Determining elements of a Boolean algebra by a set of ultrafilters
$begingroup$
Let $A$ be a Boolean algebra and let $Ult(A)$ be its Stone space. Let us say that a set $Usubseteq Ult(A)$ determines an element $ain A$ if there exists $Vsubseteq U$ such that
$$big{bin A!: (forall uin V)(bin u) wedge (forall uin Usetminus V)(bnotin u)big}={a}.$$
Let us further say that $Usubseteq Ult(A)$ is determining if it determines all elements of $A$.
Question: Is there any characterization of determining sets?
boolean-algebra filters
$endgroup$
add a comment |
$begingroup$
Let $A$ be a Boolean algebra and let $Ult(A)$ be its Stone space. Let us say that a set $Usubseteq Ult(A)$ determines an element $ain A$ if there exists $Vsubseteq U$ such that
$$big{bin A!: (forall uin V)(bin u) wedge (forall uin Usetminus V)(bnotin u)big}={a}.$$
Let us further say that $Usubseteq Ult(A)$ is determining if it determines all elements of $A$.
Question: Is there any characterization of determining sets?
boolean-algebra filters
$endgroup$
add a comment |
$begingroup$
Let $A$ be a Boolean algebra and let $Ult(A)$ be its Stone space. Let us say that a set $Usubseteq Ult(A)$ determines an element $ain A$ if there exists $Vsubseteq U$ such that
$$big{bin A!: (forall uin V)(bin u) wedge (forall uin Usetminus V)(bnotin u)big}={a}.$$
Let us further say that $Usubseteq Ult(A)$ is determining if it determines all elements of $A$.
Question: Is there any characterization of determining sets?
boolean-algebra filters
$endgroup$
Let $A$ be a Boolean algebra and let $Ult(A)$ be its Stone space. Let us say that a set $Usubseteq Ult(A)$ determines an element $ain A$ if there exists $Vsubseteq U$ such that
$$big{bin A!: (forall uin V)(bin u) wedge (forall uin Usetminus V)(bnotin u)big}={a}.$$
Let us further say that $Usubseteq Ult(A)$ is determining if it determines all elements of $A$.
Question: Is there any characterization of determining sets?
boolean-algebra filters
boolean-algebra filters
edited Jan 5 at 9:32
Peter Elias
asked Jan 5 at 8:13
Peter EliasPeter Elias
938415
938415
add a comment |
add a comment |
1 Answer
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$begingroup$
For $ain A$ denote $S(a)={uin Ult(A)!:ain u}$.
Then ${S(a)!:ain A}$ is a basis of the Stone topology on $Ult(A)$.
Claim: A set $Usubseteq Ult(A)$ is determining if and only if it is dense in the Stone topology.
Proof. Let $Usubseteq Ult(A)$ be determined, $ain A$, $aneq 0$. Then there is $Vsubseteq U$ such that $det(V,U)={a}$, where
$$det(V,U)=big{bin A!:(forall uin V)(bin u) wedge (forall uin Usetminus V)(bnotin u)big}.$$
Since $0indet(emptyset,U)$, $V$ is nonempty.
For $uin V$ we have $uin S(a)cap U$.
This proves one direction.
To prove the opposite, let $U$ be dense and $ain A$. We have to find $Vsubseteq U$ such that $det(V,U)={a}$.
Let $V={uin U!:ain U}$. Clearly, $aindet(V,U)$.
If $bneq a$ then either $awedge b'neq 0$ or $a'wedge bneq 0$.
Hence there exists $uin U$ containing exactly one of elements $a,b$, thus $bnotindet(V,U)$.
It follows that $det(V,U)={a}$.
q.e.d.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
For $ain A$ denote $S(a)={uin Ult(A)!:ain u}$.
Then ${S(a)!:ain A}$ is a basis of the Stone topology on $Ult(A)$.
Claim: A set $Usubseteq Ult(A)$ is determining if and only if it is dense in the Stone topology.
Proof. Let $Usubseteq Ult(A)$ be determined, $ain A$, $aneq 0$. Then there is $Vsubseteq U$ such that $det(V,U)={a}$, where
$$det(V,U)=big{bin A!:(forall uin V)(bin u) wedge (forall uin Usetminus V)(bnotin u)big}.$$
Since $0indet(emptyset,U)$, $V$ is nonempty.
For $uin V$ we have $uin S(a)cap U$.
This proves one direction.
To prove the opposite, let $U$ be dense and $ain A$. We have to find $Vsubseteq U$ such that $det(V,U)={a}$.
Let $V={uin U!:ain U}$. Clearly, $aindet(V,U)$.
If $bneq a$ then either $awedge b'neq 0$ or $a'wedge bneq 0$.
Hence there exists $uin U$ containing exactly one of elements $a,b$, thus $bnotindet(V,U)$.
It follows that $det(V,U)={a}$.
q.e.d.
$endgroup$
add a comment |
$begingroup$
For $ain A$ denote $S(a)={uin Ult(A)!:ain u}$.
Then ${S(a)!:ain A}$ is a basis of the Stone topology on $Ult(A)$.
Claim: A set $Usubseteq Ult(A)$ is determining if and only if it is dense in the Stone topology.
Proof. Let $Usubseteq Ult(A)$ be determined, $ain A$, $aneq 0$. Then there is $Vsubseteq U$ such that $det(V,U)={a}$, where
$$det(V,U)=big{bin A!:(forall uin V)(bin u) wedge (forall uin Usetminus V)(bnotin u)big}.$$
Since $0indet(emptyset,U)$, $V$ is nonempty.
For $uin V$ we have $uin S(a)cap U$.
This proves one direction.
To prove the opposite, let $U$ be dense and $ain A$. We have to find $Vsubseteq U$ such that $det(V,U)={a}$.
Let $V={uin U!:ain U}$. Clearly, $aindet(V,U)$.
If $bneq a$ then either $awedge b'neq 0$ or $a'wedge bneq 0$.
Hence there exists $uin U$ containing exactly one of elements $a,b$, thus $bnotindet(V,U)$.
It follows that $det(V,U)={a}$.
q.e.d.
$endgroup$
add a comment |
$begingroup$
For $ain A$ denote $S(a)={uin Ult(A)!:ain u}$.
Then ${S(a)!:ain A}$ is a basis of the Stone topology on $Ult(A)$.
Claim: A set $Usubseteq Ult(A)$ is determining if and only if it is dense in the Stone topology.
Proof. Let $Usubseteq Ult(A)$ be determined, $ain A$, $aneq 0$. Then there is $Vsubseteq U$ such that $det(V,U)={a}$, where
$$det(V,U)=big{bin A!:(forall uin V)(bin u) wedge (forall uin Usetminus V)(bnotin u)big}.$$
Since $0indet(emptyset,U)$, $V$ is nonempty.
For $uin V$ we have $uin S(a)cap U$.
This proves one direction.
To prove the opposite, let $U$ be dense and $ain A$. We have to find $Vsubseteq U$ such that $det(V,U)={a}$.
Let $V={uin U!:ain U}$. Clearly, $aindet(V,U)$.
If $bneq a$ then either $awedge b'neq 0$ or $a'wedge bneq 0$.
Hence there exists $uin U$ containing exactly one of elements $a,b$, thus $bnotindet(V,U)$.
It follows that $det(V,U)={a}$.
q.e.d.
$endgroup$
For $ain A$ denote $S(a)={uin Ult(A)!:ain u}$.
Then ${S(a)!:ain A}$ is a basis of the Stone topology on $Ult(A)$.
Claim: A set $Usubseteq Ult(A)$ is determining if and only if it is dense in the Stone topology.
Proof. Let $Usubseteq Ult(A)$ be determined, $ain A$, $aneq 0$. Then there is $Vsubseteq U$ such that $det(V,U)={a}$, where
$$det(V,U)=big{bin A!:(forall uin V)(bin u) wedge (forall uin Usetminus V)(bnotin u)big}.$$
Since $0indet(emptyset,U)$, $V$ is nonempty.
For $uin V$ we have $uin S(a)cap U$.
This proves one direction.
To prove the opposite, let $U$ be dense and $ain A$. We have to find $Vsubseteq U$ such that $det(V,U)={a}$.
Let $V={uin U!:ain U}$. Clearly, $aindet(V,U)$.
If $bneq a$ then either $awedge b'neq 0$ or $a'wedge bneq 0$.
Hence there exists $uin U$ containing exactly one of elements $a,b$, thus $bnotindet(V,U)$.
It follows that $det(V,U)={a}$.
q.e.d.
edited Jan 5 at 16:02
answered Jan 5 at 14:20
Peter EliasPeter Elias
938415
938415
add a comment |
add a comment |
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