessary, we may assume that z = θ-lim(sλ · xλ ) exists (every net in
λ
an H-closed space has a θ-convergent subnet [13]) and z 6= s0 · x0 (since
X is Urysohn). By assumption there is an element s ∈ S such that the
restriction of π to sS × X1 is jointly θ-continuous at (ss0 , x0 ) and such that
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s · z 6= ss0 · x0 . But then s · z = θ-lim[(ssλ ) · xλ ] = ss0 · x0 , contradiction.
λ
Hence the proof.
Lemma 2.13. Let S be a semigroup with identity 1, (X, T ) be an
H-closed Urysohn space and π: X × S → X, (x, s) → x · s be an action with
x · 1 = x, ∀x ∈ X. Furthemore, we assume that there exists y0 ∈ X such
that the translations πs : X → X, x → x · s, s ∈ S are equi-θ-continuous at
y0 . Then x0 is another point of equi-θ-continuity of the maps πs if for every
θ-open nbd. U of y0 there is a unit g ∈ H(1) with x0 · g ∈ U and such that
πg is θ-continuous.
Proof. Let U be the quasi-uniformity attached to X and let V be
the uniformity attached to (X, Tθ ) where Tθ is the θ-open topology on X
corresponding to T [Theorem 2.2]. Let W ∈ U be arbitrary. Now for s ∈ S
arbitrary , W [x0 · s] is an open nbd. of x0 · s in X and since X is almost
regular [8], there exists θ-open nbd. O of x0 ·s with O ⊂clX W [x0 ·s] [8]. Thus
there exists W1 ∈ V with W1 [x0 · s] = O ⊂clX W [x0 · s]. Let us now choose a
symetric W2 ∈ V with W2 ◦W2 ⊂ W1 . Since W2 [y0 ·s] is a θ-open nbd. of y0 ·s,
there exists W3 ∈ U such that clX W3 [y0 · s] ⊂ W2 [y0 · s] (by definition of θopen set). Since the maps πs are equi-θ-continuous at y0 , there exists a nbd,
U1 of y0 in X such that (clX U1 )·s ⊂clX W3 [y0 ·s] ⊂ W2 [y0 ·s], ∀s ∈ S. Let U
be a θ-open nbd. of y0 with U ⊂clX U1 [8]. Then U · s ⊂ W2 [y0 · s], ∀s ∈ S.
Let g ∈ H(1) such that x0 · g ⊂ U and πg is θ-continuous. Then there is
a nbd. V of x0 with (clX V )g ⊂ U by θ-continuity of multiplication and
definition of θ-open set. For every v ∈clX V and every s ∈ S, since x0 · g ∈ U
and v · g ∈ U , we have
(x0 · s, y0 · g −1 s) = ((x0 · g) · g −1 s, y0 · g −1 s) ∈ W2
and
(v · s, y0 · g −1 s) = ((v · g) · g −1 s, y0 · g −1 s) ∈ W2
Thus ∀v ∈clX V, ∀s ∈ S, (v · s, s0 · s) ∈ W2 ◦ W2 ∈ W1 implying (clX V ) · s ⊂
W1 [x0 · s] ⊂ O ⊂clX W [x0 , s], completing the proof.
Lemma 2.14. Let S be a locally θ-H-closed Urysohn right θ-mob
with identity 1, X be an H-closed Urysohn θ-metrizable space, X1 be an
H-set in X and π: S × X1 → X, (s, x) → s · x a right θ-continuous action.
Assume that
(a) X0 = {x ∈ X1 | πx : S → X, s → s · x is θ-continuous } is θ-dense in
X1
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S.GANGULY and T.BANDYOPADHYAY
10
(b) for every pair (x, y) ∈ X1 × X1 , x 6= y, the set of separating elements
D(x, y) = {s ∈ S | s · x 6= s · y} is of θ-2nd-category in S. Then π is jointly
θ-continuous at all (h, x), h ∈ H(1), x ∈ X1 .
Proof. Following the proof of Theorem 2.11, it is sufficient to prove
the assertion for h = 1. By Theorem 2.10, there is a θ-residual suset R1
of S such that π is jointly θ-continuous at every point (s, x) with s ∈ R1
and x ∈ X1 . The intersection of a θ-residual subset with a subset of θ2nd-category cannot be empty, so for every pair (x, y) ∈ X1 × X1 with
1 · x = x 6= y = 1 · y, there is an element s ∈ R1 ∩ D(x, y), that is, π is
θ-continuous at (s · 1, x) = (s, x) and s · x = s · (1 · x) 6= s · y. Thus Lemma
2.12 applies and yields the assertion.
Note. If π is separately θ-continuous, then {s ∈ S | s · x = s · y} =
πx−1 ({s · x}) [where πx : S → X, s → s · x] is θ-closed [πx being θ-continuous
and {s · x} being θ-closed] and hence D(x, y) = S − πx−1 {s · x} is θ-open
and also contains 1. Since a nonempty θ-open subset of a locally θ-H-closed
(and hence of an H-closed) Urysohn space is of θ-2nd category (that is, not
the union of a countable family of nowhere θ-dense sets) [8], D(x, y) is of
θ-2nd category. Also π being left θ-continuous, X0 = X1 . Thus, conditions
(a) and (b) automatically hold if π is separately θ-continuous.
Lemma 2.15. Let S0 be a countable dense subsemigroup of a right
θ-mob S, X be an H-closed Urysohn space and π: S × X → X, (s, x) → s · x
be a separately θ-continuous action. Furthermore, suppose that f : X → R
(R the spaces of reals) is a continuous function. Then
(i) the relation ∼ defined by x ∼ x0 ⇔ f (s · x) = f (s · x0 ), ∀s ∈ S0 is a θe = X/ ∼
closed equivalence relation on X and the θ-quotient space X
e
[that is, X with the largest topology for which the quotient map is
θ-continuous] is θ-metrizable.
(ii) x ∼ x0 ⇒ s · x ∼ s · x0 , ∀s ∈ S. Thus π induces a separately
e → X,
e (s, x̃) → sg
θ-continuous action π̃: S × X
· x.
(iii) Let {xλ }λ be a net in X and x ∈ X. Let x̃λ , x̃ be the image of xλ , x
e Then θ-lim x̃λ = x̃ ⇒
respectively under the θ-quotient map X → X.
λ
lim f (s · xλ ) = f (s · x), ∀s ∈ S0 .
0
Proof. Let S0 = {sn }∞
1 . For each n, define (x, x ) ∈ Rn iff f (sn ·
0
0
x) = f (sn · x ) · Rn is an equivalence ∀n. Let {(xλ , xλ )}λ ⊂ Rn such that
0
0
(x, x0 ) = θ-lim(xλ , xλ ). Thus, f (sn · xλ ) = f (sn · xλ ) ⇒ f (sn · x) = f (sn · θλ
0
lim xλ ) = f (θ-lim(sn ·xλ )) (since π is separately θ-continuous = lim f (sn ·xλ )
λ
λ
λ
11
POINTS OF JOINT θ -CONTINUITY OF MULTIPLICATION
339
(since f is continuous; R being regular, concepts of limit and θ-limit coincide)
= f (sn ·x0 ). Thus, (x, x0 ) ∈ Rn . Thus, Rn is a θ-closed equivalence on X, ∀n.
Hence, ∼= ∩Rn is a θ-closed equivalence on X.
n
Since f is continuous and hence θ-continuous, ψ: X → R defined by
ψ(x) = f (s · x) is θ-continuous, being composition of θ-continuous functions.
e → R by fs ([x]) = f (s · x) =
Further, for each s ∈ S0 , let us define fs : X
e
ψ(x). Let φ: X → (X, G) denote the θ-quotient map [that is, G is the
largest topology for which φ is θ-continuous]. Now we have the commutative
diagram:
ψ
π
S × X −→ X −→ R
|
%
φ | % fs
| %
e
(X, G)
Since ψ = fs ◦ φ is θ-continuous, by Lemma 1.3.12, fs is θ-continuous,
e
∀s ∈ S0 . Let τ be the topology induced by fs , s ∈ S0 , on X.
−1
e τ ). Then U = f (V ), for some s ∈ S0
Let U be any open set in (X,
and for some open and hence θ-open V in R. Since fs is θ-continuous, U is
e τ ). Thus, (X,
e τ ) is regular and hence almost regular.
θ-open in (X,
Next we prove that τ is coarser than G. Let U be any open (θ-open
as well) set in R and s ∈ S0 be arbitrary. Then V = fs−1 (U ) is a subbasic
element of τ . Now ψ −1 (U ) = ψ −1 fs (V ) is θ-open in X ⇒ φ−1 (V ) = ψ −1 (U )
is θ-open in X ⇒ V ∈ G. Every subbasic element of τ being in G, we have
e G) → (X,
e τ ) is continuous and
τ ⊂ G. Thus the identity function i: (X,
e G) being θhence θ-continuous. Further X being H-closed and φ: X → (X,
e G) is H-closed [since θ-continuous onto image of an H-closed
continuous, (X,
space is H-closed, [13]] and thus i is θ-open . Thus i is a θ-homeomorphism.
e Gθ ) be the topology of θ-open sets corresponding to(X,
e G). Let
Let (X,
∞
{Bn }1 be a countable basis of open sets of the real number space R with
usual topology. We next prove {i−1 fs−1 (Bn )}s∈S0 , n = 1, 2, . . . forms a basis
e Gθ ). Let O be θ-open in (X,
e G). Then i(O) is θ-open in
of open sets of (X,
e τ ) and hence fs i(O), s ∈ S0 is θ-open in R. Thus there exists n such that
(X,
e Gθ ) is 2nd countable.
Bn ⊂ fs i(O) and hence i−1 fs−1 (Bn ) ⊂ O. Thus (X,
e
e
Since (X, G) is almost regular, (X, G) is θ-metrizable [this can be proved
following Urysohn’s metrization theorem].
(ii)
Let x ∼ x0 . Then f (s1 , x) = f (s1 , x0 ), ∀s1 ∈ S0 . Let s ∈ S = θclS S0 be arbitrary. Then s = θ-lim sλ , {sλ }λ ⊂ S0 . Now f (s1 sx) = f [θλ
lim(s1 sλ x)] = f [θ-lim(s1 sλ x0 )][ ... s1 sλ ∈ S0 and x ∼ x0 ] = f (s1 sx0 ), ∀s1 ∈
λ
λ
340
S.GANGULY and T.BANDYOPADHYAY
12
S0 . Thus x ∼ x0 implies s·x ∼ s·x0 , ∀s ∈ S and hence π: S×S → X, (s, x) →
e → X,
e (s, x̃) → sg
s · x induces a separately θ-continuous action π̃: S × X
· x.
θ
e
(iii)
Since fs is θ-continuous, x̃λ → x̃ in (X, G) implies fs (x̃λ ) → fs (x̃),
that is, f (s · xλ ) → f (s · x), ∀s ∈ S0 .
Following is our long awaited generalization of the famous ’Theorem
of Ellis–Lawson’:
Theorem 2.16. Let S be a locally θ-H-closed Urysohn right θ-mob
with identity 1, X be an H-closed Urysohn space and π: S × X → X be
a separately θ-continuous action with 1 · x = x, ∀x ∈ X. Further, let
the θ-closure of any countable subsemigroup of S be H-closed. Then π is
jointly θ-continuous at every point (h0 , x0 ) where h0 ∈ H(1) and x0 ∈ X is
arbitrary.
Proof. By Lemma 2.11 , we may assume that h0 = 1. Suppose that
x0 ∈ X and π be not jointly θ-continuous at (1, x0 ). Then there is a θ-open
nbd. U [by definition of joint θ-continuity and due to almost regularity]
of x0 in X such that closure of every nbd. of (1, x0 ) in S × X contains a
point (s, x) with s · x 6∈ U . Let V be a θ-closed nbd. of x0 with V ⊂ U
[by definition of θ-set, since for an open U in a space X, clX U = θ-clX U
and since in an almost regular space, θ-clX A is θ-closed for any A ⊂ X].
Following Urysohn lemma, we can construct a θ-continuous function f : X →
[0,1] such that
1, x ∈ V
f (x) =
0, x ∈ X − U
Let W be a θ-open nbd. of 1 in S with clS W being H-closed (by localθ-H-closedness of S). We first show that it is sufficient to find sequences
{sn }n in clS W and {xn }n in X such that
(i) sn · xn 6∈ U
(ii) the set {sn }n ∪ {1} generates a semigroup S0 such that lim f (ssn · y) =
f (s · y), ∀s ∈ S0 and ∀y ∈ Y0 = {s · xn | s ∈ S0 ; n = 1, 2. . . .}.
To see this, assume that the sequences with these properties have
been found. Then Y = θ-clX Y0 is invariant under the action of S0 and
hence under T = θ-clS S0 . Also T is H-closed by assumption and hence
is a θ-topological semigroup [Let f : X → Y be θ-continuous, X, Y spaces,
f (X) ⊂ Z ⊂ Y and Z be H-closed subspace of Y . Then f : X → Z is θcontinuous [15]]. Thus, Lemma 2.15 can be applied to the induced action
0
e for the θ-quotient space X/ ∼ as defined in Lemma
π : T ×X → X. Write X
2.15 and φ: Y → Ye be the θ-quotient map. Since clS W and X are H-closed
and since every net in an H-closed space has a θ-convergent subnet, there
13
POINTS OF JOINT θ -CONTINUITY OF MULTIPLICATION
341
is a subnet {sm · xm }m of {sn · xn }n such that the θ-limits t = θ-lim sm
m
and x = θ-lim xm exists. Then, by condition (ii) above, since S is right θm
topological and since π: S×X → X is separately θ-continuous, f (st·y) = f [θlim(ssm )·y)] = f [θ-lim(ssm ·y)] = θ-lim f (ssm ·y) = f (s·y), ∀s ∈ S0 , y ∈ Y0 .
m
m
m
Since the map y → f (st · y) (being composition of θ-continuous functions is θ-continuous, hence f (st · y) = f (s · y), ∀y ∈ Y .
Now since Y0 is θ-dense in Y, φ(Y0 ), is θ-dense in φ(Y ) = Ye . Thus
∀ỹ ∈ Ye , ỹ = θ-lim ỹλ where ỹλ ∈ φ(Y0 ) and hence yλ ∈ Y0 . Now since
λ
∀s ∈ S0 , lim f (st · yλ = f (s · y), we have t · ỹ = θ-lim tg
· yλ = ỹ, ∀ỹ ∈ Ye . Now
λ
e and conclude that
we apply Lemma 2.14 to the induced action π: T × Ye → X
e
π is jointly θ-continuous at (t, x̃), t ∈ T, x̃ ∈ Y . But then θ-lim(sm ·xf
m ) = (θm
..
lim sm )·(θ-lim xf
m ) = t· x̃ = x̃ [ . {sm }m ⊂ S0 ⊂ T and xm = 1·xm ∈ Y0 ⊂ Y
m
m
] and thus lim f (1 · sm xm ) = f (1 · x) [ since 1 ∈ S0 ; by (iii), Lemma 2.15]
implying 0 = lim f (sm ·xm ) = f (x) = lim f (xm ) = 1 [... sm ·xm 6∈ U, xm ∈ V ]
, an obvious contradiction.
We now construct the required sequences. By assumption, we can find
x1 ∈ V and s1 ∈ W with s1 · x1 6∈ U . Let S1 = {1, s1 }, X1 = {x1 } and
define inductively elements sn ∈ W, xn ∈ V and finite sets Sn , Xn with
(i) sn · xn 6∈ U
(ii) sn ∈ {w ∈ W | |f (sw · y) − f (s · y)| < n1 , ∀y ∈ Yn−1 , s ∈ Sn−1 , a nbd.
of 1 }
(iii) Sn = {s · t | s, t ∈ Sn−1 ∪ {sn }}
(iv) Yn = {s · y | s ∈ Sn , y ∈ Yn−1 ∪ {xn }} .
∞
∞
S
S
Let S0 =
Sn . Then Y0 =
Yn = {s · xn | s ∈ S0 , n = 1, 2, . . .} is
n=1
n=1
invariant under S0 . Choose (s, y) ∈ S0 × Y0 . Then for all sufficiently large
n, say n > n0 , we have (s, y) ∈ Sn × Yn and therefore, by (ii), |f (ssn · y) −
f (s · y)| < n1 .
Thus, lim f (ssn · y) = f (s · y)and we conclude that condition (ii) is
satisfied.
The following ’Joint θ-continuity Theorem’ is now an easy consequence.
Theorem 2.17.
(Joint θ-continuity Theorem) Let S be an Hclosed Urysohn semi-θ-mob with identity 1. Further, let the θ-closure of
any countable subsemigroup of S [which is an H-set automatically, [4]] be
H-closed. Then the multiplication of S is jointly θ-continuous at all points
(h0 , s0 ) and (s0 , h0 ) where h0 ∈ H(1) and s0 ∈ S is arbitrary.
342
S.GANGULY and T.BANDYOPADHYAY
14
If we observe that in a compact space, the concepts of θ-closedness, Hset (H-closedness) and θ-continuity coincide respectively with the concepts of
closedness, compactness and continuity and that closed subset are automatically compact, then the famous ’Ellis–Lawson Joint continuity Theorem’
[11] comes as a corollary :
Corollary. Let S be a compact semigroup with identity 1 such that
the multiplication of S be separately continuous. Then multiplication of S
is jointly continuous at all (h0 , s0 ) and (s0 , h0 ) where h0 ∈ H(1) and s0 ∈ S
is arbitrary.
3. Some consequences of ’joint θ-continuity theorem’
Theorem 3.1.
Let S be an H-closed Urysohn semi-θ-mob with
identity 1. Let the θ-closure of any countable subsemigroup of S be Hclosed. Then H(1) is H-closed.
Proof. We first prove that inversion is θ-continuous. Let {gλ }λ be a
net in H(1) θ-converging to g ∈ H(1). Since S is H-closed, we may assume
that s = θ-lim gλ−1 exists in S. Now 1 = θ-lim(gλ gλ−1 ) = (θ-lim gλ )(θ-lim gλ−1 )
λ
λ
λ
λ
(by θ-continuity of multiplication) = gs, so that s = g −1 ∈ H(1). Thus,
the inversion H(1) → S is θ-continuous [Let f : X → Y be θ-continuous,
f (X) ⊂ Z ⊂ Y, Z be an H-closed subspace of Y . Then f : X → Z is
θ-continuous [13]]. Again, by corollary to Theorem 2.17, H(1) × S → S is θcontinuous and, since H(1) is H-closed, H(1)×H(1) → H(1) is θ-continuous.
Theorem 3.2. Let X be a semi-θ-mob. Let the θ-closure of any
subsemigroup of S be H-closed. Then every H-closed maximal subgroup
H(e) is a θ-group (that is, a group in which multiplication and inversion are
θ-continuous).
Proof.
The idempotent e is the identity of θ-clS H(e) which is,
by assumption, H-closed. Thus θ-clS H(e) → θ-clS H(e) is a semi-θ-mob.
Further, let T = θ -clθ-clS H(e) T be the θ-closure [and hence is θ-closed
since θ-clS H(e) is H-closed by assumption and thus is almost regular] of any
countable subsemigroup in θ-clS H(e). Since in an H-closed Urysohn space
every θ-closed subset is an H-set [4] and since θ-clS H(e) is H-closed, T is an
H-set in θ-clS H(e) and hence in the superspace S. Thus, T is θ-closed in S
and thus, by assumption , T = θ-clS T is H-closed. Since H(e) is H-closed,
proof is completed by applying Theorem 3.1.
Definition. A commutative semigroup all of whose elements are
idempotents is called a semilattice.
POINTS OF JOINT θ -CONTINUITY OF MULTIPLICATION
15
343
Theorem 3.3. Every H-closed Urysohn semi-θ-topological semilattice S in which every H-set subsemigroup is H-closed is a θ-mob.
Proof. Let x, y ∈ S and assume that {xλ }λ , {yλ }λ are nets in S
with θ-lim xλ = x and θ-lim yλ = y. Choosing suitable subnets if necessary
λ
λ
we may suppose using H-closedness of S that z = θ-lim(xλ yλ ) exists. Since
λ
xλ , yλ x, y, z are commuting idempotents, we have z = z 2 = zθ-lim(xλ yλ ) =
λ
θ-lim[zxλ yλ )(zxλ )]. Now z acts as an identity of the H-closed Urysohn semiλ
θ-mob zS in which every H-set subsemigroup is H-closed (after the proof of
Theorem 3.2). Thus, by Theorem 2.17, the restriction of the multiplication
to zS × zS is jointly θ-continuous at (z, zs); thus z = θ-lim[(zxλ yλ )(zxλ )] =
λ
θ-lim(zxλ yλ ) θ-lim(zxλ ) = zzx = zx. Similarly, x acts as an identity on xS
λ
λ
and therefore z = zx = θ-lim(xλ yλ xλ ) = θ-lim(xxλ ) · θ-lim(xyλ ) = x2 xy =
λ
λ
λ
xy. Hence the proof.
Definition. [8] Let X be a locally-θ-H-closed Urysohn space and
∞ 6∈ X. Then the topological space X̂ = X ∪ {∞} can be show to have
following properties when endowed with the topology : all θ-open sets in X
and the complement in X̂ of each H-set subset of X.
(1) X̂ is H-closed Urysohn
(2) U is θ-open in X ⇒ U is θ-open in X̂.
(3) X̂ is unique upto homeomorphism.
Lemma 3.4. Let S be a locally θ-H-closed Urysohn θ-group. Then
the one-point H-closization [definition above] S∞ = S ∪{∞} is a semi-θ-mob
under the multiplication s · t = st, if s, t ∈ S and = ∞ otherwise.
Proof. We need check separate θ-continuity of multiplication at ∞.
Let s ∈ S and consider s·∞ = ∞. Left θ-continuity at ∞ is clear. Let S −H
be an open nbd. of ∞ in S∞ where H is an H-set in S. Now S − s−1 H is
again an open nbd. of ∞ [since H is an H-set, multiplication in S is right
continuous, continuous image of an H-set is an H-set and since H-set in an
Urysohn space is θ-closed] such that s(S − s−1 H) = sS − ss−1 H = S − H,
proving the right continuity and thus right θ-continuity of multiplication in
S∞ .
Theorem 3.5. Every locally θ-H-closed Urysohn group S with separately θ-continuous multiplication is a θ-group if every H-set subsemigroup
of S is H-closed.
Proof. Follows from Lemma 3.4 and Theorem 3.2.
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S.GANGULY and T.BANDYOPADHYAY
16
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Received: 23.VIII.95
Dept.of Mathematics
Univ.College of Science
35, Ballygunge Circular Road
Calcutta – 700 019
INDIA