Spaces in which all closed sets are regular closed The Next CEO of Stack OverflowA space is...

Does the Idaho Potato Commission associate potato skins with healthy eating?

How to get the last not-null value in an ordered column of a huge table?

Does Germany produce more waste than the US?

Which one is the true statement?

IC has pull-down resistors on SMBus lines?

Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?

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

How did Beeri the Hittite come up with naming his daughter Yehudit?

Is a distribution that is normal, but highly skewed, considered Gaussian?

Inexact numbers as keys in Association?

New carbon wheel brake pads after use on aluminum wheel?

Prepend last line of stdin to entire stdin

AB diagonalizable then BA also diagonalizable

Reference request: Grassmannian and Plucker coordinates in type B, C, D

Computationally populating tables with probability data

Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?

Film where the government was corrupt with aliens, people sent to kill aliens are given rigged visors not showing the right aliens

What was Carter Burkes job for "the company" in "Aliens"?

Easy to read palindrome checker

Reshaping json / reparing json inside shell script (remove trailing comma)

Audio Conversion With ADS1243

Expressing the idea of having a very busy time

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

How many extra stops do monopods offer for tele photographs?



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 $A=text{cl}(text{int}(A))$




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











$endgroup$

















    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 $A=text{cl}(text{int}(A))$




    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











    $endgroup$















      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 $A=text{cl}(text{int}(A))$




      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 21 mins ago









      Eric Wofsey

      191k14216349




      191k14216349










      asked 54 mins ago









      Carlos JiménezCarlos Jiménez

      2,4341621




      2,4341621






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          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









          $endgroup$





















            2












            $begingroup$

            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
              48 mins 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
              46 mins 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









            2












            $begingroup$

            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









            $endgroup$


















              2












              $begingroup$

              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









              $endgroup$
















                2












                2








                2





                $begingroup$

                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 28 mins ago









                Eric WofseyEric Wofsey

                191k14216349




                191k14216349























                    2












                    $begingroup$

                    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
                      48 mins 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
                      46 mins ago


















                    2












                    $begingroup$

                    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
                      48 mins 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
                      46 mins ago
















                    2












                    2








                    2





                    $begingroup$

                    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 50 mins 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
                      48 mins 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
                      46 mins ago




















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


















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




                    $begingroup$
                    Is there a non trivial example? I don't mind the separation axiom.
                    $endgroup$
                    – Carlos Jiménez
                    48 mins 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
                    46 mins 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
                    46 mins 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

                    Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

                    Image
                    Tidlegare republikkarSkipingar i 1433Kommunar i SchwyzDistrikt i SchwyzNedleggingar i 1798Stader ved Vierwaldstättersee kommunekantonenSchwyzSveitskm²VierwaldstätterseeLauerz Gersau Kommunevåpenet til Gersau Plassering Styresmakter Land   Sveits Kanton kantonen Schwyz Distrikt Gersau Språk tysk Geografi Flatevidd  - kommune 23,70 km² Innbyggjarar  - kommune (desember 2010)    - folketettleik 2 094   88,4 /km² Koordinatar 46°59′0″N 8°31′0″E  Koordinatar: 46°59′0″N 8°31′0″E  Høgd over havet 435 moh Diverse annan informasjon Postnummer 6442 Nettstad: www.gersau.ch Freistaat und Republik Gersau Fristaten og republikken Gersau By-republikk og alliert protektorat i Det gamle sveitsiske eidsforundet ←   ← 1433–1798 1814–1818 →   → Hovudstad Gersau Styreform Republikk Historisk periode Tidleg moderne tid  - Kjøpte fridom frå ...
                    Read more

                    Nässjö kommun Tettstader | Kjelder | NavigasjonsmenyeVIAFISNIGeoNamesMusicBrainz (area)

                    Image
                    Nässjö kommunKommunar i Jönköpings län svenskeJönköpings länAdministrasjonssenterNässjöAneby kommunEksjö kommunVetlanda kommunSävsjö kommuneVaggeryd kommuneJönköping kommunejernbaneknutepunktBackyard BabiesEidsbergtettstader31. desember2005 Nässjö kommun kommune Kommunevåpen Land   Sverige Len Jönköpings län Landskap Småland Areal 988,49 km²  - land 930,29 km²  - vatn 58,2 km² Folketal 29 375 Wikimedia Commons: Nässjö Municipality Nässjö kommune ligg i det svenske länet Jönköpings län. Administrasjonssenter i kommunen er Nässjö. Kommunen grensar til Aneby kommun i nord, Eksjö kommun og Vetlanda kommun i aust, Sävsjö kommune og Vaggeryd kommune i sør og Jönköping kommune i vest. Kommunen er kjend som jernbaneknutepunkt, og som heimstaden til rockebandet Backyard Babies. Nässjö er venskapskommune med Eidsberg. Nässjö sentrum sett frå vasstårnet. Tettstader | Nässjö kommune har tol...
                    Read more

                    Kvitkval Innhaldsliste Taksonomi og utvikling | Utsjånad og levevis | Utbreiing | Åtferd |...

                    Image
                    Nær truga artarPattedyr i NoregNarkvalfamilien tannkvalarktiskenarkvalenmiocenkvalendelfinanetannkvalarLatinbiosonarenekkolokaliseringisbjørnenspekkhoggarinuittarArktisNew YorkNord-AmerikaEuropaJapan Kvitkval Status i verda: Nær truga Status i Noreg: Datamangel Kvitkval Systematikk Domene: Eukaryota Rike: Animalia Rekkje: Chordata Underrekkje: Vertebrata Klasse: Mammalia Orden: Cetaceae Familie: Monodontidae Slekt: Delphinapterus Art: D. leucas Vitskapleg namn Delphinapterus leucas Pallas, 1776 Kvitkval er ein tannkval som lever i arktiske og subarktiske område. Innhaldsliste 1 Taksonomi og utvikling 2 Utsjånad og levevis 3 Utbreiing 4 Åtferd 5 Bestandsstorleik og menneskeleg påverknad 6 Kjelder Taksonomi og utvikling | Kvitkvalen vart første gongen skildra av Peter Simon Pallas i 1776. Han er medlem av Monodontidae -familien saman med narkvalen. D...
                    Read more