Vietoris Topology












1












$begingroup$


‎let ‎‎$ X ‎‎‎‎$ ‎be a‎ ‎topological ‎space ‎and‎ ‎$ operatorname{‎Exp}(X‎)‎‎‎ $ ‎is ‎the set of all ‎closed ‎non-empty subsets of $X$ .‎‎



If $ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎ldots ‎V_{n}‎$ ‎are ‎the non-empty open subset ‎in ‎$ ‎X‎$‎‎‎, ‎define:‎



$$ ‎langle U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, ldots V_{‎n}‎ rangle ‎ ‎ =‎ ‎{ F‎ ‎in operatorname{‎Exp}(X‎)‎‎‎‎mid F‎subseteq‎ ‎U,‎ forall 1‎‎ leq i ‎‎‎leq n ‎‎‎:‎ F‎cap ‎V_{i} ‎neq emptyset}‎$$



families ‎$ B‎ ‎‎ $‎i‎ncludes all sets of the form $langle‎ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, ldots, ‎V_{‎n} rangle ‎$ ‎is ‎‎the basis for a topology for ‎$ ‎Exp (‎ X‎ ) ‎$‎‎
This topology is called the Vietoris topology.



My ‎question:‎



if ‎‎$ ‎X‎$ ‎is ‎‎a $T_{1}$ ‎space, ‎then ‎is the Vietoris ‎topology ‎ ‎‎$ ‎T_{1}‎$‎?‎










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    ‎let ‎‎$ X ‎‎‎‎$ ‎be a‎ ‎topological ‎space ‎and‎ ‎$ operatorname{‎Exp}(X‎)‎‎‎ $ ‎is ‎the set of all ‎closed ‎non-empty subsets of $X$ .‎‎



    If $ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎ldots ‎V_{n}‎$ ‎are ‎the non-empty open subset ‎in ‎$ ‎X‎$‎‎‎, ‎define:‎



    $$ ‎langle U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, ldots V_{‎n}‎ rangle ‎ ‎ =‎ ‎{ F‎ ‎in operatorname{‎Exp}(X‎)‎‎‎‎mid F‎subseteq‎ ‎U,‎ forall 1‎‎ leq i ‎‎‎leq n ‎‎‎:‎ F‎cap ‎V_{i} ‎neq emptyset}‎$$



    families ‎$ B‎ ‎‎ $‎i‎ncludes all sets of the form $langle‎ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, ldots, ‎V_{‎n} rangle ‎$ ‎is ‎‎the basis for a topology for ‎$ ‎Exp (‎ X‎ ) ‎$‎‎
    This topology is called the Vietoris topology.



    My ‎question:‎



    if ‎‎$ ‎X‎$ ‎is ‎‎a $T_{1}$ ‎space, ‎then ‎is the Vietoris ‎topology ‎ ‎‎$ ‎T_{1}‎$‎?‎










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      3



      $begingroup$


      ‎let ‎‎$ X ‎‎‎‎$ ‎be a‎ ‎topological ‎space ‎and‎ ‎$ operatorname{‎Exp}(X‎)‎‎‎ $ ‎is ‎the set of all ‎closed ‎non-empty subsets of $X$ .‎‎



      If $ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎ldots ‎V_{n}‎$ ‎are ‎the non-empty open subset ‎in ‎$ ‎X‎$‎‎‎, ‎define:‎



      $$ ‎langle U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, ldots V_{‎n}‎ rangle ‎ ‎ =‎ ‎{ F‎ ‎in operatorname{‎Exp}(X‎)‎‎‎‎mid F‎subseteq‎ ‎U,‎ forall 1‎‎ leq i ‎‎‎leq n ‎‎‎:‎ F‎cap ‎V_{i} ‎neq emptyset}‎$$



      families ‎$ B‎ ‎‎ $‎i‎ncludes all sets of the form $langle‎ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, ldots, ‎V_{‎n} rangle ‎$ ‎is ‎‎the basis for a topology for ‎$ ‎Exp (‎ X‎ ) ‎$‎‎
      This topology is called the Vietoris topology.



      My ‎question:‎



      if ‎‎$ ‎X‎$ ‎is ‎‎a $T_{1}$ ‎space, ‎then ‎is the Vietoris ‎topology ‎ ‎‎$ ‎T_{1}‎$‎?‎










      share|cite|improve this question











      $endgroup$




      ‎let ‎‎$ X ‎‎‎‎$ ‎be a‎ ‎topological ‎space ‎and‎ ‎$ operatorname{‎Exp}(X‎)‎‎‎ $ ‎is ‎the set of all ‎closed ‎non-empty subsets of $X$ .‎‎



      If $ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎ldots ‎V_{n}‎$ ‎are ‎the non-empty open subset ‎in ‎$ ‎X‎$‎‎‎, ‎define:‎



      $$ ‎langle U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, ldots V_{‎n}‎ rangle ‎ ‎ =‎ ‎{ F‎ ‎in operatorname{‎Exp}(X‎)‎‎‎‎mid F‎subseteq‎ ‎U,‎ forall 1‎‎ leq i ‎‎‎leq n ‎‎‎:‎ F‎cap ‎V_{i} ‎neq emptyset}‎$$



      families ‎$ B‎ ‎‎ $‎i‎ncludes all sets of the form $langle‎ U , ‎V‎_{‎1‎}‎, V‎_{‎2‎}‎, ldots, ‎V_{‎n} rangle ‎$ ‎is ‎‎the basis for a topology for ‎$ ‎Exp (‎ X‎ ) ‎$‎‎
      This topology is called the Vietoris topology.



      My ‎question:‎



      if ‎‎$ ‎X‎$ ‎is ‎‎a $T_{1}$ ‎space, ‎then ‎is the Vietoris ‎topology ‎ ‎‎$ ‎T_{1}‎$‎?‎







      general-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 31 '18 at 7:18









      Martin Sleziak

      45k10123277




      45k10123277










      asked Feb 24 '17 at 11:25









      M.OM.O

      613




      613






















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.



          I normally use a standard subbase for this, for every non-empty open set $U subset X$,
          define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
          The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.



          As to the $T_1$ question: yes.



          Suppose $A neq B$ are two different points in $H(X)$.
          So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).



          Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.



          So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.






          share|cite|improve this answer











          $endgroup$














            Your Answer








            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%2f2159322%2fvietoris-topology%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.



            I normally use a standard subbase for this, for every non-empty open set $U subset X$,
            define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
            The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.



            As to the $T_1$ question: yes.



            Suppose $A neq B$ are two different points in $H(X)$.
            So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).



            Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.



            So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.






            share|cite|improve this answer











            $endgroup$


















              5












              $begingroup$

              This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.



              I normally use a standard subbase for this, for every non-empty open set $U subset X$,
              define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
              The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.



              As to the $T_1$ question: yes.



              Suppose $A neq B$ are two different points in $H(X)$.
              So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).



              Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.



              So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.






              share|cite|improve this answer











              $endgroup$
















                5












                5








                5





                $begingroup$

                This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.



                I normally use a standard subbase for this, for every non-empty open set $U subset X$,
                define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
                The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.



                As to the $T_1$ question: yes.



                Suppose $A neq B$ are two different points in $H(X)$.
                So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).



                Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.



                So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.






                share|cite|improve this answer











                $endgroup$



                This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.



                I normally use a standard subbase for this, for every non-empty open set $U subset X$,
                define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
                The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.



                As to the $T_1$ question: yes.



                Suppose $A neq B$ are two different points in $H(X)$.
                So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).



                Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.



                So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Oct 26 '18 at 13:30

























                answered Feb 24 '17 at 17:09









                Henno BrandsmaHenno Brandsma

                117k349127




                117k349127






























                    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%2f2159322%2fvietoris-topology%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

                    mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

                    How to change which sound is reproduced for terminal bell?

                    Can I use Tabulator js library in my java Spring + Thymeleaf project?