The map $D$ is choose the correct satements












0












$begingroup$


Consider the closed interval $[0, 1]$ in the real line $mathbb{R}$ and the product space $([0, 1]^{mathbb{N}}, τ ),$



where $τ$ is a topology on $[0, 1]^mathbb{N} $. Let $ D : [0, 1] rightarrow [0, 1]^mathbb{N} $ be the map defined by
$D(x) := (x, x, · · · , x, · · ·)$ for $x in [0, 1].$
. The map $D$ is



choose the correct satements



$(a)$ not continuous if $τ$ is the box topology and also not continuous if $τ$ is the product
topology.



$(b)$ continuous if $τ$ is the product topology and also continuous if $τ$ is the box
topology.



$(c)$ continuous if $τ$ is the box topology and not continuous if $τ$ is the product
topology.



$(d)$ continuous if $τ$ is the product topology and not continuous if $τ$ is the box
topology



My attempt : i thinks option $c)$ is true because box topology is finer then product



Is its true ?



Any hints/solution






general-topology







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Consider the closed interval $[0, 1]$ in the real line $mathbb{R}$ and the product space $([0, 1]^{mathbb{N}}, τ ),$



    where $τ$ is a topology on $[0, 1]^mathbb{N} $. Let $ D : [0, 1] rightarrow [0, 1]^mathbb{N} $ be the map defined by
    $D(x) := (x, x, · · · , x, · · ·)$ for $x in [0, 1].$
    . The map $D$ is



    choose the correct satements



    $(a)$ not continuous if $τ$ is the box topology and also not continuous if $τ$ is the product
    topology.



    $(b)$ continuous if $τ$ is the product topology and also continuous if $τ$ is the box
    topology.



    $(c)$ continuous if $τ$ is the box topology and not continuous if $τ$ is the product
    topology.



    $(d)$ continuous if $τ$ is the product topology and not continuous if $τ$ is the box
    topology



    My attempt : i thinks option $c)$ is true because box topology is finer then product



    Is its true ?



    Any hints/solution






    general-topology







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Consider the closed interval $[0, 1]$ in the real line $mathbb{R}$ and the product space $([0, 1]^{mathbb{N}}, τ ),$



      where $τ$ is a topology on $[0, 1]^mathbb{N} $. Let $ D : [0, 1] rightarrow [0, 1]^mathbb{N} $ be the map defined by
      $D(x) := (x, x, · · · , x, · · ·)$ for $x in [0, 1].$
      . The map $D$ is



      choose the correct satements



      $(a)$ not continuous if $τ$ is the box topology and also not continuous if $τ$ is the product
      topology.



      $(b)$ continuous if $τ$ is the product topology and also continuous if $τ$ is the box
      topology.



      $(c)$ continuous if $τ$ is the box topology and not continuous if $τ$ is the product
      topology.



      $(d)$ continuous if $τ$ is the product topology and not continuous if $τ$ is the box
      topology



      My attempt : i thinks option $c)$ is true because box topology is finer then product



      Is its true ?



      Any hints/solution






      general-topology







      share|cite|improve this question









      $endgroup$




      Consider the closed interval $[0, 1]$ in the real line $mathbb{R}$ and the product space $([0, 1]^{mathbb{N}}, τ ),$



      where $τ$ is a topology on $[0, 1]^mathbb{N} $. Let $ D : [0, 1] rightarrow [0, 1]^mathbb{N} $ be the map defined by
      $D(x) := (x, x, · · · , x, · · ·)$ for $x in [0, 1].$
      . The map $D$ is



      choose the correct satements



      $(a)$ not continuous if $τ$ is the box topology and also not continuous if $τ$ is the product
      topology.



      $(b)$ continuous if $τ$ is the product topology and also continuous if $τ$ is the box
      topology.



      $(c)$ continuous if $τ$ is the box topology and not continuous if $τ$ is the product
      topology.



      $(d)$ continuous if $τ$ is the product topology and not continuous if $τ$ is the box
      topology



      My attempt : i thinks option $c)$ is true because box topology is finer then product



      Is its true ?



      Any hints/solution






      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 6 at 11:33









      jasminejasmine

      1,968420




      1,968420






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks u kavi sir
            $endgroup$
            – jasmine
            Jan 6 at 12:20



















          1












          $begingroup$

          Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            for box i thinks product ??
            $endgroup$
            – jasmine
            Jan 6 at 11:40










          • $begingroup$
            I do not understand your comment.
            $endgroup$
            – Mindlack
            Jan 6 at 11:41










          • $begingroup$
            @Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
            $endgroup$
            – jasmine
            Jan 6 at 11:50












          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks u kavi sir
            $endgroup$
            – jasmine
            Jan 6 at 12:20
















          1












          $begingroup$

          The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks u kavi sir
            $endgroup$
            – jasmine
            Jan 6 at 12:20














          1












          1








          1





          $begingroup$

          The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.






          share|cite|improve this answer









          $endgroup$



          The correct answer is d). Since $D^{-1}prod (-frac 1 n,frac 1 n)={0}$ is not open it follows that $D$ is not continuous for the box topology. If $x_j to x$ then $D(x_j)to D(x)$ in the product topolgy because convergence in product topolgy is coordinatewise convergence. Hence $D$ is continuous for product topology.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 12:15









          Kavi Rama MurthyKavi Rama Murthy

          73.9k53170




          73.9k53170












          • $begingroup$
            thanks u kavi sir
            $endgroup$
            – jasmine
            Jan 6 at 12:20


















          • $begingroup$
            thanks u kavi sir
            $endgroup$
            – jasmine
            Jan 6 at 12:20
















          $begingroup$
          thanks u kavi sir
          $endgroup$
          – jasmine
          Jan 6 at 12:20




          $begingroup$
          thanks u kavi sir
          $endgroup$
          – jasmine
          Jan 6 at 12:20











          1












          $begingroup$

          Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            for box i thinks product ??
            $endgroup$
            – jasmine
            Jan 6 at 11:40










          • $begingroup$
            I do not understand your comment.
            $endgroup$
            – Mindlack
            Jan 6 at 11:41










          • $begingroup$
            @Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
            $endgroup$
            – jasmine
            Jan 6 at 11:50
















          1












          $begingroup$

          Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            for box i thinks product ??
            $endgroup$
            – jasmine
            Jan 6 at 11:40










          • $begingroup$
            I do not understand your comment.
            $endgroup$
            – Mindlack
            Jan 6 at 11:41










          • $begingroup$
            @Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
            $endgroup$
            – jasmine
            Jan 6 at 11:50














          1












          1








          1





          $begingroup$

          Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?






          share|cite|improve this answer









          $endgroup$



          Hint: what is the inverse image by $D$ of open subsets from $tau$ when $tau$ is the box topology, or the product topology?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 11:37









          MindlackMindlack

          4,910211




          4,910211












          • $begingroup$
            for box i thinks product ??
            $endgroup$
            – jasmine
            Jan 6 at 11:40










          • $begingroup$
            I do not understand your comment.
            $endgroup$
            – Mindlack
            Jan 6 at 11:41










          • $begingroup$
            @Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
            $endgroup$
            – jasmine
            Jan 6 at 11:50


















          • $begingroup$
            for box i thinks product ??
            $endgroup$
            – jasmine
            Jan 6 at 11:40










          • $begingroup$
            I do not understand your comment.
            $endgroup$
            – Mindlack
            Jan 6 at 11:41










          • $begingroup$
            @Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
            $endgroup$
            – jasmine
            Jan 6 at 11:50
















          $begingroup$
          for box i thinks product ??
          $endgroup$
          – jasmine
          Jan 6 at 11:40




          $begingroup$
          for box i thinks product ??
          $endgroup$
          – jasmine
          Jan 6 at 11:40












          $begingroup$
          I do not understand your comment.
          $endgroup$
          – Mindlack
          Jan 6 at 11:41




          $begingroup$
          I do not understand your comment.
          $endgroup$
          – Mindlack
          Jan 6 at 11:41












          $begingroup$
          @Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
          $endgroup$
          – jasmine
          Jan 6 at 11:50




          $begingroup$
          @Mindblack,,,,,,can u elaborate ur answer ? just like im not getting ur answer
          $endgroup$
          – jasmine
          Jan 6 at 11:50


















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