Does the intersection of two dense, open sets have the Baire property?












0












$begingroup$


Definition: $A subseteq X$ for a metric space $X$ has the Baire Property if for any sequence of sets {$V_{n}$} for $n geq 1$ that are dense and open in $A$,
$$cl(cap V_{n}) cap A = A $$ for all $n geq 1$. That is, $cap V_{n}$ is dense in $A$.



Question:
Now suppose $G_{1}$ and $G_{2}$ are two open, dense sets of metric space $X$.
Prove that $ G_{1} cap G_{2}$ has the Baire property.



To prove this, I need to show that for any sequence, say {$V_{n}$} s.t. each $V_{n}$ is open and dense in $ G_{1} cap G_{2}$,



$cl(cap V_{n}) cap (G_{1} cap G_{2}) = (G_{1} cap G_{2}) $



Attempt $V_{n}$ dense in $G_{1} cap G_{2}$. So $cl(V_{n})cap (G_{1} cap G_{2}) = (G_{1} cap G_{2}) $. Which means $(G_{1} cap G_{2}) subseteq cl(V_{n}) $. Thus, $cl(G_{1} cap G_{2}) subseteq cl(cl(V_{n})) = cl(V_{n})$.



And since $(G_{1} cap G_{2}) subseteq cl(G_{1} cap G_{2})$, we have $(G_{1} cap G_{2}) subseteq cl(V_{n})$.



How do I proceed from here?






general-topology metric-spaces baire-category







share|cite|improve this question











$endgroup$












  • $begingroup$
    Seeing as the intersection of two dense open sets is a dense open set, all you have to do is prove that one dense open set has the Baire property.
    $endgroup$
    – bof
    Dec 6 '18 at 4:45










  • $begingroup$
    But I think your instructor's or your textbook's use of the term "Baire property" is pretty confusing. To me, a set has the property of Baire if it's the symmetric difference of an open set and a meager set.
    $endgroup$
    – bof
    Dec 6 '18 at 4:47






  • 1




    $begingroup$
    @bof Usually topologists say "a space is Baire", or "$X$ is a Baire space", while sets with the property of Baire are called just that. In measure theory we also consider the $sigma$-algebra generated by the closed (or compact) $G_delta$ sets and this is called the Baire $sigma$-algebra. Also, "the" Baire space is a name for $omega^omega$ as well. Confusing, eh?
    $endgroup$
    – Henno Brandsma
    Dec 6 '18 at 6:05
















0












$begingroup$


Definition: $A subseteq X$ for a metric space $X$ has the Baire Property if for any sequence of sets {$V_{n}$} for $n geq 1$ that are dense and open in $A$,
$$cl(cap V_{n}) cap A = A $$ for all $n geq 1$. That is, $cap V_{n}$ is dense in $A$.



Question:
Now suppose $G_{1}$ and $G_{2}$ are two open, dense sets of metric space $X$.
Prove that $ G_{1} cap G_{2}$ has the Baire property.



To prove this, I need to show that for any sequence, say {$V_{n}$} s.t. each $V_{n}$ is open and dense in $ G_{1} cap G_{2}$,



$cl(cap V_{n}) cap (G_{1} cap G_{2}) = (G_{1} cap G_{2}) $



Attempt $V_{n}$ dense in $G_{1} cap G_{2}$. So $cl(V_{n})cap (G_{1} cap G_{2}) = (G_{1} cap G_{2}) $. Which means $(G_{1} cap G_{2}) subseteq cl(V_{n}) $. Thus, $cl(G_{1} cap G_{2}) subseteq cl(cl(V_{n})) = cl(V_{n})$.



And since $(G_{1} cap G_{2}) subseteq cl(G_{1} cap G_{2})$, we have $(G_{1} cap G_{2}) subseteq cl(V_{n})$.



How do I proceed from here?






general-topology metric-spaces baire-category







share|cite|improve this question











$endgroup$












  • $begingroup$
    Seeing as the intersection of two dense open sets is a dense open set, all you have to do is prove that one dense open set has the Baire property.
    $endgroup$
    – bof
    Dec 6 '18 at 4:45










  • $begingroup$
    But I think your instructor's or your textbook's use of the term "Baire property" is pretty confusing. To me, a set has the property of Baire if it's the symmetric difference of an open set and a meager set.
    $endgroup$
    – bof
    Dec 6 '18 at 4:47






  • 1




    $begingroup$
    @bof Usually topologists say "a space is Baire", or "$X$ is a Baire space", while sets with the property of Baire are called just that. In measure theory we also consider the $sigma$-algebra generated by the closed (or compact) $G_delta$ sets and this is called the Baire $sigma$-algebra. Also, "the" Baire space is a name for $omega^omega$ as well. Confusing, eh?
    $endgroup$
    – Henno Brandsma
    Dec 6 '18 at 6:05














0












0








0





$begingroup$


Definition: $A subseteq X$ for a metric space $X$ has the Baire Property if for any sequence of sets {$V_{n}$} for $n geq 1$ that are dense and open in $A$,
$$cl(cap V_{n}) cap A = A $$ for all $n geq 1$. That is, $cap V_{n}$ is dense in $A$.



Question:
Now suppose $G_{1}$ and $G_{2}$ are two open, dense sets of metric space $X$.
Prove that $ G_{1} cap G_{2}$ has the Baire property.



To prove this, I need to show that for any sequence, say {$V_{n}$} s.t. each $V_{n}$ is open and dense in $ G_{1} cap G_{2}$,



$cl(cap V_{n}) cap (G_{1} cap G_{2}) = (G_{1} cap G_{2}) $



Attempt $V_{n}$ dense in $G_{1} cap G_{2}$. So $cl(V_{n})cap (G_{1} cap G_{2}) = (G_{1} cap G_{2}) $. Which means $(G_{1} cap G_{2}) subseteq cl(V_{n}) $. Thus, $cl(G_{1} cap G_{2}) subseteq cl(cl(V_{n})) = cl(V_{n})$.



And since $(G_{1} cap G_{2}) subseteq cl(G_{1} cap G_{2})$, we have $(G_{1} cap G_{2}) subseteq cl(V_{n})$.



How do I proceed from here?






general-topology metric-spaces baire-category







share|cite|improve this question











$endgroup$




Definition: $A subseteq X$ for a metric space $X$ has the Baire Property if for any sequence of sets {$V_{n}$} for $n geq 1$ that are dense and open in $A$,
$$cl(cap V_{n}) cap A = A $$ for all $n geq 1$. That is, $cap V_{n}$ is dense in $A$.



Question:
Now suppose $G_{1}$ and $G_{2}$ are two open, dense sets of metric space $X$.
Prove that $ G_{1} cap G_{2}$ has the Baire property.



To prove this, I need to show that for any sequence, say {$V_{n}$} s.t. each $V_{n}$ is open and dense in $ G_{1} cap G_{2}$,



$cl(cap V_{n}) cap (G_{1} cap G_{2}) = (G_{1} cap G_{2}) $



Attempt $V_{n}$ dense in $G_{1} cap G_{2}$. So $cl(V_{n})cap (G_{1} cap G_{2}) = (G_{1} cap G_{2}) $. Which means $(G_{1} cap G_{2}) subseteq cl(V_{n}) $. Thus, $cl(G_{1} cap G_{2}) subseteq cl(cl(V_{n})) = cl(V_{n})$.



And since $(G_{1} cap G_{2}) subseteq cl(G_{1} cap G_{2})$, we have $(G_{1} cap G_{2}) subseteq cl(V_{n})$.



How do I proceed from here?






general-topology metric-spaces baire-category




general-topology metric-spaces baire-category






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 4:20









Eric Wofsey

183k13211338




183k13211338










asked Dec 6 '18 at 3:27









Snop D.Snop D.

285




285












  • $begingroup$
    Seeing as the intersection of two dense open sets is a dense open set, all you have to do is prove that one dense open set has the Baire property.
    $endgroup$
    – bof
    Dec 6 '18 at 4:45










  • $begingroup$
    But I think your instructor's or your textbook's use of the term "Baire property" is pretty confusing. To me, a set has the property of Baire if it's the symmetric difference of an open set and a meager set.
    $endgroup$
    – bof
    Dec 6 '18 at 4:47






  • 1




    $begingroup$
    @bof Usually topologists say "a space is Baire", or "$X$ is a Baire space", while sets with the property of Baire are called just that. In measure theory we also consider the $sigma$-algebra generated by the closed (or compact) $G_delta$ sets and this is called the Baire $sigma$-algebra. Also, "the" Baire space is a name for $omega^omega$ as well. Confusing, eh?
    $endgroup$
    – Henno Brandsma
    Dec 6 '18 at 6:05


















  • $begingroup$
    Seeing as the intersection of two dense open sets is a dense open set, all you have to do is prove that one dense open set has the Baire property.
    $endgroup$
    – bof
    Dec 6 '18 at 4:45










  • $begingroup$
    But I think your instructor's or your textbook's use of the term "Baire property" is pretty confusing. To me, a set has the property of Baire if it's the symmetric difference of an open set and a meager set.
    $endgroup$
    – bof
    Dec 6 '18 at 4:47






  • 1




    $begingroup$
    @bof Usually topologists say "a space is Baire", or "$X$ is a Baire space", while sets with the property of Baire are called just that. In measure theory we also consider the $sigma$-algebra generated by the closed (or compact) $G_delta$ sets and this is called the Baire $sigma$-algebra. Also, "the" Baire space is a name for $omega^omega$ as well. Confusing, eh?
    $endgroup$
    – Henno Brandsma
    Dec 6 '18 at 6:05
















$begingroup$
Seeing as the intersection of two dense open sets is a dense open set, all you have to do is prove that one dense open set has the Baire property.
$endgroup$
– bof
Dec 6 '18 at 4:45




$begingroup$
Seeing as the intersection of two dense open sets is a dense open set, all you have to do is prove that one dense open set has the Baire property.
$endgroup$
– bof
Dec 6 '18 at 4:45












$begingroup$
But I think your instructor's or your textbook's use of the term "Baire property" is pretty confusing. To me, a set has the property of Baire if it's the symmetric difference of an open set and a meager set.
$endgroup$
– bof
Dec 6 '18 at 4:47




$begingroup$
But I think your instructor's or your textbook's use of the term "Baire property" is pretty confusing. To me, a set has the property of Baire if it's the symmetric difference of an open set and a meager set.
$endgroup$
– bof
Dec 6 '18 at 4:47




1




1




$begingroup$
@bof Usually topologists say "a space is Baire", or "$X$ is a Baire space", while sets with the property of Baire are called just that. In measure theory we also consider the $sigma$-algebra generated by the closed (or compact) $G_delta$ sets and this is called the Baire $sigma$-algebra. Also, "the" Baire space is a name for $omega^omega$ as well. Confusing, eh?
$endgroup$
– Henno Brandsma
Dec 6 '18 at 6:05




$begingroup$
@bof Usually topologists say "a space is Baire", or "$X$ is a Baire space", while sets with the property of Baire are called just that. In measure theory we also consider the $sigma$-algebra generated by the closed (or compact) $G_delta$ sets and this is called the Baire $sigma$-algebra. Also, "the" Baire space is a name for $omega^omega$ as well. Confusing, eh?
$endgroup$
– Henno Brandsma
Dec 6 '18 at 6:05










1 Answer
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$begingroup$

You just need to show that if $X$ is Baire, so is any open dense subspace $O$ of $X$. This is straightforward from the definition.



Then note that a finite intersection of dense open sets is still dense and open.



Note that in your statement $X$ needs to be Baire itself, or we could trivially take $X= G_1 = G_2 = mathbb{Q}$ in the usual topology and have a counterexample.






share|cite|improve this answer









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    $begingroup$

    You just need to show that if $X$ is Baire, so is any open dense subspace $O$ of $X$. This is straightforward from the definition.



    Then note that a finite intersection of dense open sets is still dense and open.



    Note that in your statement $X$ needs to be Baire itself, or we could trivially take $X= G_1 = G_2 = mathbb{Q}$ in the usual topology and have a counterexample.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You just need to show that if $X$ is Baire, so is any open dense subspace $O$ of $X$. This is straightforward from the definition.



      Then note that a finite intersection of dense open sets is still dense and open.



      Note that in your statement $X$ needs to be Baire itself, or we could trivially take $X= G_1 = G_2 = mathbb{Q}$ in the usual topology and have a counterexample.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You just need to show that if $X$ is Baire, so is any open dense subspace $O$ of $X$. This is straightforward from the definition.



        Then note that a finite intersection of dense open sets is still dense and open.



        Note that in your statement $X$ needs to be Baire itself, or we could trivially take $X= G_1 = G_2 = mathbb{Q}$ in the usual topology and have a counterexample.






        share|cite|improve this answer









        $endgroup$



        You just need to show that if $X$ is Baire, so is any open dense subspace $O$ of $X$. This is straightforward from the definition.



        Then note that a finite intersection of dense open sets is still dense and open.



        Note that in your statement $X$ needs to be Baire itself, or we could trivially take $X= G_1 = G_2 = mathbb{Q}$ in the usual topology and have a counterexample.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 6:06









        Henno BrandsmaHenno Brandsma

        107k347114




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