Let $f:Xto [0,1]$ be the function defined by $f(x_i)_{ige 1}=sum _{ige 1}x_i2^{-i}$. Choose the correct...












-1












$begingroup$


Let $X={(x_i)_{ige 1}:x_iin {0,1}$ for all $ige 1}$ with the metric $d((x_i),(y_i))=sum_{ige 1}|x_i-y_i|2^{-i}.$ Let $f:Xto [0,1]$ be the function defined by $f(x_i)_{ige 1}=sum _{ige 1}x_i2^{-i}$.



Choose the correct statements from below :



$1)$$f$ is continuous



$2)$$f$ is onto



$3)$$f$ is one-one



$4)$$f$ is open



My attempt : option $1)$ is true by triangle inequality.



$2)$ is true as it is a binary representation



im confused about option 3, 4



any hints/solution



thanks u






general-topology







share|cite|improve this question









$endgroup$



closed as off-topic by RRL, amWhy, Paul Frost, clathratus, Leucippus Jan 13 at 7:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, Paul Frost, clathratus, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.





















    -1












    $begingroup$


    Let $X={(x_i)_{ige 1}:x_iin {0,1}$ for all $ige 1}$ with the metric $d((x_i),(y_i))=sum_{ige 1}|x_i-y_i|2^{-i}.$ Let $f:Xto [0,1]$ be the function defined by $f(x_i)_{ige 1}=sum _{ige 1}x_i2^{-i}$.



    Choose the correct statements from below :



    $1)$$f$ is continuous



    $2)$$f$ is onto



    $3)$$f$ is one-one



    $4)$$f$ is open



    My attempt : option $1)$ is true by triangle inequality.



    $2)$ is true as it is a binary representation



    im confused about option 3, 4



    any hints/solution



    thanks u






    general-topology







    share|cite|improve this question









    $endgroup$



    closed as off-topic by RRL, amWhy, Paul Frost, clathratus, Leucippus Jan 13 at 7:27


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, Paul Frost, clathratus, Leucippus

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1





      $begingroup$


      Let $X={(x_i)_{ige 1}:x_iin {0,1}$ for all $ige 1}$ with the metric $d((x_i),(y_i))=sum_{ige 1}|x_i-y_i|2^{-i}.$ Let $f:Xto [0,1]$ be the function defined by $f(x_i)_{ige 1}=sum _{ige 1}x_i2^{-i}$.



      Choose the correct statements from below :



      $1)$$f$ is continuous



      $2)$$f$ is onto



      $3)$$f$ is one-one



      $4)$$f$ is open



      My attempt : option $1)$ is true by triangle inequality.



      $2)$ is true as it is a binary representation



      im confused about option 3, 4



      any hints/solution



      thanks u






      general-topology







      share|cite|improve this question









      $endgroup$




      Let $X={(x_i)_{ige 1}:x_iin {0,1}$ for all $ige 1}$ with the metric $d((x_i),(y_i))=sum_{ige 1}|x_i-y_i|2^{-i}.$ Let $f:Xto [0,1]$ be the function defined by $f(x_i)_{ige 1}=sum _{ige 1}x_i2^{-i}$.



      Choose the correct statements from below :



      $1)$$f$ is continuous



      $2)$$f$ is onto



      $3)$$f$ is one-one



      $4)$$f$ is open



      My attempt : option $1)$ is true by triangle inequality.



      $2)$ is true as it is a binary representation



      im confused about option 3, 4



      any hints/solution



      thanks u






      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 Dec 20 '18 at 8:38









      jasminejasmine

      1,766417




      1,766417




      closed as off-topic by RRL, amWhy, Paul Frost, clathratus, Leucippus Jan 13 at 7:27


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, Paul Frost, clathratus, Leucippus

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by RRL, amWhy, Paul Frost, clathratus, Leucippus Jan 13 at 7:27


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, Paul Frost, clathratus, Leucippus

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          $f(1,0,0,cdots)=f(0,1,1,cdots)$ so $f$ is not one-to one.
          ${(x_i):x_1=0}equiv {(x_i):x_1 neq 1}$ is open and its image is $[0,frac 1 2]$ which is not open. Hence $f$ is not an open map.






          share|cite|improve this answer









          $endgroup$




















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            $f(1,0,0,cdots)=f(0,1,1,cdots)$ so $f$ is not one-to one.
            ${(x_i):x_1=0}equiv {(x_i):x_1 neq 1}$ is open and its image is $[0,frac 1 2]$ which is not open. Hence $f$ is not an open map.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              $f(1,0,0,cdots)=f(0,1,1,cdots)$ so $f$ is not one-to one.
              ${(x_i):x_1=0}equiv {(x_i):x_1 neq 1}$ is open and its image is $[0,frac 1 2]$ which is not open. Hence $f$ is not an open map.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                $f(1,0,0,cdots)=f(0,1,1,cdots)$ so $f$ is not one-to one.
                ${(x_i):x_1=0}equiv {(x_i):x_1 neq 1}$ is open and its image is $[0,frac 1 2]$ which is not open. Hence $f$ is not an open map.






                share|cite|improve this answer









                $endgroup$



                $f(1,0,0,cdots)=f(0,1,1,cdots)$ so $f$ is not one-to one.
                ${(x_i):x_1=0}equiv {(x_i):x_1 neq 1}$ is open and its image is $[0,frac 1 2]$ which is not open. Hence $f$ is not an open map.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 20 '18 at 8:57









                Kavi Rama MurthyKavi Rama Murthy

                60.5k42161




                60.5k42161















                    -

                    Popular posts from this blog

                    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

                    Province de Neuquén

                    Image
                    Province de Neuquén Provincia del Neuquén .mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Héraldique Drapeau Localisation de la province de Neuquén Administration Pays   Argentine Capitale Neuquén Gouverneur Omar Gutiérrez ISO 3166-2 AR-Q Démographie Gentilé Neuquino/a Population 550 344  hab. (2010) Densité 5,8  hab./km 2 Géographie Superficie 94 078  km 2 Liens Site web http://www.neuquen.gov.ar modifier   La province de Neuquén est une province d'Argentine, qui se trouve à l'extrémité nord-ouest de la Patagonie argentine. Avec une superficie de 94 378  km 2 , elle est la plus petite province des régions australes du pays. Sa géographie s'abîme sur l'imposante Cordillère des Andes. Sommaire 1 Histoire 1.1 Avant la découverte par les Espagnols 1.2 ...
                    Read more

                    Soliste

                    Image
                    Un soliste est un interprète qui assure seul l'exécution d'une partie musicale dans un ensemble ou de manière isolée. Le soliste s'oppose donc au musicien qui, au sein d'un pupitre, exécute collectivement une partie musicale (un choriste dans une chorale, un violoniste dans un orchestre symphonique, etc.). Par exemple, dans un concerto pour piano, le soliste est le pianiste, alors que les différents musiciens de l'orchestre réalisant leurs parties musicales respectives sont des musiciens de pupitre . Les parties musicales confiées à un soliste sont d'ordinaire plus virtuoses que celles des autres musiciens : vélocité, puissance, étendue du registre, etc. En musique classique | Depuis la période baroque, le soliste s'est nettement différencié du musicien de pupitre, par une spécialisation et une technicité tout à fait particulières, encouragées ou exigées par les compositeurs eux-mêmes. En ce qui concerne la musique vocale, le soliste apparaî...
                    Read more