$partial(A)=cl(A)Longleftrightarrow int(A)=emptyset$












-2












$begingroup$


Is it write to say:



begin{align}
int(A)=emptyset& Longleftrightarrow C_{E}(int(A))=E\
&Longleftrightarrow cl(A)cap C_{E}(int(A))=cl({A})cap E=cl({A})\
&Longleftrightarrow partial(A)= cl(A)
end{align}



If I start from $partial(A)=cl(A)$ I don't know how to get $int({A})=emptyset$??






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $adh(A)$?
    $endgroup$
    – Thomas Shelby
    Jan 1 at 16:16










  • $begingroup$
    the adherence of A,
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:17










  • $begingroup$
    Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
    $endgroup$
    – DreaDk
    Jan 1 at 16:18










  • $begingroup$
    @DreaDk $Fr= partial$
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:24










  • $begingroup$
    I don't understand why -2 ????
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:38
















-2












$begingroup$


Is it write to say:



begin{align}
int(A)=emptyset& Longleftrightarrow C_{E}(int(A))=E\
&Longleftrightarrow cl(A)cap C_{E}(int(A))=cl({A})cap E=cl({A})\
&Longleftrightarrow partial(A)= cl(A)
end{align}



If I start from $partial(A)=cl(A)$ I don't know how to get $int({A})=emptyset$??






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $adh(A)$?
    $endgroup$
    – Thomas Shelby
    Jan 1 at 16:16










  • $begingroup$
    the adherence of A,
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:17










  • $begingroup$
    Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
    $endgroup$
    – DreaDk
    Jan 1 at 16:18










  • $begingroup$
    @DreaDk $Fr= partial$
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:24










  • $begingroup$
    I don't understand why -2 ????
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:38














-2












-2








-2





$begingroup$


Is it write to say:



begin{align}
int(A)=emptyset& Longleftrightarrow C_{E}(int(A))=E\
&Longleftrightarrow cl(A)cap C_{E}(int(A))=cl({A})cap E=cl({A})\
&Longleftrightarrow partial(A)= cl(A)
end{align}



If I start from $partial(A)=cl(A)$ I don't know how to get $int({A})=emptyset$??






general-topology







share|cite|improve this question











$endgroup$




Is it write to say:



begin{align}
int(A)=emptyset& Longleftrightarrow C_{E}(int(A))=E\
&Longleftrightarrow cl(A)cap C_{E}(int(A))=cl({A})cap E=cl({A})\
&Longleftrightarrow partial(A)= cl(A)
end{align}



If I start from $partial(A)=cl(A)$ I don't know how to get $int({A})=emptyset$??






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 16:39







Vrouvrou

















asked Jan 1 at 16:08









VrouvrouVrouvrou

1,9221822




1,9221822












  • $begingroup$
    What is $adh(A)$?
    $endgroup$
    – Thomas Shelby
    Jan 1 at 16:16










  • $begingroup$
    the adherence of A,
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:17










  • $begingroup$
    Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
    $endgroup$
    – DreaDk
    Jan 1 at 16:18










  • $begingroup$
    @DreaDk $Fr= partial$
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:24










  • $begingroup$
    I don't understand why -2 ????
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:38


















  • $begingroup$
    What is $adh(A)$?
    $endgroup$
    – Thomas Shelby
    Jan 1 at 16:16










  • $begingroup$
    the adherence of A,
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:17










  • $begingroup$
    Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
    $endgroup$
    – DreaDk
    Jan 1 at 16:18










  • $begingroup$
    @DreaDk $Fr= partial$
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:24










  • $begingroup$
    I don't understand why -2 ????
    $endgroup$
    – Vrouvrou
    Jan 1 at 16:38
















$begingroup$
What is $adh(A)$?
$endgroup$
– Thomas Shelby
Jan 1 at 16:16




$begingroup$
What is $adh(A)$?
$endgroup$
– Thomas Shelby
Jan 1 at 16:16












$begingroup$
the adherence of A,
$endgroup$
– Vrouvrou
Jan 1 at 16:17




$begingroup$
the adherence of A,
$endgroup$
– Vrouvrou
Jan 1 at 16:17












$begingroup$
Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
$endgroup$
– DreaDk
Jan 1 at 16:18




$begingroup$
Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
$endgroup$
– DreaDk
Jan 1 at 16:18












$begingroup$
@DreaDk $Fr= partial$
$endgroup$
– Vrouvrou
Jan 1 at 16:24




$begingroup$
@DreaDk $Fr= partial$
$endgroup$
– Vrouvrou
Jan 1 at 16:24












$begingroup$
I don't understand why -2 ????
$endgroup$
– Vrouvrou
Jan 1 at 16:38




$begingroup$
I don't understand why -2 ????
$endgroup$
– Vrouvrou
Jan 1 at 16:38










3 Answers
3






active

oldest

votes


















0












$begingroup$

Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.



      From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.



      As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.






            share|cite|improve this answer









            $endgroup$



            Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 at 16:18









            José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

            802110




            802110























                0












                $begingroup$

                Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.






                    share|cite|improve this answer









                    $endgroup$



                    Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 1 at 16:19









                    Tsemo AristideTsemo Aristide

                    58.8k11445




                    58.8k11445























                        0












                        $begingroup$

                        $partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.



                        From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.



                        As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          $partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.



                          From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.



                          As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.



                            From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.



                            As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.






                            share|cite|improve this answer









                            $endgroup$



                            $partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.



                            From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.



                            As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 1 at 22:59









                            Henno BrandsmaHenno Brandsma

                            111k348118




                            111k348118






























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