If $A cap B =emptyset $ then $A^- cap B^circ =emptyset$? [closed]

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Let $X$ be a topological space and $A,Bsubset X$ If $A cap B =emptyset $ then $A^- cap B^circ =emptyset$ ? We have that $A cap B^circ subset Acap B = emptyset$ and $A cap B^circ subset A^- cap B^circ $ ...? How can I finish it. Please






general-topology







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closed as off-topic by 5xum, user99914, Jendrik Stelzner, amWhy, José Carlos Santos Sep 3 at 23:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Jendrik Stelzner, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    What are $A^-$ and $B^circ$?
    – 5xum
    Sep 3 at 12:07










  • @5xum Probably the closure and the interior, respectively of $A$ and $B$.
    – Suzet
    Sep 3 at 12:07










  • @Suzet yes sorry. Thats it
    – user586431
    Sep 3 at 12:08














up vote
-1
down vote

favorite












Let $X$ be a topological space and $A,Bsubset X$ If $A cap B =emptyset $ then $A^- cap B^circ =emptyset$ ? We have that $A cap B^circ subset Acap B = emptyset$ and $A cap B^circ subset A^- cap B^circ $ ...? How can I finish it. Please






general-topology







share|cite|improve this question













closed as off-topic by 5xum, user99914, Jendrik Stelzner, amWhy, José Carlos Santos Sep 3 at 23:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Jendrik Stelzner, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    What are $A^-$ and $B^circ$?
    – 5xum
    Sep 3 at 12:07










  • @5xum Probably the closure and the interior, respectively of $A$ and $B$.
    – Suzet
    Sep 3 at 12:07










  • @Suzet yes sorry. Thats it
    – user586431
    Sep 3 at 12:08












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Let $X$ be a topological space and $A,Bsubset X$ If $A cap B =emptyset $ then $A^- cap B^circ =emptyset$ ? We have that $A cap B^circ subset Acap B = emptyset$ and $A cap B^circ subset A^- cap B^circ $ ...? How can I finish it. Please






general-topology







share|cite|improve this question













Let $X$ be a topological space and $A,Bsubset X$ If $A cap B =emptyset $ then $A^- cap B^circ =emptyset$ ? We have that $A cap B^circ subset Acap B = emptyset$ and $A cap B^circ subset A^- cap B^circ $ ...? How can I finish it. Please






general-topology




general-topology






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share|cite|improve this question











share|cite|improve this question




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asked Sep 3 at 12:05









user586431

295




295




closed as off-topic by 5xum, user99914, Jendrik Stelzner, amWhy, José Carlos Santos Sep 3 at 23:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Jendrik Stelzner, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by 5xum, user99914, Jendrik Stelzner, amWhy, José Carlos Santos Sep 3 at 23:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Jendrik Stelzner, amWhy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    What are $A^-$ and $B^circ$?
    – 5xum
    Sep 3 at 12:07










  • @5xum Probably the closure and the interior, respectively of $A$ and $B$.
    – Suzet
    Sep 3 at 12:07










  • @Suzet yes sorry. Thats it
    – user586431
    Sep 3 at 12:08












  • 1




    What are $A^-$ and $B^circ$?
    – 5xum
    Sep 3 at 12:07










  • @5xum Probably the closure and the interior, respectively of $A$ and $B$.
    – Suzet
    Sep 3 at 12:07










  • @Suzet yes sorry. Thats it
    – user586431
    Sep 3 at 12:08







1




1




What are $A^-$ and $B^circ$?
– 5xum
Sep 3 at 12:07




What are $A^-$ and $B^circ$?
– 5xum
Sep 3 at 12:07












@5xum Probably the closure and the interior, respectively of $A$ and $B$.
– Suzet
Sep 3 at 12:07




@5xum Probably the closure and the interior, respectively of $A$ and $B$.
– Suzet
Sep 3 at 12:07












@Suzet yes sorry. Thats it
– user586431
Sep 3 at 12:08




@Suzet yes sorry. Thats it
– user586431
Sep 3 at 12:08










2 Answers
2






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up vote
1
down vote



accepted










The other way is to use the definition of closure and interior:



Suppose $A cap B = varnothing$. Could $A^- cap B^o$ be nonempty? Suppose $x in A^- cap B^o$. Then in particular $x in B^o$. Therefore, there is a neighborhood $U$ of $x$ with $U subseteq B$. But since $A cap B = varnothing$, we have $U cap A = varnothing$. So: $x$ has a neighborhood disjoint from $A$, which means $x notin A^-$. This contradicts $x in A^- cap B^o$.






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    up vote
    1
    down vote













    Use that $A^-$ is the smallest closed set that contains $A$; also $Xsetminus B^circ$ is some closed set that contains $A$. Hence $A^-subseteq Xsetminus B^circ$.






    share|cite|improve this answer




















    • Correct. Thank you
      – user586431
      Sep 3 at 12:11

















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    The other way is to use the definition of closure and interior:



    Suppose $A cap B = varnothing$. Could $A^- cap B^o$ be nonempty? Suppose $x in A^- cap B^o$. Then in particular $x in B^o$. Therefore, there is a neighborhood $U$ of $x$ with $U subseteq B$. But since $A cap B = varnothing$, we have $U cap A = varnothing$. So: $x$ has a neighborhood disjoint from $A$, which means $x notin A^-$. This contradicts $x in A^- cap B^o$.






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      The other way is to use the definition of closure and interior:



      Suppose $A cap B = varnothing$. Could $A^- cap B^o$ be nonempty? Suppose $x in A^- cap B^o$. Then in particular $x in B^o$. Therefore, there is a neighborhood $U$ of $x$ with $U subseteq B$. But since $A cap B = varnothing$, we have $U cap A = varnothing$. So: $x$ has a neighborhood disjoint from $A$, which means $x notin A^-$. This contradicts $x in A^- cap B^o$.






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        The other way is to use the definition of closure and interior:



        Suppose $A cap B = varnothing$. Could $A^- cap B^o$ be nonempty? Suppose $x in A^- cap B^o$. Then in particular $x in B^o$. Therefore, there is a neighborhood $U$ of $x$ with $U subseteq B$. But since $A cap B = varnothing$, we have $U cap A = varnothing$. So: $x$ has a neighborhood disjoint from $A$, which means $x notin A^-$. This contradicts $x in A^- cap B^o$.






        share|cite|improve this answer












        The other way is to use the definition of closure and interior:



        Suppose $A cap B = varnothing$. Could $A^- cap B^o$ be nonempty? Suppose $x in A^- cap B^o$. Then in particular $x in B^o$. Therefore, there is a neighborhood $U$ of $x$ with $U subseteq B$. But since $A cap B = varnothing$, we have $U cap A = varnothing$. So: $x$ has a neighborhood disjoint from $A$, which means $x notin A^-$. This contradicts $x in A^- cap B^o$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 3 at 12:33









        GEdgar

        59k265165




        59k265165




















            up vote
            1
            down vote













            Use that $A^-$ is the smallest closed set that contains $A$; also $Xsetminus B^circ$ is some closed set that contains $A$. Hence $A^-subseteq Xsetminus B^circ$.






            share|cite|improve this answer




















            • Correct. Thank you
              – user586431
              Sep 3 at 12:11














            up vote
            1
            down vote













            Use that $A^-$ is the smallest closed set that contains $A$; also $Xsetminus B^circ$ is some closed set that contains $A$. Hence $A^-subseteq Xsetminus B^circ$.






            share|cite|improve this answer




















            • Correct. Thank you
              – user586431
              Sep 3 at 12:11












            up vote
            1
            down vote










            up vote
            1
            down vote









            Use that $A^-$ is the smallest closed set that contains $A$; also $Xsetminus B^circ$ is some closed set that contains $A$. Hence $A^-subseteq Xsetminus B^circ$.






            share|cite|improve this answer












            Use that $A^-$ is the smallest closed set that contains $A$; also $Xsetminus B^circ$ is some closed set that contains $A$. Hence $A^-subseteq Xsetminus B^circ$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 3 at 12:09









            hartkp

            40123




            40123











            • Correct. Thank you
              – user586431
              Sep 3 at 12:11
















            • Correct. Thank you
              – user586431
              Sep 3 at 12:11















            Correct. Thank you
            – user586431
            Sep 3 at 12:11




            Correct. Thank you
            – user586431
            Sep 3 at 12:11


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