Spaces in which all closed sets are regular closed The Next CEO of Stack OverflowA space is regular if each closed set $Z$ is the intersection of all open sets containing $Z$?Examples of topologies in which all open sets are regular?All zero dimensional spaces are completely regular.All finite Baire measures are Closed-regular?On the small first countable regular spacesShow that if $X$ and $Y$ are regular, then so is the product space $Xtimes Y$.Regular spaces and Hausdorff spaceHow to Show that Points and Closed Sets Can be Separated by Closed Sets in a T3 (Regular) SpaceSaturated sets and topological spacesOn regular closed sets which are not open-closed

Is there a way to save my career from absolute disaster?

Easy to read palindrome checker

Purpose of level-shifter with same in and out voltages

IC has pull-down resistors on SMBus lines?

Spaces in which all closed sets are regular closed

What happened in Rome, when the western empire "fell"?

Does higher Oxidation/ reduction potential translate to higher energy storage in battery?

Getting Stale Gas Out of a Gas Tank w/out Dropping the Tank

Traduction de « Life is a roller coaster »

Is there an equivalent of cd - for cp or mv

In the "Harry Potter and the Order of the Phoenix" video game, what potion is used to sabotage Umbridge's speakers?

When "be it" is at the beginning of a sentence, what kind of structure do you call it?

Won the lottery - how do I keep the money?

Lucky Feat: How can "more than one creature spend a luck point to influence the outcome of a roll"?

Physiological effects of huge anime eyes

How do I fit a non linear curve?

Are the names of these months realistic?

what's the use of '% to gdp' type of variables?

Do scriptures give a method to recognize a truly self-realized person/jivanmukta?

How many extra stops do monopods offer for tele photographs?

My ex-girlfriend uses my Apple ID to login to her iPad, do I have to give her my Apple ID password to reset it?

How to find image of a complex function with given constraints?

Why is information "lost" when it got into a black hole?

If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?



Spaces in which all closed sets are regular closed



The Next CEO of Stack OverflowA space is regular if each closed set $Z$ is the intersection of all open sets containing $Z$?Examples of topologies in which all open sets are regular?All zero dimensional spaces are completely regular.All finite Baire measures are Closed-regular?On the small first countable regular spacesShow that if $X$ and $Y$ are regular, then so is the product space $Xtimes Y$.Regular spaces and Hausdorff spaceHow to Show that Points and Closed Sets Can be Separated by Closed Sets in a T3 (Regular) SpaceSaturated sets and topological spacesOn regular closed sets which are not open-closed










2












$begingroup$


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 $A=textcl(textint(A))$




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











$endgroup$
















    2












    $begingroup$


    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 $A=textcl(textint(A))$




    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











    $endgroup$














      2












      2








      2





      $begingroup$


      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 $A=textcl(textint(A))$




      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 $A=textcl(textint(A))$




      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 48 mins ago









      Eric Wofsey

      191k14216349




      191k14216349










      asked 1 hour ago









      Carlos JiménezCarlos Jiménez

      2,4341621




      2,4341621




















          2 Answers
          2






          active

          oldest

          votes


















          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 $yinoverlinex$. Since $overliney$ 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 $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ 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












            $begingroup$

            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
              1 hour 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
              1 hour ago












            Your Answer





            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader:
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            ,
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );













            draft saved

            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169975%2fspaces-in-which-all-closed-sets-are-regular-closed%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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 $yinoverlinex$. Since $overliney$ 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 $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ 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 $yinoverlinex$. Since $overliney$ 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 $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ 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 $yinoverlinex$. Since $overliney$ 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 $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ 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 $yinoverlinex$. Since $overliney$ 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 $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ 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 55 mins 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
                      1 hour 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
                      1 hour 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
                      1 hour 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
                      1 hour 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 1 hour 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
                      1 hour 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
                      1 hour ago

















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
















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




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


















                    draft saved

                    draft discarded
















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …


                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.

                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169975%2fspaces-in-which-all-closed-sets-are-regular-closed%23new-answer', 'question_page');

                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    名間水力發電廠 目录 沿革 設施 鄰近設施 註釋 外部連結 导航菜单23°50′10″N 120°42′41″E / 23.83611°N 120.71139°E / 23.83611; 120.7113923°50′10″N 120°42′41″E / 23.83611°N 120.71139°E / 23.83611; 120.71139計畫概要原始内容臺灣第一座BOT 模式開發的水力發電廠－名間水力電廠名間水力發電廠 水利署首件BOT案原始内容《小檔案》名間電廠 首座BOT水力發電廠原始内容名間電廠BOT - 經濟部水利署中區水資源局编编

                    Image
                    東螺溪清水溪加走寮溪阿里山溪東埔蚋溪獅尾堀清水溪清水溝溪水里溪陳有蘭溪郡坑溪十八重溪那馬嘎班溪松柏坑溪卓棍溪瓠瓢坑溪益則坑溪玉崙溪丹大溪郡大溪巒大溪卡社溪栗栖溪萬大溪萬大北溪萬大南溪霧社溪塔羅灣溪馬赫坡溪萬大發電廠大觀發電廠日月潭第一發電所明潭發電廠...
                    Read more

                    Prove that NP is closed under karp reduction?Space(n) not closed under Karp reductions - what about NTime(n)?Class P is closed under rotation?Prove or disprove that $NL$ is closed under polynomial many-one reductions$mathbfNC_2$ is closed under log-space reductionOn Karp reductionwhen can I know if a class (complexity) is closed under reduction (cook/karp)Check if class $PSPACE$ is closed under polyonomially space reductionIs NPSPACE also closed under polynomial-time reduction and under log-space reduction?Prove PSPACE is closed under complement?Prove PSPACE is closed under union?

                    Image
                    Is a conference paper whose proceedings will be published in IEEE Xplore counted as a publication? Prove that NP is closed under karp reduction? Is it important to consider tone, melody, and musical form while writing a song? How is it possible to have an ability score that is less than 3? Can I make popcorn with any corn? Mathematical cryptic clues tikz: show 0 at the axis origin Minkowski space can i play a electric guitar through a bass amp? Why not use SQL instead of GraphQL? How can I prevent hyper evolved versions of regular creatures from wiping out their cousins? How old can references or sources in a thesis be? Do VLANs within a subnet need to have their own subnet for router on a stick? How to add double frame in tcolorbox? How much RAM could one put in a typical 80386 setup? "You are your self first supporter", a more proper way to say it How is the claim "I am in New York only if I am in America" the same as "If I am ...
                    Read more

                    Is my guitar’s action too high? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Strings too stiff on a recently purchased acoustic guitar | Cort AD880CEIs the action of my guitar really high?Μy little finger is too weak to play guitarWith guitar, how long should I give my fingers to strengthen / callous?When playing a fret the guitar sounds mutedPlaying (Barre) chords up the guitar neckI think my guitar strings are wound too tight and I can't play barre chordsF barre chord on an SG guitarHow to find to the right strings of a barre chord by feel?High action on higher fret on my steel acoustic guitar

                    Image
                    Is the Mordenkainen's Sword spell underpowered? Escaping $ in listings environment Short story about astronauts fertilizing soil with their own bodies No invitation for tourist visa but I want to visit Do regular languages belong to Space(1)? Can I take recommendation from someone I met at a conference? How to name indistinguishable henchmen in a screenplay? Found this skink in my tomato plant bucket. Is he trapped? Or could he leave if he wanted? How can I prevent/balance waiting and turtling as a response to cooldown mechanics Can I feed enough spin up electron to a black hole to affect it's angular momentum? Why does BitLocker not use RSA? Is my guitar’s action too high? Does GDPR cover the collection of data by websites that crawl the web and resell user data Trying to enter the Fox's den How to achieve cat-like agility? "Destructive power" carried by a B-52? Should man-made satellites feature an intelligent inverted "cow c...
                    Read more