Is it possible that $xin overlineAcap B$ for $xin A$ $B$ - open The Next CEO of Stack OverflowWhat conditions are sufficient for “Basically disconnectedness implies Extremally disconnectedness”?Is every Extremally Disconnected Hausdorff Space Regular?A topological space is extremally disconnected iff every two disjoint open sets have disjoint closuresA is a open set $ iff (forall X) (A cap overlineX subset overlineA cap X )$Let $Usubseteq X$ be open and $Ssubseteq X$ a set such that $Scap U$ is closed in $U$. Does $barScap U=Scap U$?If $overline Acap B = Acap overline B = varnothing$, $Acup B$ is disconnected.Quotients of extremally disconnected compact Hausdorff spacesExtremally disconnected space is normal$X$ is basically disconnected if every cozero-set has an open closure.Prove that If B is open, then $overlineA cap B subset overlineA cap B$

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Is it possible that $xin overlineAcap B$ for $xin A$ $B$ - open



The Next CEO of Stack OverflowWhat conditions are sufficient for “Basically disconnectedness implies Extremally disconnectedness”?Is every Extremally Disconnected Hausdorff Space Regular?A topological space is extremally disconnected iff every two disjoint open sets have disjoint closuresA is a open set $ iff (forall X) (A cap overlineX subset overlineA cap X )$Let $Usubseteq X$ be open and $Ssubseteq X$ a set such that $Scap U$ is closed in $U$. Does $barScap U=Scap U$?If $overline Acap B = Acap overline B = varnothing$, $Acup B$ is disconnected.Quotients of extremally disconnected compact Hausdorff spacesExtremally disconnected space is normal$X$ is basically disconnected if every cozero-set has an open closure.Prove that If B is open, then $overlineA cap B subset overlineA cap B$










0












$begingroup$


Is the following possible:




For a Hausdorff space (don't know if that's needed), let $xin A$ $B$ for $A,Bsubseteq_opX$. Is it possible that $xinoverlineAcap B$?




What about if $X$ is extremally disconnected (that is, the closure of each open set is open)










share|cite|improve this question











$endgroup$











  • $begingroup$
    What have you tried so far? Hint: When (if at all) is it possible that $cl(A cap B)$ contains a point outside $B$?
    $endgroup$
    – vxnture
    Mar 18 at 19:35











  • $begingroup$
    It is equivalent to proving that cl$(Acap B) subseteq$ cl$(A)cap B$
    $endgroup$
    – user3701033
    Mar 18 at 20:44















0












$begingroup$


Is the following possible:




For a Hausdorff space (don't know if that's needed), let $xin A$ $B$ for $A,Bsubseteq_opX$. Is it possible that $xinoverlineAcap B$?




What about if $X$ is extremally disconnected (that is, the closure of each open set is open)










share|cite|improve this question











$endgroup$











  • $begingroup$
    What have you tried so far? Hint: When (if at all) is it possible that $cl(A cap B)$ contains a point outside $B$?
    $endgroup$
    – vxnture
    Mar 18 at 19:35











  • $begingroup$
    It is equivalent to proving that cl$(Acap B) subseteq$ cl$(A)cap B$
    $endgroup$
    – user3701033
    Mar 18 at 20:44













0












0








0





$begingroup$


Is the following possible:




For a Hausdorff space (don't know if that's needed), let $xin A$ $B$ for $A,Bsubseteq_opX$. Is it possible that $xinoverlineAcap B$?




What about if $X$ is extremally disconnected (that is, the closure of each open set is open)










share|cite|improve this question











$endgroup$




Is the following possible:




For a Hausdorff space (don't know if that's needed), let $xin A$ $B$ for $A,Bsubseteq_opX$. Is it possible that $xinoverlineAcap B$?




What about if $X$ is extremally disconnected (that is, the closure of each open set is open)







general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 20 at 8:35









YuiTo Cheng

2,1863937




2,1863937










asked Mar 18 at 18:23









user3701033user3701033

1519




1519











  • $begingroup$
    What have you tried so far? Hint: When (if at all) is it possible that $cl(A cap B)$ contains a point outside $B$?
    $endgroup$
    – vxnture
    Mar 18 at 19:35











  • $begingroup$
    It is equivalent to proving that cl$(Acap B) subseteq$ cl$(A)cap B$
    $endgroup$
    – user3701033
    Mar 18 at 20:44
















  • $begingroup$
    What have you tried so far? Hint: When (if at all) is it possible that $cl(A cap B)$ contains a point outside $B$?
    $endgroup$
    – vxnture
    Mar 18 at 19:35











  • $begingroup$
    It is equivalent to proving that cl$(Acap B) subseteq$ cl$(A)cap B$
    $endgroup$
    – user3701033
    Mar 18 at 20:44















$begingroup$
What have you tried so far? Hint: When (if at all) is it possible that $cl(A cap B)$ contains a point outside $B$?
$endgroup$
– vxnture
Mar 18 at 19:35





$begingroup$
What have you tried so far? Hint: When (if at all) is it possible that $cl(A cap B)$ contains a point outside $B$?
$endgroup$
– vxnture
Mar 18 at 19:35













$begingroup$
It is equivalent to proving that cl$(Acap B) subseteq$ cl$(A)cap B$
$endgroup$
– user3701033
Mar 18 at 20:44




$begingroup$
It is equivalent to proving that cl$(Acap B) subseteq$ cl$(A)cap B$
$endgroup$
– user3701033
Mar 18 at 20:44










1 Answer
1






active

oldest

votes


















0












$begingroup$

Yes this is possible. In general, whenever $(A cap B)$ is dense in some neighbourhood of a point in $A setminus B$, $cl(A cap B)$ wll contain that point.



Suppose $A = X = [0,1]$. Let $q_n$ be an enumeration of the rationals in $[0,1]$. For some $epsilon < 1$, let $B_2^-nepsilon(q_n)$ be the open ball of diameter $2^-nepsilon$ around $q_n$, and let $B = bigcup_nB_2^-nepsilon(q_n)$.



$B$ is open and dense in $A$, so $cl(B) = A$. And the Lebesgue measure of $B$, $mu(B) leq sum_n geq 12^-nepsilon = epsilon < 1 = mu(A)$, so $B$ is a proper subset of $A$, i.e., $A setminus B$ is non-empty.



This shows that $A cap B = B$, but $B$ is dense in $A$ so $cl(A cap B) = A$.






share|cite|improve this answer









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    1 Answer
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    active

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    active

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    0












    $begingroup$

    Yes this is possible. In general, whenever $(A cap B)$ is dense in some neighbourhood of a point in $A setminus B$, $cl(A cap B)$ wll contain that point.



    Suppose $A = X = [0,1]$. Let $q_n$ be an enumeration of the rationals in $[0,1]$. For some $epsilon < 1$, let $B_2^-nepsilon(q_n)$ be the open ball of diameter $2^-nepsilon$ around $q_n$, and let $B = bigcup_nB_2^-nepsilon(q_n)$.



    $B$ is open and dense in $A$, so $cl(B) = A$. And the Lebesgue measure of $B$, $mu(B) leq sum_n geq 12^-nepsilon = epsilon < 1 = mu(A)$, so $B$ is a proper subset of $A$, i.e., $A setminus B$ is non-empty.



    This shows that $A cap B = B$, but $B$ is dense in $A$ so $cl(A cap B) = A$.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      Yes this is possible. In general, whenever $(A cap B)$ is dense in some neighbourhood of a point in $A setminus B$, $cl(A cap B)$ wll contain that point.



      Suppose $A = X = [0,1]$. Let $q_n$ be an enumeration of the rationals in $[0,1]$. For some $epsilon < 1$, let $B_2^-nepsilon(q_n)$ be the open ball of diameter $2^-nepsilon$ around $q_n$, and let $B = bigcup_nB_2^-nepsilon(q_n)$.



      $B$ is open and dense in $A$, so $cl(B) = A$. And the Lebesgue measure of $B$, $mu(B) leq sum_n geq 12^-nepsilon = epsilon < 1 = mu(A)$, so $B$ is a proper subset of $A$, i.e., $A setminus B$ is non-empty.



      This shows that $A cap B = B$, but $B$ is dense in $A$ so $cl(A cap B) = A$.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        Yes this is possible. In general, whenever $(A cap B)$ is dense in some neighbourhood of a point in $A setminus B$, $cl(A cap B)$ wll contain that point.



        Suppose $A = X = [0,1]$. Let $q_n$ be an enumeration of the rationals in $[0,1]$. For some $epsilon < 1$, let $B_2^-nepsilon(q_n)$ be the open ball of diameter $2^-nepsilon$ around $q_n$, and let $B = bigcup_nB_2^-nepsilon(q_n)$.



        $B$ is open and dense in $A$, so $cl(B) = A$. And the Lebesgue measure of $B$, $mu(B) leq sum_n geq 12^-nepsilon = epsilon < 1 = mu(A)$, so $B$ is a proper subset of $A$, i.e., $A setminus B$ is non-empty.



        This shows that $A cap B = B$, but $B$ is dense in $A$ so $cl(A cap B) = A$.






        share|cite|improve this answer









        $endgroup$



        Yes this is possible. In general, whenever $(A cap B)$ is dense in some neighbourhood of a point in $A setminus B$, $cl(A cap B)$ wll contain that point.



        Suppose $A = X = [0,1]$. Let $q_n$ be an enumeration of the rationals in $[0,1]$. For some $epsilon < 1$, let $B_2^-nepsilon(q_n)$ be the open ball of diameter $2^-nepsilon$ around $q_n$, and let $B = bigcup_nB_2^-nepsilon(q_n)$.



        $B$ is open and dense in $A$, so $cl(B) = A$. And the Lebesgue measure of $B$, $mu(B) leq sum_n geq 12^-nepsilon = epsilon < 1 = mu(A)$, so $B$ is a proper subset of $A$, i.e., $A setminus B$ is non-empty.



        This shows that $A cap B = B$, but $B$ is dense in $A$ so $cl(A cap B) = A$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 18 at 20:55









        vxnturevxnture

        39910




        39910



























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