The canonical open set which is equal to a set with the BP modulo meager sets is regular open (Kechris) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Question of regular openRegular open sets and semi-regularization.Exercise of comager set and the space $C([0,1])$Question about of Baire property and Baire spaceQuestion of regular openThe maximality of a family of pairwise disjoint meager open sets implies the denseness of its unionQuestion about comeager set in a Polish spaceAnalytic sets have perfect set property (Kechris)Baire Property on Baire Spaces (Kechris' book)Meagerness on an open subset with the relative topology (Kechris)Theorem $(8.29)$ (Kechris)
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The canonical open set which is equal to a set with the BP modulo meager sets is regular open (Kechris)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Question of regular openRegular open sets and semi-regularization.Exercise of comager set and the space $C([0,1])$Question about of Baire property and Baire spaceQuestion of regular openThe maximality of a family of pairwise disjoint meager open sets implies the denseness of its unionQuestion about comeager set in a Polish spaceAnalytic sets have perfect set property (Kechris)Baire Property on Baire Spaces (Kechris' book)Meagerness on an open subset with the relative topology (Kechris)Theorem $(8.29)$ (Kechris)
$begingroup$
A set $U$ in a topological space $X$ is called regular open iff $U=(overlineU)^°$.
Exercise $(8.30)$ (Kechris, "Classical Descriptive Set Theory")
Prove that
$$U(A)=bigcup U,textopenmid UVdash A$$
is regular open.
Moreover, if $X$ is a Baire space and $A$ has the BP (Baire property), then $U(A)$ is the unique regular open set $U$ with $A=^*U$.
Actually, the first part of this question has already been asked, but the answer doesn't seem to work (at least for me): indeed, what the hint allows to prove is that $A$ is comeager in $(overlineU(A))^°$, but how does the emptyness of $(overlineU(A))^°setminus U(A)$ follow?
For the second part, assume that $A=^*V$ for some open regular open set in $X$. I don't see how the assumption should help.
NOTE: I decided to ask for it once again because the user Brian M. Scott seems to be no more active on this site .
general-topology descriptive-set-theory
asked Apr 1 at 13:52
LBJFSLBJFS
374112
$endgroup$
add a comment |
$begingroup$
A set $U$ in a topological space $X$ is called regular open iff $U=(overlineU)^°$.
Exercise $(8.30)$ (Kechris, "Classical Descriptive Set Theory")
Prove that
$$U(A)=bigcup U,textopenmid UVdash A$$
is regular open.
Moreover, if $X$ is a Baire space and $A$ has the BP (Baire property), then $U(A)$ is the unique regular open set $U$ with $A=^*U$.
Actually, the first part of this question has already been asked, but the answer doesn't seem to work (at least for me): indeed, what the hint allows to prove is that $A$ is comeager in $(overlineU(A))^°$, but how does the emptyness of $(overlineU(A))^°setminus U(A)$ follow?
For the second part, assume that $A=^*V$ for some open regular open set in $X$. I don't see how the assumption should help.
NOTE: I decided to ask for it once again because the user Brian M. Scott seems to be no more active on this site .
general-topology descriptive-set-theory
asked Apr 1 at 13:52
LBJFSLBJFS
374112
$endgroup$
add a comment |
$begingroup$
A set $U$ in a topological space $X$ is called regular open iff $U=(overlineU)^°$.
Exercise $(8.30)$ (Kechris, "Classical Descriptive Set Theory")
Prove that
$$U(A)=bigcup U,textopenmid UVdash A$$
is regular open.
Moreover, if $X$ is a Baire space and $A$ has the BP (Baire property), then $U(A)$ is the unique regular open set $U$ with $A=^*U$.
Actually, the first part of this question has already been asked, but the answer doesn't seem to work (at least for me): indeed, what the hint allows to prove is that $A$ is comeager in $(overlineU(A))^°$, but how does the emptyness of $(overlineU(A))^°setminus U(A)$ follow?
For the second part, assume that $A=^*V$ for some open regular open set in $X$. I don't see how the assumption should help.
NOTE: I decided to ask for it once again because the user Brian M. Scott seems to be no more active on this site .
general-topology descriptive-set-theory
asked Apr 1 at 13:52
LBJFSLBJFS
374112
$endgroup$
A set $U$ in a topological space $X$ is called regular open iff $U=(overlineU)^°$.
Exercise $(8.30)$ (Kechris, "Classical Descriptive Set Theory")
Prove that
$$U(A)=bigcup U,textopenmid UVdash A$$
is regular open.
Moreover, if $X$ is a Baire space and $A$ has the BP (Baire property), then $U(A)$ is the unique regular open set $U$ with $A=^*U$.
Actually, the first part of this question has already been asked, but the answer doesn't seem to work (at least for me): indeed, what the hint allows to prove is that $A$ is comeager in $(overlineU(A))^°$, but how does the emptyness of $(overlineU(A))^°setminus U(A)$ follow?
For the second part, assume that $A=^*V$ for some open regular open set in $X$. I don't see how the assumption should help.
NOTE: I decided to ask for it once again because the user Brian M. Scott seems to be no more active on this site .
general-topology descriptive-set-theory
general-topology descriptive-set-theory
asked Apr 1 at 13:52
LBJFSLBJFS
374112
asked Apr 1 at 13:52
LBJFSLBJFS
374112
edited Apr 7 at 9:51
LBJFS
asked Apr 1 at 13:52
LBJFSLBJFS
374112
asked Apr 1 at 13:52
LBJFSLBJFS
374112
asked Apr 1 at 13:52
LBJFSLBJFS
374112
374112
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The aforementioned hint is that $U(A)Vdash A$. As you indicate, this allows to show $(overlineU(A))^°Vdash A$.
By the hint and its definition, $U(A)$ it is the largest open set $U$ such that $UVdash A$. On the other hand, $U(A)subseteq (overlineU(A))^°$. Hence $U(A) = (overlineU(A))^°$ and therefore it is regular open. (I fail to see which emptiness should be relevant here.)
For the second part, assume $A$ the BP and $A=^*V$ with $V$ regular open. Then $VVdash A$ and hence $Vsubseteq U(A)$.
From $A=^*V$ we also conclude $Asetminus V$ is meager, and hence $U(A)setminus A cup Asetminus V supseteq U(A) setminus V$ is meager and has empty interior. This shows that $ U(A)setminus V subseteq overlineV$, and therefore $U(A) subseteq (overlineV)^° = V$. We have both inclusions.
answered Apr 6 at 13:28
Pedro Sánchez TerrafPedro Sánchez Terraf
2,6541923
$endgroup$
$begingroup$
Thank you for your plain answer. You are right, you fail to see the emptyness because the above expression is imprecise: I meant that $(overlineU)^°setminus U$ should be empty. I apologize for this. Thank you once again.
$endgroup$
– LBJFS
Apr 6 at 15:48
$begingroup$
You're welcome!
$endgroup$
– Pedro Sánchez Terraf
Apr 6 at 16:07
add a comment |
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$begingroup$
The aforementioned hint is that $U(A)Vdash A$. As you indicate, this allows to show $(overlineU(A))^°Vdash A$.
By the hint and its definition, $U(A)$ it is the largest open set $U$ such that $UVdash A$. On the other hand, $U(A)subseteq (overlineU(A))^°$. Hence $U(A) = (overlineU(A))^°$ and therefore it is regular open. (I fail to see which emptiness should be relevant here.)
For the second part, assume $A$ the BP and $A=^*V$ with $V$ regular open. Then $VVdash A$ and hence $Vsubseteq U(A)$.
From $A=^*V$ we also conclude $Asetminus V$ is meager, and hence $U(A)setminus A cup Asetminus V supseteq U(A) setminus V$ is meager and has empty interior. This shows that $ U(A)setminus V subseteq overlineV$, and therefore $U(A) subseteq (overlineV)^° = V$. We have both inclusions.
answered Apr 6 at 13:28
Pedro Sánchez TerrafPedro Sánchez Terraf
2,6541923
$endgroup$
$begingroup$
Thank you for your plain answer. You are right, you fail to see the emptyness because the above expression is imprecise: I meant that $(overlineU)^°setminus U$ should be empty. I apologize for this. Thank you once again.
$endgroup$
– LBJFS
Apr 6 at 15:48
$begingroup$
You're welcome!
$endgroup$
– Pedro Sánchez Terraf
Apr 6 at 16:07
add a comment |
$begingroup$
The aforementioned hint is that $U(A)Vdash A$. As you indicate, this allows to show $(overlineU(A))^°Vdash A$.
By the hint and its definition, $U(A)$ it is the largest open set $U$ such that $UVdash A$. On the other hand, $U(A)subseteq (overlineU(A))^°$. Hence $U(A) = (overlineU(A))^°$ and therefore it is regular open. (I fail to see which emptiness should be relevant here.)
For the second part, assume $A$ the BP and $A=^*V$ with $V$ regular open. Then $VVdash A$ and hence $Vsubseteq U(A)$.
From $A=^*V$ we also conclude $Asetminus V$ is meager, and hence $U(A)setminus A cup Asetminus V supseteq U(A) setminus V$ is meager and has empty interior. This shows that $ U(A)setminus V subseteq overlineV$, and therefore $U(A) subseteq (overlineV)^° = V$. We have both inclusions.
answered Apr 6 at 13:28
Pedro Sánchez TerrafPedro Sánchez Terraf
2,6541923
$endgroup$
$begingroup$
Thank you for your plain answer. You are right, you fail to see the emptyness because the above expression is imprecise: I meant that $(overlineU)^°setminus U$ should be empty. I apologize for this. Thank you once again.
$endgroup$
– LBJFS
Apr 6 at 15:48
$begingroup$
You're welcome!
$endgroup$
– Pedro Sánchez Terraf
Apr 6 at 16:07
add a comment |
$begingroup$
The aforementioned hint is that $U(A)Vdash A$. As you indicate, this allows to show $(overlineU(A))^°Vdash A$.
By the hint and its definition, $U(A)$ it is the largest open set $U$ such that $UVdash A$. On the other hand, $U(A)subseteq (overlineU(A))^°$. Hence $U(A) = (overlineU(A))^°$ and therefore it is regular open. (I fail to see which emptiness should be relevant here.)
For the second part, assume $A$ the BP and $A=^*V$ with $V$ regular open. Then $VVdash A$ and hence $Vsubseteq U(A)$.
From $A=^*V$ we also conclude $Asetminus V$ is meager, and hence $U(A)setminus A cup Asetminus V supseteq U(A) setminus V$ is meager and has empty interior. This shows that $ U(A)setminus V subseteq overlineV$, and therefore $U(A) subseteq (overlineV)^° = V$. We have both inclusions.
answered Apr 6 at 13:28
Pedro Sánchez TerrafPedro Sánchez Terraf
2,6541923
$endgroup$
The aforementioned hint is that $U(A)Vdash A$. As you indicate, this allows to show $(overlineU(A))^°Vdash A$.
By the hint and its definition, $U(A)$ it is the largest open set $U$ such that $UVdash A$. On the other hand, $U(A)subseteq (overlineU(A))^°$. Hence $U(A) = (overlineU(A))^°$ and therefore it is regular open. (I fail to see which emptiness should be relevant here.)
For the second part, assume $A$ the BP and $A=^*V$ with $V$ regular open. Then $VVdash A$ and hence $Vsubseteq U(A)$.
From $A=^*V$ we also conclude $Asetminus V$ is meager, and hence $U(A)setminus A cup Asetminus V supseteq U(A) setminus V$ is meager and has empty interior. This shows that $ U(A)setminus V subseteq overlineV$, and therefore $U(A) subseteq (overlineV)^° = V$. We have both inclusions.
answered Apr 6 at 13:28
Pedro Sánchez TerrafPedro Sánchez Terraf
2,6541923
answered Apr 6 at 13:28
Pedro Sánchez TerrafPedro Sánchez Terraf
2,6541923
answered Apr 6 at 13:28
Pedro Sánchez TerrafPedro Sánchez Terraf
2,6541923
answered Apr 6 at 13:28
Pedro Sánchez TerrafPedro Sánchez Terraf
2,6541923
2,6541923
$begingroup$
Thank you for your plain answer. You are right, you fail to see the emptyness because the above expression is imprecise: I meant that $(overlineU)^°setminus U$ should be empty. I apologize for this. Thank you once again.
$endgroup$
– LBJFS
Apr 6 at 15:48
$begingroup$
You're welcome!
$endgroup$
– Pedro Sánchez Terraf
Apr 6 at 16:07
add a comment |
$begingroup$
Thank you for your plain answer. You are right, you fail to see the emptyness because the above expression is imprecise: I meant that $(overlineU)^°setminus U$ should be empty. I apologize for this. Thank you once again.
$endgroup$
– LBJFS
Apr 6 at 15:48
$begingroup$
You're welcome!
$endgroup$
– Pedro Sánchez Terraf
Apr 6 at 16:07
$begingroup$
Thank you for your plain answer. You are right, you fail to see the emptyness because the above expression is imprecise: I meant that $(overlineU)^°setminus U$ should be empty. I apologize for this. Thank you once again.
$endgroup$
– LBJFS
Apr 6 at 15:48
$begingroup$
Thank you for your plain answer. You are right, you fail to see the emptyness because the above expression is imprecise: I meant that $(overlineU)^°setminus U$ should be empty. I apologize for this. Thank you once again.
$endgroup$
– LBJFS
Apr 6 at 15:48
$begingroup$
You're welcome!
$endgroup$
– Pedro Sánchez Terraf
Apr 6 at 16:07
$begingroup$
You're welcome!
$endgroup$
– Pedro Sánchez Terraf
Apr 6 at 16:07
add a comment |
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