Characterization of interior in metric spaces












0












$begingroup$


I stumbled upon a theorem in my topology book that didn't have a proof. It says:



For every $x in X$ and $A subset X$,



$x in text{Int} A$ iff $d(x, A^C) > 0$



Here Int denotes interior, A^C is complement of A and d is a metric on X.



If x is in interior of A, then there is an open ball contained in A that contains x. Now, I don't have any idea how to use this to prove that distance between x and XA is positive.



The other direction seems even more difficult.



Any help would be appreciated.






general-topology metric-spaces







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    0












    $begingroup$


    I stumbled upon a theorem in my topology book that didn't have a proof. It says:



    For every $x in X$ and $A subset X$,



    $x in text{Int} A$ iff $d(x, A^C) > 0$



    Here Int denotes interior, A^C is complement of A and d is a metric on X.



    If x is in interior of A, then there is an open ball contained in A that contains x. Now, I don't have any idea how to use this to prove that distance between x and XA is positive.



    The other direction seems even more difficult.



    Any help would be appreciated.






    general-topology metric-spaces







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I stumbled upon a theorem in my topology book that didn't have a proof. It says:



      For every $x in X$ and $A subset X$,



      $x in text{Int} A$ iff $d(x, A^C) > 0$



      Here Int denotes interior, A^C is complement of A and d is a metric on X.



      If x is in interior of A, then there is an open ball contained in A that contains x. Now, I don't have any idea how to use this to prove that distance between x and XA is positive.



      The other direction seems even more difficult.



      Any help would be appreciated.






      general-topology metric-spaces







      share|cite|improve this question









      $endgroup$




      I stumbled upon a theorem in my topology book that didn't have a proof. It says:



      For every $x in X$ and $A subset X$,



      $x in text{Int} A$ iff $d(x, A^C) > 0$



      Here Int denotes interior, A^C is complement of A and d is a metric on X.



      If x is in interior of A, then there is an open ball contained in A that contains x. Now, I don't have any idea how to use this to prove that distance between x and XA is positive.



      The other direction seems even more difficult.



      Any help would be appreciated.






      general-topology metric-spaces




      general-topology metric-spaces






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      share|cite|improve this question











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      asked Nov 25 '18 at 22:05









      windircursewindircurse

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          $begingroup$

          If for some $r>0$ we have $B(x,r) subseteq A$,
          then let $y$ be any point in $A^complement$.



          We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.



          This holds for all $y$ in the complement so



          $$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$



          because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.



          So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.



          The reverse also holds:



          Suppose that $r = d(x,A^complement) > 0$.
          Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula



          $$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.






            share|cite|improve this answer









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              $begingroup$

              If for some $r>0$ we have $B(x,r) subseteq A$,
              then let $y$ be any point in $A^complement$.



              We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.



              This holds for all $y$ in the complement so



              $$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$



              because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.



              So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.



              The reverse also holds:



              Suppose that $r = d(x,A^complement) > 0$.
              Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula



              $$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                If for some $r>0$ we have $B(x,r) subseteq A$,
                then let $y$ be any point in $A^complement$.



                We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.



                This holds for all $y$ in the complement so



                $$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$



                because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.



                So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.



                The reverse also holds:



                Suppose that $r = d(x,A^complement) > 0$.
                Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula



                $$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  If for some $r>0$ we have $B(x,r) subseteq A$,
                  then let $y$ be any point in $A^complement$.



                  We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.



                  This holds for all $y$ in the complement so



                  $$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$



                  because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.



                  So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.



                  The reverse also holds:



                  Suppose that $r = d(x,A^complement) > 0$.
                  Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula



                  $$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.






                  share|cite|improve this answer









                  $endgroup$



                  If for some $r>0$ we have $B(x,r) subseteq A$,
                  then let $y$ be any point in $A^complement$.



                  We know that $d(x,y) ge r$, because otherwise $y in B(x,r)$ and so $y in A$, contradiction, as $y in A^complement$.



                  This holds for all $y$ in the complement so



                  $$d(x,A^complement) = inf {d(x,y): y in A^complement }ge r$$



                  because $r$ is (as I showed) a lower bound for the set of distances and the inf is the greatest lower bound of the set of distances.



                  So if $x in operatorname{int}(A)$ then $d(x,A^complement) > 0$.



                  The reverse also holds:



                  Suppose that $r = d(x,A^complement) > 0$.
                  Let $y in X$ be such that $d(x,y) < r$ (so $y in B(X,r)$, really). Then $y$ cannot be in $A^complement$ because then $d(x,y)$ would be a number $<r$ on the right in the formula



                  $$d(x,A^complement) = inf {d(x,z): z in A^complement }$$ and so we'd have $d(x,A^complement) le d(x,y) < r$ which is a contradiction wih the definition of $r$. So $y notin A^complement$ so $y in A$ and we have shown $$B(x,r) subseteq A$$ as required and so $x in operatorname{int}(A)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 '18 at 22:18









                  Henno BrandsmaHenno Brandsma

                  106k347114




                  106k347114























                      1












                      $begingroup$

                      For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












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                          1





                          $begingroup$

                          For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.






                          share|cite|improve this answer









                          $endgroup$



                          For all $xin text{Int}A$ there is $r>0$ with $B(r,x) subset A$. If $d(x,A^c)=0$ there exists some $yin A^c$ s.t. $d(x,y)<r$: a contradiction. Conversely, if $d(x,y)ge r>0$ for all $yin A^c$ then $B(r,x)subset A$, otherwise, there would be some $yin A^ccap B(r,x)$, which contradicts $d(x,y)ge r$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 25 '18 at 22:18









                          Guacho PerezGuacho Perez

                          3,91411132




                          3,91411132






























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