Need help proving the product of closed sets is closed in the product topology












-1












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Let ${X_{alpha}}$ be a family of topological spaces and ${A_{alpha}}$ be a family of sets such that $A_{alpha} subseteq X_{alpha}$ and $A_{alpha}$ is closed in $X_{alpha}$.
How do you prove that $prod A_{alpha}$ is closed in $prod X_{alpha}$ with the product topology? Is it also the case for the box topology in $prod X_{alpha}$? Use this to prove $overline{prod_{}^{} {A_{alpha}}} = prod_{}^{}{overline{A_{alpha}}}$
in $ prod_{}^{} {X_{alpha}}$ with the product topology. Is this also true in the box topology?






general-topology







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  • $begingroup$
    Have you forgotten to accept the answer? This is the minimum you should do to show your appreciation for the answerer. Just klick on the checkmark of your preferred answer.
    $endgroup$
    – callculus
    Jan 14 at 14:07


















-1












$begingroup$


Let ${X_{alpha}}$ be a family of topological spaces and ${A_{alpha}}$ be a family of sets such that $A_{alpha} subseteq X_{alpha}$ and $A_{alpha}$ is closed in $X_{alpha}$.
How do you prove that $prod A_{alpha}$ is closed in $prod X_{alpha}$ with the product topology? Is it also the case for the box topology in $prod X_{alpha}$? Use this to prove $overline{prod_{}^{} {A_{alpha}}} = prod_{}^{}{overline{A_{alpha}}}$
in $ prod_{}^{} {X_{alpha}}$ with the product topology. Is this also true in the box topology?






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you forgotten to accept the answer? This is the minimum you should do to show your appreciation for the answerer. Just klick on the checkmark of your preferred answer.
    $endgroup$
    – callculus
    Jan 14 at 14:07
















-1












-1








-1





$begingroup$


Let ${X_{alpha}}$ be a family of topological spaces and ${A_{alpha}}$ be a family of sets such that $A_{alpha} subseteq X_{alpha}$ and $A_{alpha}$ is closed in $X_{alpha}$.
How do you prove that $prod A_{alpha}$ is closed in $prod X_{alpha}$ with the product topology? Is it also the case for the box topology in $prod X_{alpha}$? Use this to prove $overline{prod_{}^{} {A_{alpha}}} = prod_{}^{}{overline{A_{alpha}}}$
in $ prod_{}^{} {X_{alpha}}$ with the product topology. Is this also true in the box topology?






general-topology







share|cite|improve this question











$endgroup$




Let ${X_{alpha}}$ be a family of topological spaces and ${A_{alpha}}$ be a family of sets such that $A_{alpha} subseteq X_{alpha}$ and $A_{alpha}$ is closed in $X_{alpha}$.
How do you prove that $prod A_{alpha}$ is closed in $prod X_{alpha}$ with the product topology? Is it also the case for the box topology in $prod X_{alpha}$? Use this to prove $overline{prod_{}^{} {A_{alpha}}} = prod_{}^{}{overline{A_{alpha}}}$
in $ prod_{}^{} {X_{alpha}}$ with the product topology. Is this also true in the box topology?






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 21:34







Sepehr23

















asked Dec 15 '18 at 20:49









Sepehr23Sepehr23

123




123












  • $begingroup$
    Have you forgotten to accept the answer? This is the minimum you should do to show your appreciation for the answerer. Just klick on the checkmark of your preferred answer.
    $endgroup$
    – callculus
    Jan 14 at 14:07




















  • $begingroup$
    Have you forgotten to accept the answer? This is the minimum you should do to show your appreciation for the answerer. Just klick on the checkmark of your preferred answer.
    $endgroup$
    – callculus
    Jan 14 at 14:07


















$begingroup$
Have you forgotten to accept the answer? This is the minimum you should do to show your appreciation for the answerer. Just klick on the checkmark of your preferred answer.
$endgroup$
– callculus
Jan 14 at 14:07






$begingroup$
Have you forgotten to accept the answer? This is the minimum you should do to show your appreciation for the answerer. Just klick on the checkmark of your preferred answer.
$endgroup$
– callculus
Jan 14 at 14:07












2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint:



Let $pi_{alpha}$ be projection on the $X_{alpha}$ then
$$cap_{alpha } ~pi_{alpha}^{-1} (A_{alpha}) = prod A_{alpha}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's only helpful for finite index sets, isn't it?
    $endgroup$
    – YoungMath
    Dec 16 '18 at 10:12






  • 1




    $begingroup$
    @YoungMath Intersection of arbitrary numbers of closed sets is closed
    $endgroup$
    – Red shoes
    Dec 16 '18 at 17:33










  • $begingroup$
    Of course! You're right! Then, your approach is more concise.
    $endgroup$
    – YoungMath
    Dec 16 '18 at 17:36










  • $begingroup$
    @sepehr could you please accept my answer by clicking on the check mark?
    $endgroup$
    – Red shoes
    Dec 22 '18 at 0:03






  • 1




    $begingroup$
    It's not my question. xD
    $endgroup$
    – YoungMath
    Jan 14 at 13:56



















0












$begingroup$

Assume all $A_alpha$ are closed. Let $(x_gamma)$ be a net of $Pi_alpha A_alpha$ converging towards $x in X:=Pi_alpha X_alpha$. Since the projections $pi_alpha: X to X_alpha$ are continous by the definition of the product topology, we have $$pi_alpha(x_gamma) xrightarrow{gamma} pi_alpha(x).$$
Now, the $A_alpha$ are closed by assumption so $pi_alpha(x) in A_alpha$ for every $alpha$ in the index set. From this follows $x in Pi_alpha A_alpha$ and hence $Pi_alpha A_alpha$ is closed.



I admit, this approach needs some knowledge about nets.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint:



    Let $pi_{alpha}$ be projection on the $X_{alpha}$ then
    $$cap_{alpha } ~pi_{alpha}^{-1} (A_{alpha}) = prod A_{alpha}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That's only helpful for finite index sets, isn't it?
      $endgroup$
      – YoungMath
      Dec 16 '18 at 10:12






    • 1




      $begingroup$
      @YoungMath Intersection of arbitrary numbers of closed sets is closed
      $endgroup$
      – Red shoes
      Dec 16 '18 at 17:33










    • $begingroup$
      Of course! You're right! Then, your approach is more concise.
      $endgroup$
      – YoungMath
      Dec 16 '18 at 17:36










    • $begingroup$
      @sepehr could you please accept my answer by clicking on the check mark?
      $endgroup$
      – Red shoes
      Dec 22 '18 at 0:03






    • 1




      $begingroup$
      It's not my question. xD
      $endgroup$
      – YoungMath
      Jan 14 at 13:56
















    1












    $begingroup$

    Hint:



    Let $pi_{alpha}$ be projection on the $X_{alpha}$ then
    $$cap_{alpha } ~pi_{alpha}^{-1} (A_{alpha}) = prod A_{alpha}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That's only helpful for finite index sets, isn't it?
      $endgroup$
      – YoungMath
      Dec 16 '18 at 10:12






    • 1




      $begingroup$
      @YoungMath Intersection of arbitrary numbers of closed sets is closed
      $endgroup$
      – Red shoes
      Dec 16 '18 at 17:33










    • $begingroup$
      Of course! You're right! Then, your approach is more concise.
      $endgroup$
      – YoungMath
      Dec 16 '18 at 17:36










    • $begingroup$
      @sepehr could you please accept my answer by clicking on the check mark?
      $endgroup$
      – Red shoes
      Dec 22 '18 at 0:03






    • 1




      $begingroup$
      It's not my question. xD
      $endgroup$
      – YoungMath
      Jan 14 at 13:56














    1












    1








    1





    $begingroup$

    Hint:



    Let $pi_{alpha}$ be projection on the $X_{alpha}$ then
    $$cap_{alpha } ~pi_{alpha}^{-1} (A_{alpha}) = prod A_{alpha}$$






    share|cite|improve this answer









    $endgroup$



    Hint:



    Let $pi_{alpha}$ be projection on the $X_{alpha}$ then
    $$cap_{alpha } ~pi_{alpha}^{-1} (A_{alpha}) = prod A_{alpha}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 15 '18 at 21:47









    Red shoesRed shoes

    4,761721




    4,761721












    • $begingroup$
      That's only helpful for finite index sets, isn't it?
      $endgroup$
      – YoungMath
      Dec 16 '18 at 10:12






    • 1




      $begingroup$
      @YoungMath Intersection of arbitrary numbers of closed sets is closed
      $endgroup$
      – Red shoes
      Dec 16 '18 at 17:33










    • $begingroup$
      Of course! You're right! Then, your approach is more concise.
      $endgroup$
      – YoungMath
      Dec 16 '18 at 17:36










    • $begingroup$
      @sepehr could you please accept my answer by clicking on the check mark?
      $endgroup$
      – Red shoes
      Dec 22 '18 at 0:03






    • 1




      $begingroup$
      It's not my question. xD
      $endgroup$
      – YoungMath
      Jan 14 at 13:56


















    • $begingroup$
      That's only helpful for finite index sets, isn't it?
      $endgroup$
      – YoungMath
      Dec 16 '18 at 10:12






    • 1




      $begingroup$
      @YoungMath Intersection of arbitrary numbers of closed sets is closed
      $endgroup$
      – Red shoes
      Dec 16 '18 at 17:33










    • $begingroup$
      Of course! You're right! Then, your approach is more concise.
      $endgroup$
      – YoungMath
      Dec 16 '18 at 17:36










    • $begingroup$
      @sepehr could you please accept my answer by clicking on the check mark?
      $endgroup$
      – Red shoes
      Dec 22 '18 at 0:03






    • 1




      $begingroup$
      It's not my question. xD
      $endgroup$
      – YoungMath
      Jan 14 at 13:56
















    $begingroup$
    That's only helpful for finite index sets, isn't it?
    $endgroup$
    – YoungMath
    Dec 16 '18 at 10:12




    $begingroup$
    That's only helpful for finite index sets, isn't it?
    $endgroup$
    – YoungMath
    Dec 16 '18 at 10:12




    1




    1




    $begingroup$
    @YoungMath Intersection of arbitrary numbers of closed sets is closed
    $endgroup$
    – Red shoes
    Dec 16 '18 at 17:33




    $begingroup$
    @YoungMath Intersection of arbitrary numbers of closed sets is closed
    $endgroup$
    – Red shoes
    Dec 16 '18 at 17:33












    $begingroup$
    Of course! You're right! Then, your approach is more concise.
    $endgroup$
    – YoungMath
    Dec 16 '18 at 17:36




    $begingroup$
    Of course! You're right! Then, your approach is more concise.
    $endgroup$
    – YoungMath
    Dec 16 '18 at 17:36












    $begingroup$
    @sepehr could you please accept my answer by clicking on the check mark?
    $endgroup$
    – Red shoes
    Dec 22 '18 at 0:03




    $begingroup$
    @sepehr could you please accept my answer by clicking on the check mark?
    $endgroup$
    – Red shoes
    Dec 22 '18 at 0:03




    1




    1




    $begingroup$
    It's not my question. xD
    $endgroup$
    – YoungMath
    Jan 14 at 13:56




    $begingroup$
    It's not my question. xD
    $endgroup$
    – YoungMath
    Jan 14 at 13:56











    0












    $begingroup$

    Assume all $A_alpha$ are closed. Let $(x_gamma)$ be a net of $Pi_alpha A_alpha$ converging towards $x in X:=Pi_alpha X_alpha$. Since the projections $pi_alpha: X to X_alpha$ are continous by the definition of the product topology, we have $$pi_alpha(x_gamma) xrightarrow{gamma} pi_alpha(x).$$
    Now, the $A_alpha$ are closed by assumption so $pi_alpha(x) in A_alpha$ for every $alpha$ in the index set. From this follows $x in Pi_alpha A_alpha$ and hence $Pi_alpha A_alpha$ is closed.



    I admit, this approach needs some knowledge about nets.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Assume all $A_alpha$ are closed. Let $(x_gamma)$ be a net of $Pi_alpha A_alpha$ converging towards $x in X:=Pi_alpha X_alpha$. Since the projections $pi_alpha: X to X_alpha$ are continous by the definition of the product topology, we have $$pi_alpha(x_gamma) xrightarrow{gamma} pi_alpha(x).$$
      Now, the $A_alpha$ are closed by assumption so $pi_alpha(x) in A_alpha$ for every $alpha$ in the index set. From this follows $x in Pi_alpha A_alpha$ and hence $Pi_alpha A_alpha$ is closed.



      I admit, this approach needs some knowledge about nets.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Assume all $A_alpha$ are closed. Let $(x_gamma)$ be a net of $Pi_alpha A_alpha$ converging towards $x in X:=Pi_alpha X_alpha$. Since the projections $pi_alpha: X to X_alpha$ are continous by the definition of the product topology, we have $$pi_alpha(x_gamma) xrightarrow{gamma} pi_alpha(x).$$
        Now, the $A_alpha$ are closed by assumption so $pi_alpha(x) in A_alpha$ for every $alpha$ in the index set. From this follows $x in Pi_alpha A_alpha$ and hence $Pi_alpha A_alpha$ is closed.



        I admit, this approach needs some knowledge about nets.






        share|cite|improve this answer











        $endgroup$



        Assume all $A_alpha$ are closed. Let $(x_gamma)$ be a net of $Pi_alpha A_alpha$ converging towards $x in X:=Pi_alpha X_alpha$. Since the projections $pi_alpha: X to X_alpha$ are continous by the definition of the product topology, we have $$pi_alpha(x_gamma) xrightarrow{gamma} pi_alpha(x).$$
        Now, the $A_alpha$ are closed by assumption so $pi_alpha(x) in A_alpha$ for every $alpha$ in the index set. From this follows $x in Pi_alpha A_alpha$ and hence $Pi_alpha A_alpha$ is closed.



        I admit, this approach needs some knowledge about nets.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 12:06

























        answered Dec 16 '18 at 11:53









        YoungMathYoungMath

        197111




        197111






























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