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












2














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



























    2














    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

























      2












      2








      2







      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













      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 2 days ago









      Gilles Castel

      1269




      1269






















          1 Answer
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          3














          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





















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            active

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            active

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            3














            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


























              3














              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
























                3












                3








                3






                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












                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 2 days ago









                Henno Brandsma

                105k347114




                105k347114






























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