Translation of a dense set in a normed vector space












1












$begingroup$


I would like to prove that if $X$ is, let's say, a normed vector space, and $A$ is a dense subset of $X$, and $x_0 in X$, that ${x_0} + A$ is also dense in $X$.



I know that translation by a point is a homeomorphism, but not translation by a set. I have also tried using sequences, and using closure arguments. Is there any easy way to do this problem?






general-topology functional-analysis







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    1












    $begingroup$


    I would like to prove that if $X$ is, let's say, a normed vector space, and $A$ is a dense subset of $X$, and $x_0 in X$, that ${x_0} + A$ is also dense in $X$.



    I know that translation by a point is a homeomorphism, but not translation by a set. I have also tried using sequences, and using closure arguments. Is there any easy way to do this problem?






    general-topology functional-analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I would like to prove that if $X$ is, let's say, a normed vector space, and $A$ is a dense subset of $X$, and $x_0 in X$, that ${x_0} + A$ is also dense in $X$.



      I know that translation by a point is a homeomorphism, but not translation by a set. I have also tried using sequences, and using closure arguments. Is there any easy way to do this problem?






      general-topology functional-analysis







      share|cite|improve this question









      $endgroup$




      I would like to prove that if $X$ is, let's say, a normed vector space, and $A$ is a dense subset of $X$, and $x_0 in X$, that ${x_0} + A$ is also dense in $X$.



      I know that translation by a point is a homeomorphism, but not translation by a set. I have also tried using sequences, and using closure arguments. Is there any easy way to do this problem?






      general-topology functional-analysis




      general-topology functional-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Sep 23 '15 at 2:36









      Johnny AppleJohnny Apple

      1,3091817




      1,3091817






















          2 Answers
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          3












          $begingroup$

          A set is dense if it meets every non-empty open set. $V$ meets $x_0+A$ iff $V-x_0$ meets $A$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Of course. Thanks!
            $endgroup$
            – Johnny Apple
            Sep 23 '15 at 2:52










          • $begingroup$
            @Johnny: You're welcome!
            $endgroup$
            – Brian M. Scott
            Sep 23 '15 at 3:00



















          1












          $begingroup$

          An elementary proof, using sequences, is the following:



          Let $xin X$.



          $A$ is dense and therefore there exists a sequence ${x_n}$ in $A$ such that $ x_nto x-x_0$. So the sequence $y_n=x_0 + x_n$ is in $left { x_0 right }+A$ and $y_nto x_0+x-x_0=x$. So $xinoverline{ left { x_0 right }+A}$ and therefore $left { x_0 right }+A$ is dense.






          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









            3












            $begingroup$

            A set is dense if it meets every non-empty open set. $V$ meets $x_0+A$ iff $V-x_0$ meets $A$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Of course. Thanks!
              $endgroup$
              – Johnny Apple
              Sep 23 '15 at 2:52










            • $begingroup$
              @Johnny: You're welcome!
              $endgroup$
              – Brian M. Scott
              Sep 23 '15 at 3:00
















            3












            $begingroup$

            A set is dense if it meets every non-empty open set. $V$ meets $x_0+A$ iff $V-x_0$ meets $A$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Of course. Thanks!
              $endgroup$
              – Johnny Apple
              Sep 23 '15 at 2:52










            • $begingroup$
              @Johnny: You're welcome!
              $endgroup$
              – Brian M. Scott
              Sep 23 '15 at 3:00














            3












            3








            3





            $begingroup$

            A set is dense if it meets every non-empty open set. $V$ meets $x_0+A$ iff $V-x_0$ meets $A$.






            share|cite|improve this answer









            $endgroup$



            A set is dense if it meets every non-empty open set. $V$ meets $x_0+A$ iff $V-x_0$ meets $A$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 23 '15 at 2:42









            Brian M. ScottBrian M. Scott

            456k38507908




            456k38507908












            • $begingroup$
              Of course. Thanks!
              $endgroup$
              – Johnny Apple
              Sep 23 '15 at 2:52










            • $begingroup$
              @Johnny: You're welcome!
              $endgroup$
              – Brian M. Scott
              Sep 23 '15 at 3:00


















            • $begingroup$
              Of course. Thanks!
              $endgroup$
              – Johnny Apple
              Sep 23 '15 at 2:52










            • $begingroup$
              @Johnny: You're welcome!
              $endgroup$
              – Brian M. Scott
              Sep 23 '15 at 3:00
















            $begingroup$
            Of course. Thanks!
            $endgroup$
            – Johnny Apple
            Sep 23 '15 at 2:52




            $begingroup$
            Of course. Thanks!
            $endgroup$
            – Johnny Apple
            Sep 23 '15 at 2:52












            $begingroup$
            @Johnny: You're welcome!
            $endgroup$
            – Brian M. Scott
            Sep 23 '15 at 3:00




            $begingroup$
            @Johnny: You're welcome!
            $endgroup$
            – Brian M. Scott
            Sep 23 '15 at 3:00











            1












            $begingroup$

            An elementary proof, using sequences, is the following:



            Let $xin X$.



            $A$ is dense and therefore there exists a sequence ${x_n}$ in $A$ such that $ x_nto x-x_0$. So the sequence $y_n=x_0 + x_n$ is in $left { x_0 right }+A$ and $y_nto x_0+x-x_0=x$. So $xinoverline{ left { x_0 right }+A}$ and therefore $left { x_0 right }+A$ is dense.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              An elementary proof, using sequences, is the following:



              Let $xin X$.



              $A$ is dense and therefore there exists a sequence ${x_n}$ in $A$ such that $ x_nto x-x_0$. So the sequence $y_n=x_0 + x_n$ is in $left { x_0 right }+A$ and $y_nto x_0+x-x_0=x$. So $xinoverline{ left { x_0 right }+A}$ and therefore $left { x_0 right }+A$ is dense.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                An elementary proof, using sequences, is the following:



                Let $xin X$.



                $A$ is dense and therefore there exists a sequence ${x_n}$ in $A$ such that $ x_nto x-x_0$. So the sequence $y_n=x_0 + x_n$ is in $left { x_0 right }+A$ and $y_nto x_0+x-x_0=x$. So $xinoverline{ left { x_0 right }+A}$ and therefore $left { x_0 right }+A$ is dense.






                share|cite|improve this answer











                $endgroup$



                An elementary proof, using sequences, is the following:



                Let $xin X$.



                $A$ is dense and therefore there exists a sequence ${x_n}$ in $A$ such that $ x_nto x-x_0$. So the sequence $y_n=x_0 + x_n$ is in $left { x_0 right }+A$ and $y_nto x_0+x-x_0=x$. So $xinoverline{ left { x_0 right }+A}$ and therefore $left { x_0 right }+A$ is dense.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 5 '18 at 23:35

























                answered Apr 2 '17 at 23:20









                richarddedekindricharddedekind

                711316




                711316






























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