If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.












0












$begingroup$



Let $X$ and $Y$ be subsets of a metric space. If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.




I know that $int(overline{Xcup Y}) = int(overline{X}cup overline{Y})$ and so I'm trying to prove the more general statement
$$
int(X) = int(Y) = emptyset implies int(Xcup Y) = emptyset
$$
which seems true to me but I not getting the proof.



What I tried so far is to consider an $xin int(Xcup Y)$, so there is a open ball $B_x$ that is contained in $Xcup Y$ but it cannot be contained in $X$ or $Y$ because these have empty interior. But I don't know what to do next.



Thanks in advance for any help.






general-topology metric-spaces







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
    $endgroup$
    – Daniel Schepler
    Dec 8 '18 at 0:54
















0












$begingroup$



Let $X$ and $Y$ be subsets of a metric space. If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.




I know that $int(overline{Xcup Y}) = int(overline{X}cup overline{Y})$ and so I'm trying to prove the more general statement
$$
int(X) = int(Y) = emptyset implies int(Xcup Y) = emptyset
$$
which seems true to me but I not getting the proof.



What I tried so far is to consider an $xin int(Xcup Y)$, so there is a open ball $B_x$ that is contained in $Xcup Y$ but it cannot be contained in $X$ or $Y$ because these have empty interior. But I don't know what to do next.



Thanks in advance for any help.






general-topology metric-spaces







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
    $endgroup$
    – Daniel Schepler
    Dec 8 '18 at 0:54














0












0








0





$begingroup$



Let $X$ and $Y$ be subsets of a metric space. If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.




I know that $int(overline{Xcup Y}) = int(overline{X}cup overline{Y})$ and so I'm trying to prove the more general statement
$$
int(X) = int(Y) = emptyset implies int(Xcup Y) = emptyset
$$
which seems true to me but I not getting the proof.



What I tried so far is to consider an $xin int(Xcup Y)$, so there is a open ball $B_x$ that is contained in $Xcup Y$ but it cannot be contained in $X$ or $Y$ because these have empty interior. But I don't know what to do next.



Thanks in advance for any help.






general-topology metric-spaces







share|cite|improve this question









$endgroup$





Let $X$ and $Y$ be subsets of a metric space. If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.




I know that $int(overline{Xcup Y}) = int(overline{X}cup overline{Y})$ and so I'm trying to prove the more general statement
$$
int(X) = int(Y) = emptyset implies int(Xcup Y) = emptyset
$$
which seems true to me but I not getting the proof.



What I tried so far is to consider an $xin int(Xcup Y)$, so there is a open ball $B_x$ that is contained in $Xcup Y$ but it cannot be contained in $X$ or $Y$ because these have empty interior. But I don't know what to do next.



Thanks in advance for any help.






general-topology metric-spaces




general-topology metric-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 0:48









AnalyticHarmonyAnalyticHarmony

642313




642313








  • 4




    $begingroup$
    The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
    $endgroup$
    – Daniel Schepler
    Dec 8 '18 at 0:54














  • 4




    $begingroup$
    The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
    $endgroup$
    – Daniel Schepler
    Dec 8 '18 at 0:54








4




4




$begingroup$
The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:54




$begingroup$
The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:54










2 Answers
2






active

oldest

votes


















1












$begingroup$

Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:




Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).



Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.



Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.



Corollary 2. The intersection of two dense open subsets is also dense and open.




If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    And all of this is valid in any kind of topological space....+1
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 10:07






  • 1




    $begingroup$
    Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
    $endgroup$
    – ΑΘΩ
    Dec 9 '18 at 10:10










  • $begingroup$
    Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 18:19





















0












$begingroup$

$X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.



As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.






share|cite|improve this answer











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:




    Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).



    Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.



    Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.



    Corollary 2. The intersection of two dense open subsets is also dense and open.




    If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      And all of this is valid in any kind of topological space....+1
      $endgroup$
      – DanielWainfleet
      Dec 9 '18 at 10:07






    • 1




      $begingroup$
      Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
      $endgroup$
      – ΑΘΩ
      Dec 9 '18 at 10:10










    • $begingroup$
      Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
      $endgroup$
      – DanielWainfleet
      Dec 9 '18 at 18:19


















    1












    $begingroup$

    Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:




    Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).



    Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.



    Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.



    Corollary 2. The intersection of two dense open subsets is also dense and open.




    If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      And all of this is valid in any kind of topological space....+1
      $endgroup$
      – DanielWainfleet
      Dec 9 '18 at 10:07






    • 1




      $begingroup$
      Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
      $endgroup$
      – ΑΘΩ
      Dec 9 '18 at 10:10










    • $begingroup$
      Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
      $endgroup$
      – DanielWainfleet
      Dec 9 '18 at 18:19
















    1












    1








    1





    $begingroup$

    Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:




    Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).



    Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.



    Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.



    Corollary 2. The intersection of two dense open subsets is also dense and open.




    If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.






    share|cite|improve this answer











    $endgroup$



    Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:




    Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).



    Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.



    Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.



    Corollary 2. The intersection of two dense open subsets is also dense and open.




    If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 8 '18 at 22:22









    Henno Brandsma

    107k347114




    107k347114










    answered Dec 8 '18 at 8:58









    ΑΘΩΑΘΩ

    2463




    2463












    • $begingroup$
      And all of this is valid in any kind of topological space....+1
      $endgroup$
      – DanielWainfleet
      Dec 9 '18 at 10:07






    • 1




      $begingroup$
      Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
      $endgroup$
      – ΑΘΩ
      Dec 9 '18 at 10:10










    • $begingroup$
      Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
      $endgroup$
      – DanielWainfleet
      Dec 9 '18 at 18:19




















    • $begingroup$
      And all of this is valid in any kind of topological space....+1
      $endgroup$
      – DanielWainfleet
      Dec 9 '18 at 10:07






    • 1




      $begingroup$
      Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
      $endgroup$
      – ΑΘΩ
      Dec 9 '18 at 10:10










    • $begingroup$
      Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
      $endgroup$
      – DanielWainfleet
      Dec 9 '18 at 18:19


















    $begingroup$
    And all of this is valid in any kind of topological space....+1
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 10:07




    $begingroup$
    And all of this is valid in any kind of topological space....+1
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 10:07




    1




    1




    $begingroup$
    Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
    $endgroup$
    – ΑΘΩ
    Dec 9 '18 at 10:10




    $begingroup$
    Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
    $endgroup$
    – ΑΘΩ
    Dec 9 '18 at 10:10












    $begingroup$
    Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 18:19






    $begingroup$
    Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
    $endgroup$
    – DanielWainfleet
    Dec 9 '18 at 18:19













    0












    $begingroup$

    $X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.



    As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      $X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.



      As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        $X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.



        As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.






        share|cite|improve this answer











        $endgroup$



        $X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.



        As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 8 '18 at 8:19

























        answered Dec 8 '18 at 6:21









        Henno BrandsmaHenno Brandsma

        107k347114




        107k347114






























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