A special open subset of a completely regular space












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Let $X $ be a completely regular topological space and let $U $ be an open subset of $X $. I am looking for an equivalent condition on $U $ such that if $C_1$ and $C_2$ are two closed subsets of $X $ with $Usubseteq C_1cup C_2$, then $Usubseteq C_1$ or $Usubseteq C_2$.



For example, isolated points have that property.






general-topology







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    $begingroup$


    Let $X $ be a completely regular topological space and let $U $ be an open subset of $X $. I am looking for an equivalent condition on $U $ such that if $C_1$ and $C_2$ are two closed subsets of $X $ with $Usubseteq C_1cup C_2$, then $Usubseteq C_1$ or $Usubseteq C_2$.



    For example, isolated points have that property.






    general-topology







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Let $X $ be a completely regular topological space and let $U $ be an open subset of $X $. I am looking for an equivalent condition on $U $ such that if $C_1$ and $C_2$ are two closed subsets of $X $ with $Usubseteq C_1cup C_2$, then $Usubseteq C_1$ or $Usubseteq C_2$.



      For example, isolated points have that property.






      general-topology







      share|cite|improve this question











      $endgroup$




      Let $X $ be a completely regular topological space and let $U $ be an open subset of $X $. I am looking for an equivalent condition on $U $ such that if $C_1$ and $C_2$ are two closed subsets of $X $ with $Usubseteq C_1cup C_2$, then $Usubseteq C_1$ or $Usubseteq C_2$.



      For example, isolated points have that property.






      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 8 at 22:07









      Eric Wofsey

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      183k13211338










      asked Jan 8 at 17:10









      K.ZK.Z

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          Very generally, a topological space with this property (that it cannot be written as a union of two proper closed sets) is called irreducible. So, what you are asking for is for $U$ to be irreducible as a subspace of $X$. A Hausdorff space with more than one point is never irreducible: pick distinct points $x,y$ and disjoint open sets containing them, and then the complements of these open sets will be closed proper subsets whose union is the whole space.



          So, if your definition of "completely regular" includes that $X$ is $T_0$ (and thus Hausdorff), the only such open subsets $U$ are the empty set and singletons. (If $X$ is not required to be $T_0$, its $T_0$ quotient will still be Hausdorff, and so the image of $U$ in the $T_0$ quotient has at most one point; equivalently, $U$ must have the indiscrete topology.)






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            $begingroup$

            Very generally, a topological space with this property (that it cannot be written as a union of two proper closed sets) is called irreducible. So, what you are asking for is for $U$ to be irreducible as a subspace of $X$. A Hausdorff space with more than one point is never irreducible: pick distinct points $x,y$ and disjoint open sets containing them, and then the complements of these open sets will be closed proper subsets whose union is the whole space.



            So, if your definition of "completely regular" includes that $X$ is $T_0$ (and thus Hausdorff), the only such open subsets $U$ are the empty set and singletons. (If $X$ is not required to be $T_0$, its $T_0$ quotient will still be Hausdorff, and so the image of $U$ in the $T_0$ quotient has at most one point; equivalently, $U$ must have the indiscrete topology.)






            share|cite|improve this answer









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              $begingroup$

              Very generally, a topological space with this property (that it cannot be written as a union of two proper closed sets) is called irreducible. So, what you are asking for is for $U$ to be irreducible as a subspace of $X$. A Hausdorff space with more than one point is never irreducible: pick distinct points $x,y$ and disjoint open sets containing them, and then the complements of these open sets will be closed proper subsets whose union is the whole space.



              So, if your definition of "completely regular" includes that $X$ is $T_0$ (and thus Hausdorff), the only such open subsets $U$ are the empty set and singletons. (If $X$ is not required to be $T_0$, its $T_0$ quotient will still be Hausdorff, and so the image of $U$ in the $T_0$ quotient has at most one point; equivalently, $U$ must have the indiscrete topology.)






              share|cite|improve this answer









              $endgroup$
















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                3








                3





                $begingroup$

                Very generally, a topological space with this property (that it cannot be written as a union of two proper closed sets) is called irreducible. So, what you are asking for is for $U$ to be irreducible as a subspace of $X$. A Hausdorff space with more than one point is never irreducible: pick distinct points $x,y$ and disjoint open sets containing them, and then the complements of these open sets will be closed proper subsets whose union is the whole space.



                So, if your definition of "completely regular" includes that $X$ is $T_0$ (and thus Hausdorff), the only such open subsets $U$ are the empty set and singletons. (If $X$ is not required to be $T_0$, its $T_0$ quotient will still be Hausdorff, and so the image of $U$ in the $T_0$ quotient has at most one point; equivalently, $U$ must have the indiscrete topology.)






                share|cite|improve this answer









                $endgroup$



                Very generally, a topological space with this property (that it cannot be written as a union of two proper closed sets) is called irreducible. So, what you are asking for is for $U$ to be irreducible as a subspace of $X$. A Hausdorff space with more than one point is never irreducible: pick distinct points $x,y$ and disjoint open sets containing them, and then the complements of these open sets will be closed proper subsets whose union is the whole space.



                So, if your definition of "completely regular" includes that $X$ is $T_0$ (and thus Hausdorff), the only such open subsets $U$ are the empty set and singletons. (If $X$ is not required to be $T_0$, its $T_0$ quotient will still be Hausdorff, and so the image of $U$ in the $T_0$ quotient has at most one point; equivalently, $U$ must have the indiscrete topology.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 22:17









                Eric WofseyEric Wofsey

                183k13211338




                183k13211338






























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