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)










2












$begingroup$


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 .










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    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 .










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      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 .










      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 7 at 9:51







      LBJFS

















      asked Apr 1 at 13:52









      LBJFSLBJFS

      374112




      374112




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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











          Your Answer








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          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            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













          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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
















          • $begingroup$
            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

















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