Frontier properties of a subset of $mathbb{R}$












0












$begingroup$


I am supposed to prove or disprove the following claims:





  1. If $A subset mathbb{R}$ arbitrary, then $(Fr(Fr(A))^{mathbb{o}} = emptyset$



  2. If $A, B subset mathbb{R}$ arbitrary, then $Fr(A times B) = (Fr(A) times B) cup (A times Fr(B))$


It seems the latter claim is correct, but I cannot prove it. It's not hard to visualize. On the other hand, I have no idea about the first claim.






general-topology







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$endgroup$












  • $begingroup$
    What is the frontier of a set?
    $endgroup$
    – Arthur
    Dec 9 '18 at 10:37










  • $begingroup$
    I suppose other notations for $Fr(A)$ are $Bd(A)$ or $partial A$.
    $endgroup$
    – Nemanja Beric
    Dec 9 '18 at 10:38






  • 1




    $begingroup$
    So it's what I would call the boundary? Would you look at that, the article even mentions that some authors use "frontier". I learned something today.
    $endgroup$
    – Arthur
    Dec 9 '18 at 10:39












  • $begingroup$
    So you are not using the other definition of frontier $(partial A)-A$? The second claim is false if $A,B$ are not closed.
    $endgroup$
    – user10354138
    Dec 9 '18 at 10:39








  • 1




    $begingroup$
    I'll give you a hint to the first one: prove that $partialpartial Asubseteqpartial A$ for all $A$. (Note that $partialpartial A$ is not necessarily empty, e.g., Kuratowski's closure-complement problem).
    $endgroup$
    – user10354138
    Dec 9 '18 at 11:07
















0












$begingroup$


I am supposed to prove or disprove the following claims:





  1. If $A subset mathbb{R}$ arbitrary, then $(Fr(Fr(A))^{mathbb{o}} = emptyset$



  2. If $A, B subset mathbb{R}$ arbitrary, then $Fr(A times B) = (Fr(A) times B) cup (A times Fr(B))$


It seems the latter claim is correct, but I cannot prove it. It's not hard to visualize. On the other hand, I have no idea about the first claim.






general-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the frontier of a set?
    $endgroup$
    – Arthur
    Dec 9 '18 at 10:37










  • $begingroup$
    I suppose other notations for $Fr(A)$ are $Bd(A)$ or $partial A$.
    $endgroup$
    – Nemanja Beric
    Dec 9 '18 at 10:38






  • 1




    $begingroup$
    So it's what I would call the boundary? Would you look at that, the article even mentions that some authors use "frontier". I learned something today.
    $endgroup$
    – Arthur
    Dec 9 '18 at 10:39












  • $begingroup$
    So you are not using the other definition of frontier $(partial A)-A$? The second claim is false if $A,B$ are not closed.
    $endgroup$
    – user10354138
    Dec 9 '18 at 10:39








  • 1




    $begingroup$
    I'll give you a hint to the first one: prove that $partialpartial Asubseteqpartial A$ for all $A$. (Note that $partialpartial A$ is not necessarily empty, e.g., Kuratowski's closure-complement problem).
    $endgroup$
    – user10354138
    Dec 9 '18 at 11:07














0












0








0





$begingroup$


I am supposed to prove or disprove the following claims:





  1. If $A subset mathbb{R}$ arbitrary, then $(Fr(Fr(A))^{mathbb{o}} = emptyset$



  2. If $A, B subset mathbb{R}$ arbitrary, then $Fr(A times B) = (Fr(A) times B) cup (A times Fr(B))$


It seems the latter claim is correct, but I cannot prove it. It's not hard to visualize. On the other hand, I have no idea about the first claim.






general-topology







share|cite|improve this question









$endgroup$




I am supposed to prove or disprove the following claims:





  1. If $A subset mathbb{R}$ arbitrary, then $(Fr(Fr(A))^{mathbb{o}} = emptyset$



  2. If $A, B subset mathbb{R}$ arbitrary, then $Fr(A times B) = (Fr(A) times B) cup (A times Fr(B))$


It seems the latter claim is correct, but I cannot prove it. It's not hard to visualize. On the other hand, I have no idea about the first claim.






general-topology




general-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 10:36









Nemanja BericNemanja Beric

34118




34118












  • $begingroup$
    What is the frontier of a set?
    $endgroup$
    – Arthur
    Dec 9 '18 at 10:37










  • $begingroup$
    I suppose other notations for $Fr(A)$ are $Bd(A)$ or $partial A$.
    $endgroup$
    – Nemanja Beric
    Dec 9 '18 at 10:38






  • 1




    $begingroup$
    So it's what I would call the boundary? Would you look at that, the article even mentions that some authors use "frontier". I learned something today.
    $endgroup$
    – Arthur
    Dec 9 '18 at 10:39












  • $begingroup$
    So you are not using the other definition of frontier $(partial A)-A$? The second claim is false if $A,B$ are not closed.
    $endgroup$
    – user10354138
    Dec 9 '18 at 10:39








  • 1




    $begingroup$
    I'll give you a hint to the first one: prove that $partialpartial Asubseteqpartial A$ for all $A$. (Note that $partialpartial A$ is not necessarily empty, e.g., Kuratowski's closure-complement problem).
    $endgroup$
    – user10354138
    Dec 9 '18 at 11:07


















  • $begingroup$
    What is the frontier of a set?
    $endgroup$
    – Arthur
    Dec 9 '18 at 10:37










  • $begingroup$
    I suppose other notations for $Fr(A)$ are $Bd(A)$ or $partial A$.
    $endgroup$
    – Nemanja Beric
    Dec 9 '18 at 10:38






  • 1




    $begingroup$
    So it's what I would call the boundary? Would you look at that, the article even mentions that some authors use "frontier". I learned something today.
    $endgroup$
    – Arthur
    Dec 9 '18 at 10:39












  • $begingroup$
    So you are not using the other definition of frontier $(partial A)-A$? The second claim is false if $A,B$ are not closed.
    $endgroup$
    – user10354138
    Dec 9 '18 at 10:39








  • 1




    $begingroup$
    I'll give you a hint to the first one: prove that $partialpartial Asubseteqpartial A$ for all $A$. (Note that $partialpartial A$ is not necessarily empty, e.g., Kuratowski's closure-complement problem).
    $endgroup$
    – user10354138
    Dec 9 '18 at 11:07
















$begingroup$
What is the frontier of a set?
$endgroup$
– Arthur
Dec 9 '18 at 10:37




$begingroup$
What is the frontier of a set?
$endgroup$
– Arthur
Dec 9 '18 at 10:37












$begingroup$
I suppose other notations for $Fr(A)$ are $Bd(A)$ or $partial A$.
$endgroup$
– Nemanja Beric
Dec 9 '18 at 10:38




$begingroup$
I suppose other notations for $Fr(A)$ are $Bd(A)$ or $partial A$.
$endgroup$
– Nemanja Beric
Dec 9 '18 at 10:38




1




1




$begingroup$
So it's what I would call the boundary? Would you look at that, the article even mentions that some authors use "frontier". I learned something today.
$endgroup$
– Arthur
Dec 9 '18 at 10:39






$begingroup$
So it's what I would call the boundary? Would you look at that, the article even mentions that some authors use "frontier". I learned something today.
$endgroup$
– Arthur
Dec 9 '18 at 10:39














$begingroup$
So you are not using the other definition of frontier $(partial A)-A$? The second claim is false if $A,B$ are not closed.
$endgroup$
– user10354138
Dec 9 '18 at 10:39






$begingroup$
So you are not using the other definition of frontier $(partial A)-A$? The second claim is false if $A,B$ are not closed.
$endgroup$
– user10354138
Dec 9 '18 at 10:39






1




1




$begingroup$
I'll give you a hint to the first one: prove that $partialpartial Asubseteqpartial A$ for all $A$. (Note that $partialpartial A$ is not necessarily empty, e.g., Kuratowski's closure-complement problem).
$endgroup$
– user10354138
Dec 9 '18 at 11:07




$begingroup$
I'll give you a hint to the first one: prove that $partialpartial Asubseteqpartial A$ for all $A$. (Note that $partialpartial A$ is not necessarily empty, e.g., Kuratowski's closure-complement problem).
$endgroup$
– user10354138
Dec 9 '18 at 11:07










1 Answer
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$begingroup$


  • the boundary of a closed set or an open set has empty interior.


  • the boundary of any set is closed.



Together this implies that $operatorname{int}(partial partial A) = emptyset$ for any $A$, the first boundary operation makes it closed and then the boundary of a closed set has empty interior.



Fact 2 is refuted by $A = B =mathbb{Q}$, as $(mathbb{R} times mathbb{Q}) cup (mathbb{Q} times mathbb{R}) neq mathbb{R} times mathbb{R}$. IIRC, we need a term $partial A times partial B$ in the union as well.






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    $begingroup$


    • the boundary of a closed set or an open set has empty interior.


    • the boundary of any set is closed.



    Together this implies that $operatorname{int}(partial partial A) = emptyset$ for any $A$, the first boundary operation makes it closed and then the boundary of a closed set has empty interior.



    Fact 2 is refuted by $A = B =mathbb{Q}$, as $(mathbb{R} times mathbb{Q}) cup (mathbb{Q} times mathbb{R}) neq mathbb{R} times mathbb{R}$. IIRC, we need a term $partial A times partial B$ in the union as well.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$


      • the boundary of a closed set or an open set has empty interior.


      • the boundary of any set is closed.



      Together this implies that $operatorname{int}(partial partial A) = emptyset$ for any $A$, the first boundary operation makes it closed and then the boundary of a closed set has empty interior.



      Fact 2 is refuted by $A = B =mathbb{Q}$, as $(mathbb{R} times mathbb{Q}) cup (mathbb{Q} times mathbb{R}) neq mathbb{R} times mathbb{R}$. IIRC, we need a term $partial A times partial B$ in the union as well.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$


        • the boundary of a closed set or an open set has empty interior.


        • the boundary of any set is closed.



        Together this implies that $operatorname{int}(partial partial A) = emptyset$ for any $A$, the first boundary operation makes it closed and then the boundary of a closed set has empty interior.



        Fact 2 is refuted by $A = B =mathbb{Q}$, as $(mathbb{R} times mathbb{Q}) cup (mathbb{Q} times mathbb{R}) neq mathbb{R} times mathbb{R}$. IIRC, we need a term $partial A times partial B$ in the union as well.






        share|cite|improve this answer











        $endgroup$




        • the boundary of a closed set or an open set has empty interior.


        • the boundary of any set is closed.



        Together this implies that $operatorname{int}(partial partial A) = emptyset$ for any $A$, the first boundary operation makes it closed and then the boundary of a closed set has empty interior.



        Fact 2 is refuted by $A = B =mathbb{Q}$, as $(mathbb{R} times mathbb{Q}) cup (mathbb{Q} times mathbb{R}) neq mathbb{R} times mathbb{R}$. IIRC, we need a term $partial A times partial B$ in the union as well.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 16:13

























        answered Dec 9 '18 at 15:14









        Henno BrandsmaHenno Brandsma

        106k347114




        106k347114






























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