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Borel $sigma$-algebra of the order topology $textit{vs.}$ $sigma$-algebra generated by the topological...
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This is a proof-verification (or more like counterexample-verification) request.
Let $Omega$ be a non-empty set and $geq$ a linear order (a complete, transitive, antisymmetric order) on it. Define
begin{align*}
mathscr Lequiv&;big{{psiinOmega,|,psi<omega},big|,omegainOmegabig},\
mathscr Uequiv&;big{{psiinOmega,|,psi>omega},big|,omegainOmegabig}
end{align*}
to be the family of lower sets and that of upper sets, respectively.
The order topology $tau$ on $Omega$ is the topology which has $mathscr Lcupmathscr U$ as its topological subbasis; it is the coarsest topology according to which all lower and upper sets are open. From this topology, one can generate the Borel $sigma$-algebra $mathscr Bequivsigma(tau)$, which is the smallest $sigma$-algebra according to which sets open according to the order topology are measurable.
My goal is to show that, in general,
begin{align*}
mathscr Bneqsigma(mathscr Lcupmathscr U);tag{$clubsuit$}
end{align*}
that is, the $sigma$-algebra generated by the lower and upper sets is not large enough to give the Borel $sigma$-algebra.
Remark: In some special cases, $mathscr B$ and $sigma(mathscr Lcup mathscr U)$ do coincide—for example, when $Omega=mathbb R$ with the usual order, in which case the order topology is the standard Euclidean topology, whose Borel $sigma$-algebra is well-known to be generated by $mathscr Lcupmathscr U$ (and even by $mathscr L$ alone or by $mathscr U$ alone). More generally, if the order topology $tau$ is second-countable, then $mathscr Lcupmathscr U$ does generate $mathscr B$; the details can be found here. However, as a rule, the two $sigma$-algebras $mathscr B$ and $sigma(mathscr Lcupmathscr U)$ may be different.
In particular, I have the following supposed counterexample in mind. Let $(Omega,geq)$ be the set of countable ordinals. This is a linearly ordered set such that
$Omega$ is uncountable; but- each lower set in $mathscr L$ is countable; furthermore,
- every non-empty subset of $Omega$ has a least element with respect to the order $geq$.
The first step is to show that
begin{align*}
sigma(mathscr Lcupmathscr U)=mathscr C,
end{align*}
where $mathscr C$ is the $sigma$-algebra of countable and co-countable sets:
begin{align*}
mathscr Cequiv{AsubseteqOmega,|,text{$A$ is countable or $A^{mathsf c}$ is countable}}.
end{align*}
Clearly, each set in $mathscr L$ is countable and each set in $mathscr U$ is co-countable, so that $sigma(mathscr Lcupmathscr U)subseteqmathscr C$. Conversely, note that for each $omegainOmega$, the singleton ${omega}$ is the intersection of the complement of a set from $mathscr L$ and the complement of another set from $mathscr U$. Therefore, the singletons are in $sigma(mathscr Lcupmathscr U)$, which implies that $mathscr C$ is included in $sigma(mathscr Lcupmathscr U)$.
Therefore, in order to prove ($clubsuit$), it is sufficient to exhibit a set that is open in the order topology (and hence a fortiori Borel measurable) but is neither countable nor co-countable. To this end, define the immediate-successor function $S:OmegatoOmega$ as follows: for each $omegainOmega$, $S(omega)inOmega$ is the (unique) element of $Omega$ such that
$S(omega)>omega$; and- there exists no $psiinOmega$ such that $S(omega)>psi>omega$.
More formally, $S(omega)$ is the least element of the upper set ${psiinOmega,|,psi>omega}$.
It is easily seen that for each $omegainOmega$, one has
begin{align*}
{S(omega)}={psiinOmega,|,psi<S(S(omega))}cap{psiinOmega,|,psi>omega},
end{align*}
so that the set
begin{align*}
Uequiv S(Omega)=bigcup_{omegainOmega}{S(omega)}
end{align*}
is open in the order topology. What is more, since $U$ is a union of open singletons, any subset of $U$ is open, too. In addition, the function $S$ is injective, so that $U$ is uncountable. Therefore, $U$ can be partitioned into two disjoint uncountable subsets, say $V$ and $W$. Hence, $V$ is an uncountable open set and since $Wsubseteq V^{mathsf c}$, the complement of $V$ is also uncountable. It follows that $Vnotinsigma(mathscr Lcupmathscr U)$. This observation concludes the construction of the desired counterexample.
Any comments and suggestions are greatly appreciated.
general-topology measure-theory proof-verification order-theory
asked 14 hours ago
triple_sec
15.5k21850
add a comment |
up vote
1
down vote
favorite
This is a proof-verification (or more like counterexample-verification) request.
Let $Omega$ be a non-empty set and $geq$ a linear order (a complete, transitive, antisymmetric order) on it. Define
begin{align*}
mathscr Lequiv&;big{{psiinOmega,|,psi<omega},big|,omegainOmegabig},\
mathscr Uequiv&;big{{psiinOmega,|,psi>omega},big|,omegainOmegabig}
end{align*}
to be the family of lower sets and that of upper sets, respectively.
The order topology $tau$ on $Omega$ is the topology which has $mathscr Lcupmathscr U$ as its topological subbasis; it is the coarsest topology according to which all lower and upper sets are open. From this topology, one can generate the Borel $sigma$-algebra $mathscr Bequivsigma(tau)$, which is the smallest $sigma$-algebra according to which sets open according to the order topology are measurable.
My goal is to show that, in general,
begin{align*}
mathscr Bneqsigma(mathscr Lcupmathscr U);tag{$clubsuit$}
end{align*}
that is, the $sigma$-algebra generated by the lower and upper sets is not large enough to give the Borel $sigma$-algebra.
Remark: In some special cases, $mathscr B$ and $sigma(mathscr Lcup mathscr U)$ do coincide—for example, when $Omega=mathbb R$ with the usual order, in which case the order topology is the standard Euclidean topology, whose Borel $sigma$-algebra is well-known to be generated by $mathscr Lcupmathscr U$ (and even by $mathscr L$ alone or by $mathscr U$ alone). More generally, if the order topology $tau$ is second-countable, then $mathscr Lcupmathscr U$ does generate $mathscr B$; the details can be found here. However, as a rule, the two $sigma$-algebras $mathscr B$ and $sigma(mathscr Lcupmathscr U)$ may be different.
In particular, I have the following supposed counterexample in mind. Let $(Omega,geq)$ be the set of countable ordinals. This is a linearly ordered set such that
$Omega$ is uncountable; but- each lower set in $mathscr L$ is countable; furthermore,
- every non-empty subset of $Omega$ has a least element with respect to the order $geq$.
The first step is to show that
begin{align*}
sigma(mathscr Lcupmathscr U)=mathscr C,
end{align*}
where $mathscr C$ is the $sigma$-algebra of countable and co-countable sets:
begin{align*}
mathscr Cequiv{AsubseteqOmega,|,text{$A$ is countable or $A^{mathsf c}$ is countable}}.
end{align*}
Clearly, each set in $mathscr L$ is countable and each set in $mathscr U$ is co-countable, so that $sigma(mathscr Lcupmathscr U)subseteqmathscr C$. Conversely, note that for each $omegainOmega$, the singleton ${omega}$ is the intersection of the complement of a set from $mathscr L$ and the complement of another set from $mathscr U$. Therefore, the singletons are in $sigma(mathscr Lcupmathscr U)$, which implies that $mathscr C$ is included in $sigma(mathscr Lcupmathscr U)$.
Therefore, in order to prove ($clubsuit$), it is sufficient to exhibit a set that is open in the order topology (and hence a fortiori Borel measurable) but is neither countable nor co-countable. To this end, define the immediate-successor function $S:OmegatoOmega$ as follows: for each $omegainOmega$, $S(omega)inOmega$ is the (unique) element of $Omega$ such that
$S(omega)>omega$; and- there exists no $psiinOmega$ such that $S(omega)>psi>omega$.
More formally, $S(omega)$ is the least element of the upper set ${psiinOmega,|,psi>omega}$.
It is easily seen that for each $omegainOmega$, one has
begin{align*}
{S(omega)}={psiinOmega,|,psi<S(S(omega))}cap{psiinOmega,|,psi>omega},
end{align*}
so that the set
begin{align*}
Uequiv S(Omega)=bigcup_{omegainOmega}{S(omega)}
end{align*}
is open in the order topology. What is more, since $U$ is a union of open singletons, any subset of $U$ is open, too. In addition, the function $S$ is injective, so that $U$ is uncountable. Therefore, $U$ can be partitioned into two disjoint uncountable subsets, say $V$ and $W$. Hence, $V$ is an uncountable open set and since $Wsubseteq V^{mathsf c}$, the complement of $V$ is also uncountable. It follows that $Vnotinsigma(mathscr Lcupmathscr U)$. This observation concludes the construction of the desired counterexample.
Any comments and suggestions are greatly appreciated.
general-topology measure-theory proof-verification order-theory
asked 14 hours ago
triple_sec
15.5k21850
You could stop with $U=S(Omega)$ because $U$ is neither countable nor co-countable.
– bof
9 hours ago
@bof Thank you, I corrected the misplaced terminology. I couldn’t prove that $U^{textsf c}$ (the set of limit ordinals) is not countable because I am not familiar with cardinal/ordinal arithmetic, so I took a small detour.
– triple_sec
8 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is a proof-verification (or more like counterexample-verification) request.
Let $Omega$ be a non-empty set and $geq$ a linear order (a complete, transitive, antisymmetric order) on it. Define
begin{align*}
mathscr Lequiv&;big{{psiinOmega,|,psi<omega},big|,omegainOmegabig},\
mathscr Uequiv&;big{{psiinOmega,|,psi>omega},big|,omegainOmegabig}
end{align*}
to be the family of lower sets and that of upper sets, respectively.
The order topology $tau$ on $Omega$ is the topology which has $mathscr Lcupmathscr U$ as its topological subbasis; it is the coarsest topology according to which all lower and upper sets are open. From this topology, one can generate the Borel $sigma$-algebra $mathscr Bequivsigma(tau)$, which is the smallest $sigma$-algebra according to which sets open according to the order topology are measurable.
My goal is to show that, in general,
begin{align*}
mathscr Bneqsigma(mathscr Lcupmathscr U);tag{$clubsuit$}
end{align*}
that is, the $sigma$-algebra generated by the lower and upper sets is not large enough to give the Borel $sigma$-algebra.
Remark: In some special cases, $mathscr B$ and $sigma(mathscr Lcup mathscr U)$ do coincide—for example, when $Omega=mathbb R$ with the usual order, in which case the order topology is the standard Euclidean topology, whose Borel $sigma$-algebra is well-known to be generated by $mathscr Lcupmathscr U$ (and even by $mathscr L$ alone or by $mathscr U$ alone). More generally, if the order topology $tau$ is second-countable, then $mathscr Lcupmathscr U$ does generate $mathscr B$; the details can be found here. However, as a rule, the two $sigma$-algebras $mathscr B$ and $sigma(mathscr Lcupmathscr U)$ may be different.
In particular, I have the following supposed counterexample in mind. Let $(Omega,geq)$ be the set of countable ordinals. This is a linearly ordered set such that
$Omega$ is uncountable; but- each lower set in $mathscr L$ is countable; furthermore,
- every non-empty subset of $Omega$ has a least element with respect to the order $geq$.
The first step is to show that
begin{align*}
sigma(mathscr Lcupmathscr U)=mathscr C,
end{align*}
where $mathscr C$ is the $sigma$-algebra of countable and co-countable sets:
begin{align*}
mathscr Cequiv{AsubseteqOmega,|,text{$A$ is countable or $A^{mathsf c}$ is countable}}.
end{align*}
Clearly, each set in $mathscr L$ is countable and each set in $mathscr U$ is co-countable, so that $sigma(mathscr Lcupmathscr U)subseteqmathscr C$. Conversely, note that for each $omegainOmega$, the singleton ${omega}$ is the intersection of the complement of a set from $mathscr L$ and the complement of another set from $mathscr U$. Therefore, the singletons are in $sigma(mathscr Lcupmathscr U)$, which implies that $mathscr C$ is included in $sigma(mathscr Lcupmathscr U)$.
Therefore, in order to prove ($clubsuit$), it is sufficient to exhibit a set that is open in the order topology (and hence a fortiori Borel measurable) but is neither countable nor co-countable. To this end, define the immediate-successor function $S:OmegatoOmega$ as follows: for each $omegainOmega$, $S(omega)inOmega$ is the (unique) element of $Omega$ such that
$S(omega)>omega$; and- there exists no $psiinOmega$ such that $S(omega)>psi>omega$.
More formally, $S(omega)$ is the least element of the upper set ${psiinOmega,|,psi>omega}$.
It is easily seen that for each $omegainOmega$, one has
begin{align*}
{S(omega)}={psiinOmega,|,psi<S(S(omega))}cap{psiinOmega,|,psi>omega},
end{align*}
so that the set
begin{align*}
Uequiv S(Omega)=bigcup_{omegainOmega}{S(omega)}
end{align*}
is open in the order topology. What is more, since $U$ is a union of open singletons, any subset of $U$ is open, too. In addition, the function $S$ is injective, so that $U$ is uncountable. Therefore, $U$ can be partitioned into two disjoint uncountable subsets, say $V$ and $W$. Hence, $V$ is an uncountable open set and since $Wsubseteq V^{mathsf c}$, the complement of $V$ is also uncountable. It follows that $Vnotinsigma(mathscr Lcupmathscr U)$. This observation concludes the construction of the desired counterexample.
Any comments and suggestions are greatly appreciated.
general-topology measure-theory proof-verification order-theory
asked 14 hours ago
triple_sec
15.5k21850
This is a proof-verification (or more like counterexample-verification) request.
Let $Omega$ be a non-empty set and $geq$ a linear order (a complete, transitive, antisymmetric order) on it. Define
begin{align*}
mathscr Lequiv&;big{{psiinOmega,|,psi<omega},big|,omegainOmegabig},\
mathscr Uequiv&;big{{psiinOmega,|,psi>omega},big|,omegainOmegabig}
end{align*}
to be the family of lower sets and that of upper sets, respectively.
The order topology $tau$ on $Omega$ is the topology which has $mathscr Lcupmathscr U$ as its topological subbasis; it is the coarsest topology according to which all lower and upper sets are open. From this topology, one can generate the Borel $sigma$-algebra $mathscr Bequivsigma(tau)$, which is the smallest $sigma$-algebra according to which sets open according to the order topology are measurable.
My goal is to show that, in general,
begin{align*}
mathscr Bneqsigma(mathscr Lcupmathscr U);tag{$clubsuit$}
end{align*}
that is, the $sigma$-algebra generated by the lower and upper sets is not large enough to give the Borel $sigma$-algebra.
Remark: In some special cases, $mathscr B$ and $sigma(mathscr Lcup mathscr U)$ do coincide—for example, when $Omega=mathbb R$ with the usual order, in which case the order topology is the standard Euclidean topology, whose Borel $sigma$-algebra is well-known to be generated by $mathscr Lcupmathscr U$ (and even by $mathscr L$ alone or by $mathscr U$ alone). More generally, if the order topology $tau$ is second-countable, then $mathscr Lcupmathscr U$ does generate $mathscr B$; the details can be found here. However, as a rule, the two $sigma$-algebras $mathscr B$ and $sigma(mathscr Lcupmathscr U)$ may be different.
In particular, I have the following supposed counterexample in mind. Let $(Omega,geq)$ be the set of countable ordinals. This is a linearly ordered set such that
$Omega$ is uncountable; but- each lower set in $mathscr L$ is countable; furthermore,
- every non-empty subset of $Omega$ has a least element with respect to the order $geq$.
The first step is to show that
begin{align*}
sigma(mathscr Lcupmathscr U)=mathscr C,
end{align*}
where $mathscr C$ is the $sigma$-algebra of countable and co-countable sets:
begin{align*}
mathscr Cequiv{AsubseteqOmega,|,text{$A$ is countable or $A^{mathsf c}$ is countable}}.
end{align*}
Clearly, each set in $mathscr L$ is countable and each set in $mathscr U$ is co-countable, so that $sigma(mathscr Lcupmathscr U)subseteqmathscr C$. Conversely, note that for each $omegainOmega$, the singleton ${omega}$ is the intersection of the complement of a set from $mathscr L$ and the complement of another set from $mathscr U$. Therefore, the singletons are in $sigma(mathscr Lcupmathscr U)$, which implies that $mathscr C$ is included in $sigma(mathscr Lcupmathscr U)$.
Therefore, in order to prove ($clubsuit$), it is sufficient to exhibit a set that is open in the order topology (and hence a fortiori Borel measurable) but is neither countable nor co-countable. To this end, define the immediate-successor function $S:OmegatoOmega$ as follows: for each $omegainOmega$, $S(omega)inOmega$ is the (unique) element of $Omega$ such that
$S(omega)>omega$; and- there exists no $psiinOmega$ such that $S(omega)>psi>omega$.
More formally, $S(omega)$ is the least element of the upper set ${psiinOmega,|,psi>omega}$.
It is easily seen that for each $omegainOmega$, one has
begin{align*}
{S(omega)}={psiinOmega,|,psi<S(S(omega))}cap{psiinOmega,|,psi>omega},
end{align*}
so that the set
begin{align*}
Uequiv S(Omega)=bigcup_{omegainOmega}{S(omega)}
end{align*}
is open in the order topology. What is more, since $U$ is a union of open singletons, any subset of $U$ is open, too. In addition, the function $S$ is injective, so that $U$ is uncountable. Therefore, $U$ can be partitioned into two disjoint uncountable subsets, say $V$ and $W$. Hence, $V$ is an uncountable open set and since $Wsubseteq V^{mathsf c}$, the complement of $V$ is also uncountable. It follows that $Vnotinsigma(mathscr Lcupmathscr U)$. This observation concludes the construction of the desired counterexample.
Any comments and suggestions are greatly appreciated.
general-topology measure-theory proof-verification order-theory
general-topology measure-theory proof-verification order-theory
asked 14 hours ago
triple_sec
15.5k21850
asked 14 hours ago
triple_sec
15.5k21850
edited 8 hours ago
asked 14 hours ago
triple_sec
15.5k21850
asked 14 hours ago
triple_sec
15.5k21850
asked 14 hours ago
triple_sec
15.5k21850
15.5k21850
You could stop with $U=S(Omega)$ because $U$ is neither countable nor co-countable.
– bof
9 hours ago
@bof Thank you, I corrected the misplaced terminology. I couldn’t prove that $U^{textsf c}$ (the set of limit ordinals) is not countable because I am not familiar with cardinal/ordinal arithmetic, so I took a small detour.
– triple_sec
8 hours ago
add a comment |
You could stop with $U=S(Omega)$ because $U$ is neither countable nor co-countable.
– bof
9 hours ago
@bof Thank you, I corrected the misplaced terminology. I couldn’t prove that $U^{textsf c}$ (the set of limit ordinals) is not countable because I am not familiar with cardinal/ordinal arithmetic, so I took a small detour.
– triple_sec
8 hours ago
You could stop with $U=S(Omega)$ because $U$ is neither countable nor co-countable.
– bof
9 hours ago
You could stop with $U=S(Omega)$ because $U$ is neither countable nor co-countable.
– bof
9 hours ago
@bof Thank you, I corrected the misplaced terminology. I couldn’t prove that $U^{textsf c}$ (the set of limit ordinals) is not countable because I am not familiar with cardinal/ordinal arithmetic, so I took a small detour.
– triple_sec
8 hours ago
@bof Thank you, I corrected the misplaced terminology. I couldn’t prove that $U^{textsf c}$ (the set of limit ordinals) is not countable because I am not familiar with cardinal/ordinal arithmetic, so I took a small detour.
– triple_sec
8 hours ago
add a comment |
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You could stop with $U=S(Omega)$ because $U$ is neither countable nor co-countable.
– bof
9 hours ago
@bof Thank you, I corrected the misplaced terminology. I couldn’t prove that $U^{textsf c}$ (the set of limit ordinals) is not countable because I am not familiar with cardinal/ordinal arithmetic, so I took a small detour.
– triple_sec
8 hours ago