This paper explores the structure of ultrafilters on measurable semigroups, demonstrating that the collection of ultrafilters forms a compact right topological semigroup with certain algebraic properties.
Contribution
It introduces the concept of ultrafilters on measurable semigroups and proves their structure as a compact right topological semigroup.
Findings
01
m^eta is a compact right topological semigroup.
02
Ultrafilters on measurable semigroups have specific algebraic properties.
03
The paper establishes foundational properties of m^eta.
Abstract
Let (S,⋅) be a semigroup and m be a σ-algebra on S. We say (S,⋅,m) is a measurable semigroup if π:S×S⟶S by π(x,y)=x⋅y is a measurable function. In this paper , we consider to mβ as the collection of all ultrafilters on m. We show that mβ is a compact right topological semigroup respect to generated topology by σ-algebra m on mβ. Also we study some elementary properties of the algebraic structure of mβ.
Equations49
A={C∈m:thereissomeB∈pwithA∩B⊆C}.
A={C∈m:thereissomeB∈pwithA∩B⊆C}.
(A∪B)∩(D∩C)
(A∪B)∩(D∩C)
⊆(A∩C)∪(B∩D)=∅.
x^={A∈m:x∈A}.
x^={A∈m:x∈A}.
A={p∈mβ:A∈p}.
A={p∈mβ:A∈p}.
g(p)={A∈m2:φ−1(A)∈p}.
g(p)={A∈m2:φ−1(A)∈p}.
g(e(x))
g(e(x))
={A∈m2:x∈φ−1(A)}
={A∈m2:φ(x)∈A}
p⊗q={A∈m⊗m:{s∈X:As∈q}∈p}
p⊗q={A∈m⊗m:{s∈X:As∈q}∈p}
Rq−1(A×B)
Rq−1(A×B)
={s∈S:A×B∈s⊗q}
={s∈S:{u:{t:(u,t)∈A×B}∈q}∈s}
={s∈S:{t:(s,t)∈A×B}∈q}
={s∈S:s∈A,B∈q}=A∈m
{s∈S:ρq(s)∈A}=ρq−1(A)∈p.
{s∈S:ρq(s)∈A}=ρq−1(A)∈p.
{s∈S:s−1A∈q}
{s∈S:s−1A∈q}
={s∈S:A∈ρq(s)}
={s∈S:s∈ρq−1(A)}
{x∈X:f(x)>0}
{x∈X:f(x)>0}
L∘={A∈B:λ(A)=0}
L∘={A∈B:λ(A)=0}
Lk={A∈B:λ(A)>k}.
Lk={A∈B:λ(A)>k}.
L∘β={p∈Lβ:∃A∈pλ(A)=0},
L∘β={p∈Lβ:∃A∈pλ(A)=0},
Lkβ={p∈Lβ:∀A∈pλ(A)>k}
Lkβ={p∈Lβ:∀A∈pλ(A)>k}
L+β={p∈Lβ:∀A∈pλ(A)>0}
L+β={p∈Lβ:∀A∈pλ(A)>0}
{x∈S:−x+A∈p}∈q.
{x∈S:−x+A∈p}∈q.
x∗={p∈Lβ:x∈A∈p⋂clSA}.
x∗={p∈Lβ:x∈A∈p⋂clSA}.
∞∗={p∈Lβ:∀x>0,(x,+∞)∈p}.
∞∗={p∈Lβ:∀x>0,(x,+∞)∈p}.
(0,+∞)=t∈G⋃−t+A.
(0,+∞)=t∈G⋃−t+A.
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Taxonomy
TopicsAdvanced Operator Algebra Research · Advanced Banach Space Theory · Advanced Topics in Algebra
Let (S,⋅) be a semigroup and m be a σ-algebra on S. We say (S,⋅,m) is a measurable semigroup if π:S×S⟶S by π(x,y)=x⋅y is a measurable function.
In this paper , we consider to mβ as the collection of all ultrafilters on m.
We show that mβ is a compact right topological semigroup respect to generated topology by σ-algebra m on mβ. Also we study some elementary properties of the algebraic structure of mβ.
A semigroup S which is
also a Hausdorff topological space is called a semitopological semigroup, if for each s∈S, λs:S→S and rs:S→S are
continuous, where for each x∈S, λs(x)=sx and
rs(x)=xs. If just rs, for each s∈S, is continuous, S is called a right topological semigroup.
A pair (ψ,X) is a semigroup compactification of S if X is a Hausdorff compact right topological
semigroup and ψ:S→X is continuous homomorphism with dense image
such that for all s∈S, the mapping
x↦ψ(s)x:X→X is continuous. For more details
see [6, Section 2].
All semigroup compactifications of a semitopological semigroup as a
collection of z-ultrafilters or e-ultrafilters have been described, see [1], [2], [3] and [4] for more details and some applications. This approach gives us a new light on studying this
kind of compactifications. It seems that the methods presented in
[1], [2], [3] and
[4] can serve as a valuable tool in the study of semigroup compactifications and also of topological compactifications. But this methods depend on concepts of topology on semigroups. For a clear example, the Stone-Cˇech compactification of real additive numbers equipped to the natural topology is not semigroup compactification.
Let (S,⋅) be a semitopological semigroup. In Preliminary for a nonempty set X, we define ultrafiler on a σ-algebra m, is called m-ultrafilter. Also, we study mβ as a collection of all m-ultrafilters on m respect to σ-algebra generated by {A:A∈m}, where A={p∈mβ:A∈p}.
In Section 3, for two measurable spaces (X,m) and (Y,n) we show that if f:(X,m)→(Y,n) is measurable function then there exists a unique measurable extension of f.
In Section 4, we define measurable semigroup and we extend "⋅" naturally to "∗" on mβ. Also, some elementary algebraic properties of (mβ,∗) as extension of measurable semigroup (S,⋅,m) is stated.
In Section 5, we concentrate on the Lebesgue measurable subsets od S=(0,+∞).
2. Preliminary
Let X be a nonempty set and m be an infinite σ-algebra on X. Then (X,m) is called measurable space. We say that m separate X if for each x,y∈X there exist A,B∈m such that x∈A, y∈B and A∩B=∅.
If μ is a measure on m, then (X,m,μ) is a measure space. The outer measure μ∗ can be defined
for every nonnegative measure μ. The collection of μ∗−measurable sets is denoted by mμ. Also μ∗ is a measure on
mμ, (X,mμ,μ∗) is a Lebesgue extension or the Lebesgue completion of the measure μ, and m⊆mμ, see Theorem 1.5.6 in [7].
Let (X,m1) and (X,m2) be measurable spaces. The function f:X→Y is an
(m1,m2)-measurable function if f−1(A)∈m1 for every A∈m2.
Let (Y,τ) be a topological space. The σ-algebra generated by τ is called the Borel σ-algebra and is denoted by Bτ.
The function f:(X,m)→(Y,τ) is an
m-measurable function if f−1(A)∈m for every A∈τ.
Let (X,A) and (Y,B) be two measurable spaces. For X×Y, we consider the collection of all sets of the from A×B , where A∈A and B∈B, called measurable rectangle. The σ-algebra generated by all measurable rectangles is called product of the σ-algebras A and B, and is denoted by A⨂B. Let μ and ν be measures on A and B, respectively. Define μ×ν(A×B)=μ(A)ν(B) for each measurable rectangles A×B.
Theorem 2.1**.**
The set function μ×ν extends to a countably additive measure, denoted by μ⨂ν, on A⨂B.
Let (X1,A1) and (X2,A2) be measurable spaces. Then the set
Ax1={x2∈X2:(x1,x2)∈A}∈A2 for every A∈A1⊗A2 and x1∈X1.
Also the set
Ax2={x1∈X1:(x1,x2)∈A}∈A1 for every A∈A1⊗A2 and x2∈X2.
Let X be a nonempty set and m be an infinite σ-algebra on X. We say p⊆m is an m-filter if
(a)∅∈/p and X∈p.
(b) If A,B∈m, A∈p, and A⊆B, then B∈p.
(c) If A,B∈p, then A∩B∈p.
An m-ultrafilter is an m-filter which is not properly contained in
any other m-filter. Zorn’s Lemma guarantees that there exist mβ-ultrafilters.
Lemma 2.4**.**
Let p be an m-filter and A∈m. Either
(a)* there is some B∈p such that A∩B=∅ or*
(b)* {C∈m:thereissomeB∈pwithA∩B⊆C} is an m-filter.*
Proof.
If there is (a), then the statement is obvious.
If (a) is false, then A∩B=∅ for every B∈p. Let
[TABLE]
It is obvious that A∈A, ∅∈/A, and X∈A. If C∈A and C⊆C′∈m, then C′∈A. Also if C,D∈A, then there exist B1,B2∈p such that B1∩A⊆C and B2∩A⊆D. So
(B1∩B2)∩A⊆D∩C and hence B1∩B2∈p Since p is an m-filter. So D∩C∈A and this implies
that A is an m-filter.
∎
Theorem 2.5**.**
Let m be an infinite σ-algebra on X and p⊆m be an m-filter. The following statements are
equivalent:
(a)* p is an m-ultrafilter.*
(b)* for all A,B∈m, if A∪B∈p then A∈p or B∈p.*
(c)* for all A∈m either A∈p or Ac∈p.*
(d)* for each A∈m∖p there is some B∈p such that A∩B=∅.*
Proof.
(a) implies (b). Let A,B∈m such that A∪B∈p. If B∈/p and A∈/p, then there are
C,D∈p such that A∩C=∅ and by Lemma 2.4, B∩D=∅. So
[TABLE]
Then ∅∈p, is contradiction.
(b) implies (c). It is obvious.
(c) implies (d). Let A∈m such that A∈/p. Then Ac∈p so A∩Ac=∅.
(d) implies (a). Let p be an m-filter and let A be an m-ultrafilter such that p⊆A.
Pick A∈A∖p. Then A∈/p so there is a some B∈p such that A∩B=∅.
But ∅=A∩B∈A, a contradiction.
∎
Definition 2.6**.**
Let X be a nonempty set and m be an infinite σ-algebra on X. For x∈X, we define
[TABLE]
and x^ is called m-principal ultrafilter.
It is obvious that p is the m-principal ultrafilter if and only if ⋂p=∅.
Theorem 2.7**.**
Let X be a nonempty set, m be an infinite σ-algebra on X, and A⊆m has the finite
intersection property. Then there is an m-ultrafilter p such that A⊆p.
Proof.
By Zorn’s Lemma is obvious.
∎
Definition 2.8**.**
Let X be a nonempty set, let m be an infinite σ-algebra on X, and let R be a nonempty set such that R⊆m. We say that R is partition regular
if and only if whenever A,B∈m and A∪B∈R, there exists C∈R
such that C⊆A or C⊆B.
Theorem 2.9**.**
Let X be a nonempty set, let m be an infinite σ-algebra on X, and let R be a nonempty set such that R⊆m and assume that ∅∈/R.
Let R↑={B∈m:A⊆B\mboxforsomeA∈R}.
The following statements
are equivalent:
(a)* R is partition regular.*
(b)* Whenever A⊆m has the property that every finite nonempty subfamily of
A has an intersection which is in R↑, there is an m-ultrafilter p⊆m such that
A⊆p⊆R↑.*
(c)* Whenever A∈R, there is some m-ultrafilter p such that A∈p⊆R↑.*
Proof.
(a) implies (b). Let B={A∈m:∀B∈R,A∩B=∅} and note that
B=∅. We may assume that A=∅, since {X} has the
hypothesized property. Let C=A∪B. We claim that C has the finite intersection
property. To see this it suffices (since A and B are nonempty) to let F∈Pf(A) and
G∈Pf(B) and show that ⋂F∩⋂G=∅. So suppose instead that we have such F
and G with ⋂F∩⋂G=∅. Pick B∈R such that B⊆⋂F.
Then B∩⋂G=∅
and so B=⋃A∈G(B∖A). Pick A∈G and C∈R such that C⊆B∖A. Then
A∩C=∅, contradicting the fact that A∈B.
By Theorem 2.7, there is an m-ultrafilter p such that C⊆p. Given C∈p,
X∖C∈/B (since C∩(X∖C)=∅∈/p). So pick some B∈R such that B∩(X∖C)=∅.
That is, B⊆C.
(b) implies (c). Let A={A}.
(c) implies (a). Let F be a finite set of elements of m with ⋃F∈R and let p be an
m-ultrafilter such that ⋃F∈p and for each C∈p there is some B∈R such
that B⊆C. Pick by Theorem 2.5 some A∈F∩p.
∎
Corollary 2.10**.**
Let (X,m) be a measurable space. Let A⊆m be an arbitrary
family. If the intersection of every finite subfamily of A is infinite, then A is contained in an m-ultrafilter
all of whose members are infinite. More generally, if κ is an infinite cardinal
and if the intersection of every finite subfamily of A has cardinality at least κ, then there exists
m-ultrafilter p such that ∣A∣≥κ for every A∈p.
Proof.
The proof is obvious.
∎
Let (X,m) be a measurable space. The collection of all m-ultrafilters
is denoted by mβ. For each A∈m, we define
[TABLE]
Lemma 2.11**.**
Let X be a nonempty set, let m be an infinite σ-algebra on X, and let A,B∈m.
(a)* A∩B=A∩B;*
(b)* A∪B=A∪B;*
(c)* X∖A=mβ∖A;*
(d)* A=∅ if and only if A=∅;*
(e)* A=mβ if and only if A=X;*
(f)* A=B if and only if A=B.*
Proof.
The proof is obvious.
∎
The collection {A:A∈m} is a basis for topology on mβ.
Theorem 2.12**.**
Let X be a nonempty set, let m be an infinite σ-algebra on X. Then mβ is a compact and Hausdorff space.
Proof.
Suppose that p and q are distinct elements of mβ. If A∈p∖q, then
X∖A∈q. So A and X∖A are disjoint open subsets of mβ containing p and q, respectively. Thus mβ is Hausdorff.
To show that mβ is compact, we shall consider a family
A of sets of the form A with the finite intersection property and show that A has a
nonempty intersection. Let B={A∈m:A∈A}. If F∈Pf(B), then there is
some p∈⋂A∈FA and so ⋂F∈p and thus ⋂F=∅. That is, B has the finite
intersection property, so by Theorem 2.7 pick q∈mβ with B⊆q. Then q∈⋂A.
∎
Theorem 2.13**.**
Let X be a nonempty set and let m be an infinite σ-algebra on X. Then the sets of the form A are the clopen
subsets of mβ.
Proof.
Each set A is closed as well as open.
Suppose that C is any clopen subset of mβ. Let A={A:A∈mandA⊆C}.
Since C is open, A is an open cover of C. Since C is closed, it is compact by Theorem 2.5(b). Now by Theorem 2.12, pick a finite subfamily F of m such that C=⋃A∈FA. Then by Lemma 2.11, C=⋃F.
∎
Theorem 2.14**.**
Let X be a nonempty set and let m be a σ-algebra on X containing Pf(X).
(a)* For every A∈m, A=clmβe(A) where e:X→mβ is defined by e(x)=x^.*
(b)* For any A∈m and any p∈mβ, p∈clmβe(A) if and only if A∈p.*
Proof.
The proof is obvious.
∎
Theorem 2.15**.**
Let X be a nonempty set and let m be a σ-algebra on X and m separate X. Let e:X→mβ is defined by e(x)=x^.
(a)* The mapping e is m-measurable.*
(b)* If (X,m) is measurable space and let Pf(X)⊆m, then the mapping e is injective.*
(c)* e(X) is a dense subset of mβ.*
Proof.
(a) Since A=e−1(A) for every A∈m, so the mapping e is m-measurable.
The proofs (b) and (c) are obvious.
∎
Theorem 2.16**.**
Let X be a nonempty set and let m be a σ-algebra on X containing Pf(X). If U is an open subset of mβ,
clmβU is also open.
Proof.
If U=∅, the conclusion is trivial and so we assume that U=∅. Put
A=e−1[U]. We claim first that U⊆clmβe(A). So let p∈U and let B be a basic
neighborhood of p. Then U∩B is a nonempty open set and so by Theorem 2.15(c), U∩B∩e(X)=∅.
So pick b∈B with e(b)∈U. Then e(b)∈B∩e(A) and so B∩e(A)=∅.
Also e[A]⊆U and hence U⊆clmβe(A)⊆clmβU. By Theorem 2.14(a),
clmβU=clmβe(A)=A,
and so clmβU is open in mβ.
∎
Definition 2.17**.**
Let X be a nonempty set, let m be a σ-algebra on X containing Pf(X), and let A be an m-filter.
We define A={p∈mβ:A⊆p}.
Theorem 2.18**.**
Let X be a nonempty set and let m be a σ-algebra on X containing Pf(X).
(a)* If A is an m-filter, then A is a closed subset of mβ.*
(b)* If ∅=A⊆mβ and A=⋂A, then A is an m-filter and A=clmβA.*
Proof.
(a) Let p∈mβ∖A. Pick B∈A∖p. Then X∖B is a neighborhood of p
which misses A.
(b)A is the intersection of a set of m-filters, so A is an m-filter. Further, for each p∈A,
A⊆p so A⊆A and thus by (a),
clmβA⊆A. To see that A⊆clmβA,
let p∈A and let
B∈p. Suppose B∩A=∅. Then for each q∈A, X∖B∈q so X∖B∈A⊆p,
a contradiction.
∎
Theorem 2.19**.**
Let X be a nonempty set, let m be a σ-algebra on X containing Pf(X), p∈mβ, and U⊆mβ.
If U is a neighborhood of
p in mβ, then e−1[U]∈p.
Proof.
If U is a neighborhood of p, there is a basic open subset A of mβ for which
p∈A⊆U. This implies that A∈p and so e−1[U]∈p, because A⊆e−1[U].
∎
3. **More Topology of mβ **
Now let (X,m1) and (Y,m2) be two measurable spaces such that Pf(X)⊆m1 and Pf(Y)⊆m2. Also let φ:X→Y be an
(m1,m2)-measurable function. In Lemma 3.1, we show that φ has a unique extension.
Lemma 3.1**.**
Let (X,m1) and (Y,m2) be measurable spaces such that Pf(X)⊆m1 and Pf(Y)⊆m2. Let φ:(X,m1)→(Y,m2) be an
(m1,m2)-measurable function. Then there exists a continuous function
φβ:m1β→m2β such that
(a)* φβ(p)={A∈m2:φ−1(A)∈p}.*
(b)* eY∘φ = φβ∘eX.*
(c)* If A∈p∈m1β, then φ(A)∈φβ(p) and if B∈φβ(p), then φ−1(B)∈p.*
Proof.
(a) It is obvious that {A∈m2:φ−1(A)∈p} is an m2-ultrafilter. For every p∈m1β, let
[TABLE]
Then for every x∈X,
[TABLE]
So g(e(x))=φ(x).
To see that g is continuous, let A∈m2. Then
(g∘e)−1(A)=φ−1(A).
Since g is a continuous extension
of φ, we have φ=g.
The proofs (b) and (c) are obvious.
∎
Theorem 3.2**.**
Let (X,m) be a measurable space, such that Pf(X)⊆m.
Then every Gδ-subset of X∗ has nonempty interior in X∗.
Proof.
Choose p∈⋂n∈N(Un∩X∗), where Un is an open subset of
mβ for each n∈N. Choose a subset An of X such that
p∈An∗⊆Un and An+1⊆An for each n∈N. Now choose an infinite sequence
{an}n=1∞ of distinct points of X such that an∈An. Let A={an:n∈N}.
It is obvious that A∗ is an open subset of X∗ and
A∗⊆Un for each n∈N.
∎
Theorem 3.3**.**
Let (X,m) be a measurable space, such that Pf(X)⊆m.
Then every countable union of nowhere dense subsets of X∗ is nowhere dense in X∗.
Proof.
Let An be a nowhere dense subset of X∗ for each n∈N. We show that
B=⋃n∈NclAn is nowhere dense. Suppose instead that U=intX∗clB=∅. By
the Baire Category Theorem X∗∖B is dense in X∗ so U∖B=∅. Thus U∖B is a
nonempty Gδ-set which thus by Theorem 3.2, has nonempty interior. This is a contradiction.
∎
4. ** Measurable Semigroups**
In this section we extend the operation of a measurable semigroup.
Definition 4.1**.**
(a) A right measurable semigroup is a triple (S,⋅,m) where (S,⋅) is a semigroup, (S,m) is a measurable space, and ρx:S→S is (m,m)-measurable function for each x∈S where ρx(y)=yx for each y∈S.
(b) A left measurable semigroup is a triple (S,⋅,m) where (S,⋅) is a semigroup, (S,m) is a measurable space, and λx:S→S is (m,m)-measurable function for each x∈S where λx(y)=xy for each y∈S.
(c) A measurable semigroup is a triple (S,⋅,m) where (S,⋅) is a semigroup, (S,m) is a measurable space, and π:S×S→S is (m⊗m,m)-measurable function where π(x,y)=xy.
In this section we assume that Pf(S)⊆m, so {s} belongs to m for each s∈S.
Definition 4.2**.**
Let (X,m) be a measurable space.
The tensor product of two m-ultrafilters p and q in mβ is denoted by p⊗q and is defined as following
[TABLE]
where As={t∈S:(s,t)∈A}.
Lemma 4.3**.**
If p,q∈mβ, then p⊗q∈(m⊗m)β.
Proof.
The proof is obvious.
∎
Lemma 4.4**.**
The following statements hold:
(a)* S⊗S is dense in (m⊗m)β.*
(b)* lims→plimt→qs⊗t=p⊗q for every p,q∈mβ.*
Proof.
The proof is obvious.
∎
Lemma 4.5**.**
The mapping Rq:S→(m⊗m)β by Rq(s)=s⊗q is a
(m,m⊗m)-measurable function where q∈mβ.
Proof.
Since m×m generates m⊗m, then
[TABLE]
where A,B∈m.
∎
Lemma 4.6**.**
Let s∈S and q∈mβ. Then πβ∘Rq(s)=πβ(s⊗q)=s∗q
Proof.
The proof is obvious.
∎
Theorem 4.7**.**
Let (S,⋅,m) be a measurable semigroup. Then there is a unique binary
operation ∗:mβ×mβ→mβ satisfying the following three
conditions:
(a)* For every s,t∈S, s∗t=s⋅t.*
(b)* For each q∈mβ, the function ρq:mβ→mβ is continuous,
where ρq(p)=p∗q.*
(c)* For each s∈S, the function λs:mβ→mβ is continuous,
where λs(q)=s∗q.*
Proof.
Given any s∈S, define
ls:S→S by ls(t)=st, then ls is a measurable function. So there is a unique continuous function λs:mβ→mβ such that λs(e(t))=ls(t) for each t∈S. If s∈S and q∈mβ, we define
s∗q=λs(q). Then (c) holds and so does (a), because λs extends ls.
Now we extend ∗ to the rest of mβ×mβ. Given q∈mβ, define
rq=πβ∘Rq:S→mβ. The mapping rq is measurable, so there is a unique continuous extension ρq:mβ→mβ. If p,q∈mβ, we define
p∗q=ρq(p). This is the only possible definition
which satisfies the required conditions.
∎
Theorem 4.8**.**
Let (S,⋅,m) be a measurable semigroup. Then the extended operation on mβ is associative.
Proof.
The proof is obvious.
∎
Theorem 4.9**.**
Let (S,⋅,m) be a measurable semigroup. Then (mβ,∗) is a compact right semigroup.
Proof.
The proof is obvious.
∎
Theorem 4.10**.**
Let (S,⋅,m) be a measurable semigroup and A∈m.
(a)* For any s∈S and q∈mβ, A∈s⋅q if and only if s−1A∈q.*
(b)* For any p,q∈mβ, A∈p⋅q if and only if {s∈S:s−1A∈q}∈p.*
Proof.
(a) Necessity. Let A∈s⋅q. Then A∈λs(q) and hence by Lemma 3.1, s−1A=λs−1(A)∈q.
Sufficiency. Assume s−1A∈q and suppose that A∈/s⋅q. Then S∖A∈s⋅q.
So, by the already established necessity, s−1(S∖A)∈q. This is a contradiction since
s−1A∩s−1(S∖A)=∅.
a) Let L denote the collection of all Lebesgue measurable subsets od real numbers. Then
(Lβ,+) and (Lβ,⋅) are measurable semigroup compactification of (R,L).
b)Every vector space has a Hamel basis, i.e. a maximal linearly independent subset. Since Real numbers as a vector space on rational numbers has infinite dimension, therefore there exists an uncountable maximal linearly independent I⊆[0,1]. Now choice a countable subset A⊆I. Then the generated vector space by I∖A is not closed. Define i(x)=x when x∈A and i(x)=0 for x∈I∖A. We denote i as to extension i to R as a linear transformation. It is obvious that i is Borel measurable but it is not continuous.
Now define ∗i:R×R→R by ∗i(x,y)=i(x+y) for each x,y∈R.
Then (R,∗i) is measurable semigroup, but it is not semitopological semigroup.
Theorem 4.14**.**
Let (S,⋅,m) be a measurable semigroup. Then S∗=mβ∖S is a
subsemigroup of mβ if and only
if for any A∈Pf(S) and for any infinite subset B of m there exists F∈Pf(B) such
that ⋂x∈Fx−1A is finite.
Proof.
Necessity. Let A∈Pf(S) and an infinite subset B∈m be
given. Suppose that for each F∈Pf(B), ⋂x∈Fx−1A is infinite. Then {x−1A:x∈B}
has the property that all of its finite intersections are infinite so by Corollary 2.10
we may pick p∈S∗ such that {x−1A:x∈B}⊆p. Pick q∈S∗ such that B∈q.
Then A∈q⋅p and A is finite so q⋅p∈S, a contradiction.
Sufficiency. Let p,q∈S∗ be given and suppose that q⋅p=y∈S, (that is,
precisely, that q⋅p is the m-principal ultrafilter generated by y). Let A={y} and let
B={x∈S:x−1A∈p}. Then B∈p while for each F∈Pf(B), one has
⋂x∈Fx−1A∈p so that ⋂x∈Fx−1A is infinite, a contradiction.
∎
Definition 4.15**.**
Let S be a measurable semigroup. We say that
S is weakly left cancellative if and only if λs−1({y}) is finite for every x,y∈S.
Theorem 4.16**.**
Let (S,⋅,m) be a measurable semigroup. Then S∗ is a left ideal of mβ if and
only if S is weakly left cancellative.
Proof.
Necessity. Let x,y∈S be given, let A=λs−1({y}) and suppose that A is
infinite. Pick p∈S∗∩A. Then y⋅p=x, a contradiction.
Sufficiency. Since S is infinite, S∗=∅. Let p∈S∗, let q∈mβ and suppose
that q⋅p=x∈S. Then {x}∈q⋅p so {y∈S:y−1{x}∈p}∈q and is
hence nonempty. So pick y∈S such that y−1{x}∈p. But y−1{x}=λy−1({x}) so
λy−1({x}) is infinite, a contradiction.
∎
Theorem 4.17**.**
Let (S,⋅,m) be a measurable semigroup. The following statements are equivalent:
(a)* S∗ is a right ideal of mβ.*
(b)* Given any finite subset A∈m, any sequence ⟨zn⟩n=1∞ in S, and any one-to-one
sequence ⟨xn⟩n=1∞ in S, there exist n<m in N such that xn⋅zm∈/A.*
(c)* Given any a∈S, any sequence ⟨zn⟩n=1∞ in S, and any one-to-one sequence
⟨xn⟩n=1∞ in S, there exist n<m in N such that xn⋅zm=a.*
Recall a σ-algebra on a topological space X is the Borel σ-algebra
generated by all open sets; it is denoted by B(X). A set in B(X) is called the Borel set in the space X.
A σ-algebra on a topological space X is generated by all sets of the form
[TABLE]
where f is a real valued continuous function on X, is called Baire
σ-algebra and denoted by Ba(X). So A∈Ba(X) is called a Baire set. By Corollary 5.3.5 in [7],
when X is a metric space then
A⊆X is Borel set if and only if A is a Baire set.
Let L denote the collection of all Lebesgue measurable sets on S=(0,+∞), and let λ be the Lebesgue measure on real numbers.
It is obvious that B(S)⊆L. For each p,q∈Lβ, p+q and p⋅q are well define, see Theorem 4.10. So Lβ is a multiplicative and additive semigroup.
Now define
[TABLE]
and for k∈(0,1] define
[TABLE]
We define L+=⋃k>∘Lk. It is obvious that L0, Lk and Lk are partition regular,
and also L0, Lk and L+ are invariant under translate, where A+x is
the translate of A by x∈S. Therefore the below collections
[TABLE]
[TABLE]
and
[TABLE]
are non-empty sets, by Theorem 2.9.
Let m be a σ-algebra on X and let A∈m, then mA={T∩A:T∈m}
is a σ-algebra and we can write mAβ⊆mβ, because
for each pA∈mAβ there exists a unique p∈m such that pA⊆p.
Lemma 5.1**.**
*a) Lkβ and L+β are left ideals of (Lβ,+). Also L+β is left ideal of (Lβ,⋅).
b) L∘β is a multiplicative and additive subsemigroup of Lβ.*
Proof.
a) Pick p∈Lkβ and q∈Lβ. Let A∈q+p, so
[TABLE]
Since p∈Lkβ so λ(A)=λ(−x+A)>k. This implies q+p∈Lkβ.
By similar way, L+β is left ideal of (Lβ,+).
b) Pick p,q∈L∘β, so there exist Ap∈p and Aq∈q
such that Ap∩Aq=∅ and λ(Ap)=λ(Aq)=0. Now Let A=Ap∪Aq, and define
p∘={T∩A:T∈p} and q∘={T∩A:T∈q}. It is obvious that p∘ and q∘
are L-ultrafilters on A, p∘⊆p and q∘⊆q. It is obvious that
p∘+q∘⊆p+q. By similar way, we can show that L∘β is a multiplicative subsemigroup.
∎
For x∈S, we define
[TABLE]
It is obvious that x∗={p∈Lβ:∀y>0,(x−y,x+y)∈p}, and x∈x∗. We say p∈x∗ is
a near point to x. We define B(S)=∪x∈Sx∗ and ∞∗=Lβ−B(S). An element p∈B(S) is called a bounded ultrafilter and p∈∞∗ is called an unbounded ultrafilter. It is obvious that
By above Lemma, it is obvious that 0∗ is a compact multiplicative and additive subsemigroup of Lβ.
Lemma 5.3**.**
∞∗* is a multiplicative and additive subsemigroup of Lβ.*
Proof.
Let p,q∈∞∗ and let pq∈x∗ for some x∈S. Therefore (a,b)∈pq for some a,b∈S, and so for some t∈S, t−1(a,b)∈q is a contradiction. So ∞∗ is a multiplicative subsemigroup. By similar way, ∞∗ is an additive semigroup.
∎
We say that A⊆S is meager if and only if A is the countable union of
nowhere dense sets. Now we state a relation between Baire and meager sets.
Definition 5.4**.**
a) A Lebesgue measurable set A⊆S is called Baire large at ∞ if and only if
for each x>0, A∩(x,+∞) is not meager.
The collection of all Baire large sets at ∞ is denoted by BL(∞).
b) A Lebesgue measurable set A⊆S is called Baire large at [math] if and only if for each x>0, A∩(0,x) is not meager. The collection of
all Baire large sets at [math] is denoted by BL(0).
Lemma 5.5**.**
*a) BL(∞) and BL(0) are partition regular.
b) BL(0)={p∈Lβ:∀A∈p,A∈BL(0)} is left ideal in (0∗,⋅).
c) BL(∞)={p∈Lβ:∀A∈p,A∈BL(∞)} is left ideal in (∞∗,⋅).
d) BL(∞) is left ideal in (∞∗,+).*
Proof.
a) It is obvious.
b) By a) and Theorem 2.9 implies BL(0) is non empty. Now similar Lemma 17.39 in [9], let p∈BL(0), q∈0∗ and let
A∈qp. Pick x∈(0,1) such that x−1A∈p. Let x−1A∩(0,ϵ) is not meager for some ϵ>0, so
x(x−1A∩(0,ϵ)) is not meager and x(x−1A∩(0,ϵ))⊆A∩(0,ϵ) .
c) and d). The proof is similar to b).
∎
Definition 5.6**.**
a) A⊆(0,+∞) is called syndetic if and only if there exists some G∈Pf(0,+∞) such that
[TABLE]
b) A⊆(0,+∞) is called thick if and only if for each F∈Pf((0,+∞)) there
exists x∈A such that x+F⊆A.
c) A⊆(0,+∞) is called piecewise syndetic if and only if ⋃t∈G−t+A is thick for some G∈Pf(0,+∞).
Theorem 5.7**.**
a) Let p∈Lβ. Then p∈K(Lβ,+) if and only if {x>0:−x+A∈p} is syndetic for each A∈p.
b) Let A⊆(0,+∞). Then K(Lβ,+)∩A=∅ if and only if A is piecewise syndetic.
Proof.
The proof is similar to the proof of Theorem 4.39 and 4.40 in [9].
∎
The following Theorem has been stated as Theorem 1.3 in [8].
Theorem 5.8**.**
The closure of the minimal ideal K(Lβ,+) is a left ideal of (Lβ,⋅). In particular there is
a multiplicative idempotent in clLβK(Lβ).
Proof.
Let p∈clLβK(Lβ,+) and q∈Lβ. We show that qp∈clLβK(Lβ,+). So let A∈qp.
Then {x>0:x−1A∈p}∈q. Pick x>0 such that x−1A∈p. Since p∈clLβK(Lβ), so x−1A is piecewise syndetic.
Since x−1A is a neighborhood of p, so there exists an η∈x−1A∩K(Lβ,+). This implies that
x−1A∈η, and hence A∈xη. Now Let lx:(0,∞)→(0,+∞) is defined by lx(y)=xy. It is obvious that lx is an additive isomorphism. So lx has a unique continuous isomorphism extension lxβ:Lβ→Lβ. Therefore xK(Lβ,+)=K(Lβ,+) and so A∈xη is piecewise syndetic. This implies that A∩K(Lβ,+)=∅ for each A∈qp. So qp∈clLβK(Lβ,+).
∎
Bibliography9
The reference list from the paper itself. Each links out to its DOI / PubMed record.
1[1] T. Alaste, 𝒰 𝒰 \mathcal{U} -filters and uniform compactification, Studia Math. 211 , 215-229 (2012).
2[2] T. Alaste, A note on κ 𝜅 \kappa -uniform points of G ℒ 𝒰 𝒞 superscript 𝐺 ℒ 𝒰 𝒞 G^{\mathcal{LUC}} , Semigroup Forum, 87 , 489-493,(2013).
3[3] A. Akbari Tootkaboni, Lmc-compactification of a semitopological semigroup as a space of e-ultrafilters, New York J. Math 19 (2013), 669-688.
4[4] M. Akbari Tootkaboni and A. Riazi, Ultrafilters on Semitopological Semigroup, Semigroup Forum 70 (3), (2005), 317-328.
5[5] A. Akbari Tootkaboni and T. Vahed The semigroup of ultrafilters near an idempotent of a semitopological semigroup, Topology and its Application, 159 (2012), 3494-3503.
6[6] J. F. Berglund, H. D. Junghenn and P. Milnes, Analysis on Semigroups: Function Spaces, Compactifications, Representations, Wiley, New York., 1989.
7[7] V.I. Bogachev, Measure Theory , Springer-Verlag Berlin Heidelber, (2007).
8[8] A. T. Lisan , Ultrafilters on a discrete set with two binary operations, Semigroup Forum 43 , (1991), 77-81.