Spaces in which all closed sets are regular closed












3












$begingroup$


I was reading about the regular closed sets. The definition is




Let $X$ be a topological space and $Asubseteq X$. We say that $A$ is a regular closed if $A=text{cl}(text{int}(A))$




Then, one question came to my mind: is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set?



Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists?



Thanks in advance.






general-topology examples-counterexamples







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$endgroup$

















    3












    $begingroup$


    I was reading about the regular closed sets. The definition is




    Let $X$ be a topological space and $Asubseteq X$. We say that $A$ is a regular closed if $A=text{cl}(text{int}(A))$




    Then, one question came to my mind: is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set?



    Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists?



    Thanks in advance.






    general-topology examples-counterexamples







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      I was reading about the regular closed sets. The definition is




      Let $X$ be a topological space and $Asubseteq X$. We say that $A$ is a regular closed if $A=text{cl}(text{int}(A))$




      Then, one question came to my mind: is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set?



      Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists?



      Thanks in advance.






      general-topology examples-counterexamples







      share|cite|improve this question











      $endgroup$




      I was reading about the regular closed sets. The definition is




      Let $X$ be a topological space and $Asubseteq X$. We say that $A$ is a regular closed if $A=text{cl}(text{int}(A))$




      Then, one question came to my mind: is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set?



      Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists?



      Thanks in advance.






      general-topology examples-counterexamples




      general-topology examples-counterexamples






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      Eric Wofsey

      191k14216349




      191k14216349










      asked 2 hours ago









      Carlos JiménezCarlos Jiménez

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      2,4391621






















          2 Answers
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          $begingroup$

          Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)



          I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverline{{x}}$. Since $overline{{y}}$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since ${y}$ is dense in $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ is open.



          So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.






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            3












            $begingroup$

            You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.



            But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Is there a non trivial example? I don't mind the separation axiom.
              $endgroup$
              – Carlos Jiménez
              2 hours ago










            • $begingroup$
              @CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
              $endgroup$
              – Moishe Kohan
              2 hours ago














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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            votes






            active

            oldest

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            3












            $begingroup$

            Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)



            I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverline{{x}}$. Since $overline{{y}}$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since ${y}$ is dense in $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ is open.



            So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)



              I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverline{{x}}$. Since $overline{{y}}$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since ${y}$ is dense in $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ is open.



              So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)



                I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverline{{x}}$. Since $overline{{y}}$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since ${y}$ is dense in $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ is open.



                So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.






                share|cite|improve this answer









                $endgroup$



                Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)



                I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverline{{x}}$. Since $overline{{y}}$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since ${y}$ is dense in $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ is open.



                So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Eric WofseyEric Wofsey

                191k14216349




                191k14216349























                    3












                    $begingroup$

                    You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.



                    But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Is there a non trivial example? I don't mind the separation axiom.
                      $endgroup$
                      – Carlos Jiménez
                      2 hours ago










                    • $begingroup$
                      @CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
                      $endgroup$
                      – Moishe Kohan
                      2 hours ago


















                    3












                    $begingroup$

                    You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.



                    But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Is there a non trivial example? I don't mind the separation axiom.
                      $endgroup$
                      – Carlos Jiménez
                      2 hours ago










                    • $begingroup$
                      @CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
                      $endgroup$
                      – Moishe Kohan
                      2 hours ago
















                    3












                    3








                    3





                    $begingroup$

                    You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.



                    But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.






                    share|cite|improve this answer









                    $endgroup$



                    You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.



                    But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    Moishe KohanMoishe Kohan

                    48.4k344110




                    48.4k344110












                    • $begingroup$
                      Is there a non trivial example? I don't mind the separation axiom.
                      $endgroup$
                      – Carlos Jiménez
                      2 hours ago










                    • $begingroup$
                      @CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
                      $endgroup$
                      – Moishe Kohan
                      2 hours ago




















                    • $begingroup$
                      Is there a non trivial example? I don't mind the separation axiom.
                      $endgroup$
                      – Carlos Jiménez
                      2 hours ago










                    • $begingroup$
                      @CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
                      $endgroup$
                      – Moishe Kohan
                      2 hours ago


















                    $begingroup$
                    Is there a non trivial example? I don't mind the separation axiom.
                    $endgroup$
                    – Carlos Jiménez
                    2 hours ago




                    $begingroup$
                    Is there a non trivial example? I don't mind the separation axiom.
                    $endgroup$
                    – Carlos Jiménez
                    2 hours ago












                    $begingroup$
                    @CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
                    $endgroup$
                    – Moishe Kohan
                    2 hours ago






                    $begingroup$
                    @CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
                    $endgroup$
                    – Moishe Kohan
                    2 hours ago




















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