Rudin functional analysis, section 3.8 (Topological preliminaries)












2












$begingroup$


A quote from section 3.8. of Rudin's functional analysis:




Suppose next that $X$ is a set and $mathcal{F}$ is a nonempty family of mappings $f: X to Y_f$, where each $Y_f$ is a topological space (In many important cases $Y_f$ is the same for all $f in mathcal{F}$. Let $tau$ be the collection of all unions of finite intersections of sets $f^{-1}(V)$, with $f in mathcal{F}$ and $V$ open in $Y_f$. Then $tau$ is a topology on $X$, and it is in fact the weakest topology on $X$ that makes every $f in mathcal{F}$ continuous.




Why is such topology the weakest?



Also consider in the very same section the following proposition




If $mathcal{F}$ is a family of mappings $f : X to Y_f$ where $X$ is a set and each $Y_f$ is a Hausdorff space, and if $mathcal{F}$ separates points on $X$, then the $mathcal{F}$-topology of $X$ is a Hausdorff topology.




What is the meaning of the terminology $mathcal{F}$-topology on $X$ in this context? Is it either 1) a topology on $X$ constructed using the family $mathcal{F}$ or b) a topology on the set $mathcal{F}$? From the very short proof below it seems to me is the former, but I wanna be sure.






functional-analysis topological-vector-spaces weak-topology







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    A quote from section 3.8. of Rudin's functional analysis:




    Suppose next that $X$ is a set and $mathcal{F}$ is a nonempty family of mappings $f: X to Y_f$, where each $Y_f$ is a topological space (In many important cases $Y_f$ is the same for all $f in mathcal{F}$. Let $tau$ be the collection of all unions of finite intersections of sets $f^{-1}(V)$, with $f in mathcal{F}$ and $V$ open in $Y_f$. Then $tau$ is a topology on $X$, and it is in fact the weakest topology on $X$ that makes every $f in mathcal{F}$ continuous.




    Why is such topology the weakest?



    Also consider in the very same section the following proposition




    If $mathcal{F}$ is a family of mappings $f : X to Y_f$ where $X$ is a set and each $Y_f$ is a Hausdorff space, and if $mathcal{F}$ separates points on $X$, then the $mathcal{F}$-topology of $X$ is a Hausdorff topology.




    What is the meaning of the terminology $mathcal{F}$-topology on $X$ in this context? Is it either 1) a topology on $X$ constructed using the family $mathcal{F}$ or b) a topology on the set $mathcal{F}$? From the very short proof below it seems to me is the former, but I wanna be sure.






    functional-analysis topological-vector-spaces weak-topology







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      A quote from section 3.8. of Rudin's functional analysis:




      Suppose next that $X$ is a set and $mathcal{F}$ is a nonempty family of mappings $f: X to Y_f$, where each $Y_f$ is a topological space (In many important cases $Y_f$ is the same for all $f in mathcal{F}$. Let $tau$ be the collection of all unions of finite intersections of sets $f^{-1}(V)$, with $f in mathcal{F}$ and $V$ open in $Y_f$. Then $tau$ is a topology on $X$, and it is in fact the weakest topology on $X$ that makes every $f in mathcal{F}$ continuous.




      Why is such topology the weakest?



      Also consider in the very same section the following proposition




      If $mathcal{F}$ is a family of mappings $f : X to Y_f$ where $X$ is a set and each $Y_f$ is a Hausdorff space, and if $mathcal{F}$ separates points on $X$, then the $mathcal{F}$-topology of $X$ is a Hausdorff topology.




      What is the meaning of the terminology $mathcal{F}$-topology on $X$ in this context? Is it either 1) a topology on $X$ constructed using the family $mathcal{F}$ or b) a topology on the set $mathcal{F}$? From the very short proof below it seems to me is the former, but I wanna be sure.






      functional-analysis topological-vector-spaces weak-topology







      share|cite|improve this question









      $endgroup$




      A quote from section 3.8. of Rudin's functional analysis:




      Suppose next that $X$ is a set and $mathcal{F}$ is a nonempty family of mappings $f: X to Y_f$, where each $Y_f$ is a topological space (In many important cases $Y_f$ is the same for all $f in mathcal{F}$. Let $tau$ be the collection of all unions of finite intersections of sets $f^{-1}(V)$, with $f in mathcal{F}$ and $V$ open in $Y_f$. Then $tau$ is a topology on $X$, and it is in fact the weakest topology on $X$ that makes every $f in mathcal{F}$ continuous.




      Why is such topology the weakest?



      Also consider in the very same section the following proposition




      If $mathcal{F}$ is a family of mappings $f : X to Y_f$ where $X$ is a set and each $Y_f$ is a Hausdorff space, and if $mathcal{F}$ separates points on $X$, then the $mathcal{F}$-topology of $X$ is a Hausdorff topology.




      What is the meaning of the terminology $mathcal{F}$-topology on $X$ in this context? Is it either 1) a topology on $X$ constructed using the family $mathcal{F}$ or b) a topology on the set $mathcal{F}$? From the very short proof below it seems to me is the former, but I wanna be sure.






      functional-analysis topological-vector-spaces weak-topology




      functional-analysis topological-vector-spaces weak-topology






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 9:52









      user8469759user8469759

      1,4031618




      1,4031618






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          (1) If you want all the $f$ be continuous, then all the $f^{-1}(V)$ must be open. Also, by definition of topology, all unions of finite intersections of such sets must be open. But this family is a topology, as you can check.



          (2) Is the weakest topology that makes continuous all the mappings $finmathcal{F}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            For (2). Why is such topology the weakest?
            $endgroup$
            – user8469759
            Dec 10 '18 at 11:22






          • 2




            $begingroup$
            @user8469759 Because any topology that has those maps be continuous must have those sets as open sets, hence their finite intersections be open, and unions of open sets must also be open. So any topology that has these open sets must contain the topology they generate, so by definition, it is the weakest topology making those maps continuous.
            $endgroup$
            – Aaron
            Dec 10 '18 at 11:26











          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%2f3033712%2frudin-functional-analysis-section-3-8-topological-preliminaries%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









          2












          $begingroup$

          (1) If you want all the $f$ be continuous, then all the $f^{-1}(V)$ must be open. Also, by definition of topology, all unions of finite intersections of such sets must be open. But this family is a topology, as you can check.



          (2) Is the weakest topology that makes continuous all the mappings $finmathcal{F}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            For (2). Why is such topology the weakest?
            $endgroup$
            – user8469759
            Dec 10 '18 at 11:22






          • 2




            $begingroup$
            @user8469759 Because any topology that has those maps be continuous must have those sets as open sets, hence their finite intersections be open, and unions of open sets must also be open. So any topology that has these open sets must contain the topology they generate, so by definition, it is the weakest topology making those maps continuous.
            $endgroup$
            – Aaron
            Dec 10 '18 at 11:26
















          2












          $begingroup$

          (1) If you want all the $f$ be continuous, then all the $f^{-1}(V)$ must be open. Also, by definition of topology, all unions of finite intersections of such sets must be open. But this family is a topology, as you can check.



          (2) Is the weakest topology that makes continuous all the mappings $finmathcal{F}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            For (2). Why is such topology the weakest?
            $endgroup$
            – user8469759
            Dec 10 '18 at 11:22






          • 2




            $begingroup$
            @user8469759 Because any topology that has those maps be continuous must have those sets as open sets, hence their finite intersections be open, and unions of open sets must also be open. So any topology that has these open sets must contain the topology they generate, so by definition, it is the weakest topology making those maps continuous.
            $endgroup$
            – Aaron
            Dec 10 '18 at 11:26














          2












          2








          2





          $begingroup$

          (1) If you want all the $f$ be continuous, then all the $f^{-1}(V)$ must be open. Also, by definition of topology, all unions of finite intersections of such sets must be open. But this family is a topology, as you can check.



          (2) Is the weakest topology that makes continuous all the mappings $finmathcal{F}$.






          share|cite|improve this answer









          $endgroup$



          (1) If you want all the $f$ be continuous, then all the $f^{-1}(V)$ must be open. Also, by definition of topology, all unions of finite intersections of such sets must be open. But this family is a topology, as you can check.



          (2) Is the weakest topology that makes continuous all the mappings $finmathcal{F}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 11:18









          Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

          34.2k42871




          34.2k42871












          • $begingroup$
            For (2). Why is such topology the weakest?
            $endgroup$
            – user8469759
            Dec 10 '18 at 11:22






          • 2




            $begingroup$
            @user8469759 Because any topology that has those maps be continuous must have those sets as open sets, hence their finite intersections be open, and unions of open sets must also be open. So any topology that has these open sets must contain the topology they generate, so by definition, it is the weakest topology making those maps continuous.
            $endgroup$
            – Aaron
            Dec 10 '18 at 11:26


















          • $begingroup$
            For (2). Why is such topology the weakest?
            $endgroup$
            – user8469759
            Dec 10 '18 at 11:22






          • 2




            $begingroup$
            @user8469759 Because any topology that has those maps be continuous must have those sets as open sets, hence their finite intersections be open, and unions of open sets must also be open. So any topology that has these open sets must contain the topology they generate, so by definition, it is the weakest topology making those maps continuous.
            $endgroup$
            – Aaron
            Dec 10 '18 at 11:26
















          $begingroup$
          For (2). Why is such topology the weakest?
          $endgroup$
          – user8469759
          Dec 10 '18 at 11:22




          $begingroup$
          For (2). Why is such topology the weakest?
          $endgroup$
          – user8469759
          Dec 10 '18 at 11:22




          2




          2




          $begingroup$
          @user8469759 Because any topology that has those maps be continuous must have those sets as open sets, hence their finite intersections be open, and unions of open sets must also be open. So any topology that has these open sets must contain the topology they generate, so by definition, it is the weakest topology making those maps continuous.
          $endgroup$
          – Aaron
          Dec 10 '18 at 11:26




          $begingroup$
          @user8469759 Because any topology that has those maps be continuous must have those sets as open sets, hence their finite intersections be open, and unions of open sets must also be open. So any topology that has these open sets must contain the topology they generate, so by definition, it is the weakest topology making those maps continuous.
          $endgroup$
          – Aaron
          Dec 10 '18 at 11:26


















          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%2f3033712%2frudin-functional-analysis-section-3-8-topological-preliminaries%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

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
          Read more

          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

          Image
          0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
          Read more