Show that $tau := {X setminus phi (A) mid A subset X }$ is a topology on $X$.












2












$begingroup$


Let $X$ be a set and $mathcal P (X)$ the powerset of $X$. Suppose $phi: mathcal P(X) to mathcal P(X)$ is a function such that





  • $phi(emptyset) = emptyset$,

  • For all $A subset X$: $A subset phi(A)$,

  • For all $A subset X$: $phi(phi(A)) = phi(A)$ and

  • For all $A, B in mathcal P(X)$: $phi(A cup B) = phi(A) cup phi(B)$.


Show that $tau := { X setminus phi(A) mid A subset X }$ is a topology on $X$.



Progress I've made:





  • $emptyset = X setminus phi(X) in tau$.


  • $X = X setminus phi(emptyset) in tau$.

  • Finite intersections: $(X setminus phi(A)) cap (X setminus phi(B)) = X setminus (phi(A) cup phi(B)) = X setminus phi(A cup B) in tau$.

  • Infinite unions: This is what I'm struggling with. How do I complete the proof?






general-topology







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $X$ be a set and $mathcal P (X)$ the powerset of $X$. Suppose $phi: mathcal P(X) to mathcal P(X)$ is a function such that





    • $phi(emptyset) = emptyset$,

    • For all $A subset X$: $A subset phi(A)$,

    • For all $A subset X$: $phi(phi(A)) = phi(A)$ and

    • For all $A, B in mathcal P(X)$: $phi(A cup B) = phi(A) cup phi(B)$.


    Show that $tau := { X setminus phi(A) mid A subset X }$ is a topology on $X$.



    Progress I've made:





    • $emptyset = X setminus phi(X) in tau$.


    • $X = X setminus phi(emptyset) in tau$.

    • Finite intersections: $(X setminus phi(A)) cap (X setminus phi(B)) = X setminus (phi(A) cup phi(B)) = X setminus phi(A cup B) in tau$.

    • Infinite unions: This is what I'm struggling with. How do I complete the proof?






    general-topology







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $X$ be a set and $mathcal P (X)$ the powerset of $X$. Suppose $phi: mathcal P(X) to mathcal P(X)$ is a function such that





      • $phi(emptyset) = emptyset$,

      • For all $A subset X$: $A subset phi(A)$,

      • For all $A subset X$: $phi(phi(A)) = phi(A)$ and

      • For all $A, B in mathcal P(X)$: $phi(A cup B) = phi(A) cup phi(B)$.


      Show that $tau := { X setminus phi(A) mid A subset X }$ is a topology on $X$.



      Progress I've made:





      • $emptyset = X setminus phi(X) in tau$.


      • $X = X setminus phi(emptyset) in tau$.

      • Finite intersections: $(X setminus phi(A)) cap (X setminus phi(B)) = X setminus (phi(A) cup phi(B)) = X setminus phi(A cup B) in tau$.

      • Infinite unions: This is what I'm struggling with. How do I complete the proof?






      general-topology







      share|cite|improve this question









      $endgroup$




      Let $X$ be a set and $mathcal P (X)$ the powerset of $X$. Suppose $phi: mathcal P(X) to mathcal P(X)$ is a function such that





      • $phi(emptyset) = emptyset$,

      • For all $A subset X$: $A subset phi(A)$,

      • For all $A subset X$: $phi(phi(A)) = phi(A)$ and

      • For all $A, B in mathcal P(X)$: $phi(A cup B) = phi(A) cup phi(B)$.


      Show that $tau := { X setminus phi(A) mid A subset X }$ is a topology on $X$.



      Progress I've made:





      • $emptyset = X setminus phi(X) in tau$.


      • $X = X setminus phi(emptyset) in tau$.

      • Finite intersections: $(X setminus phi(A)) cap (X setminus phi(B)) = X setminus (phi(A) cup phi(B)) = X setminus phi(A cup B) in tau$.

      • Infinite unions: This is what I'm struggling with. How do I complete the proof?






      general-topology




      general-topology






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 3 at 19:26









      Gilles CastelGilles Castel

      1269




      1269






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:



          $C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED



          Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.



          Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.



          Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:



          $$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$



          as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.






          share|cite|improve this answer









          $endgroup$














            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%2f3060906%2fshow-that-tau-x-setminus-phi-a-mid-a-subset-x-is-a-topology-on%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









            3












            $begingroup$

            Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:



            $C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED



            Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.



            Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.



            Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:



            $$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$



            as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:



              $C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED



              Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.



              Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.



              Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:



              $$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$



              as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:



                $C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED



                Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.



                Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.



                Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:



                $$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$



                as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.






                share|cite|improve this answer









                $endgroup$



                Easy fact from the axioms: $C subseteq D$ implies $phi(C) subseteq phi(D)$:



                $C subseteq D$ iff $C cup D=D$ which implies $phi(C) cup phi(D) = phi(C cup D) = phi(D)$ which holds iff $phi(C) subseteq phi(D)$. QED



                Suppose the subsets $B_i, i in I$ of $X$ obey $phi(B_i) = B_i$ for all $i$.



                Then define $B=bigcap_{i in I} B_i$ and note that by the second axiom $B subseteq phi(B)$. On the other hand, for all $i$ $B subseteq B_i$ so that by the first fact we have $forall i: phi(B) subseteq phi(B_i) = B_i$ so $phi(B) subseteq bigcap_{i in I} B_i = B$ and so $phi(B) = B$.



                Now if $Xsetminus phi(A_i), i in I$ are in $tau$ then:



                $$bigcup_i (Xsetminus phi(A_i)) = Xsetminus bigcap_{i in I} phi(A_i)$$ and as $phi(phi(A_i)) = phi(A_i)$ we apply the last paragraph's fact (with $B_i = phi(A_i)$ on intersections to see that $bigcap_{i in I} phi(A_i) = phileft(bigcap_{i in I} phi(A_i)right)$ and so $$bigcup_i (Xsetminus phi(A_i)) = X setminus phileft(bigcap_{i in I} phi(A_i)right) in tau$$



                as required. The proof idea is clear: sets $phi(A)$ are just the closed sets of the topology and thus ought to be closed under all intersections.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 3 at 22:52









                Henno BrandsmaHenno Brandsma

                115k349125




                115k349125






























                    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%2f3060906%2fshow-that-tau-x-setminus-phi-a-mid-a-subset-x-is-a-topology-on%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

                    Probability when a professor distributes a quiz and homework assignment to a class of n students.

                    Image
                    0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                    Read more

                    Aardman Animations

                    Image
                    Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                    Read more

                    Are they similar matrix

                    Image
                    0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                    Read more