Open linear subspace of a Hilbert space.












4












$begingroup$


Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.



Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed:
Take the Hilbert space H = C[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.






functional-analysis







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Linear span of a ball = ?
    $endgroup$
    – metamorphy
    Jan 2 at 5:08






  • 1




    $begingroup$
    @metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
    $endgroup$
    – epsilon_delta
    Jan 2 at 5:18








  • 1




    $begingroup$
    I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
    $endgroup$
    – metamorphy
    Jan 2 at 5:21






  • 2




    $begingroup$
    Sorry, please disregard my close vote; I misread the question.
    $endgroup$
    – angryavian
    Jan 2 at 5:27






  • 1




    $begingroup$
    Nice question. Glad you didn't give up on it. :) +1.
    $endgroup$
    – DisintegratingByParts
    Jan 2 at 6:02


















4












$begingroup$


Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.



Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed:
Take the Hilbert space H = C[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.






functional-analysis







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Linear span of a ball = ?
    $endgroup$
    – metamorphy
    Jan 2 at 5:08






  • 1




    $begingroup$
    @metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
    $endgroup$
    – epsilon_delta
    Jan 2 at 5:18








  • 1




    $begingroup$
    I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
    $endgroup$
    – metamorphy
    Jan 2 at 5:21






  • 2




    $begingroup$
    Sorry, please disregard my close vote; I misread the question.
    $endgroup$
    – angryavian
    Jan 2 at 5:27






  • 1




    $begingroup$
    Nice question. Glad you didn't give up on it. :) +1.
    $endgroup$
    – DisintegratingByParts
    Jan 2 at 6:02
















4












4








4


1



$begingroup$


Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.



Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed:
Take the Hilbert space H = C[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.






functional-analysis







share|cite|improve this question











$endgroup$




Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.



Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed:
Take the Hilbert space H = C[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.






functional-analysis




functional-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 20:40







epsilon_delta

















asked Jan 2 at 5:02









epsilon_deltaepsilon_delta

286




286








  • 3




    $begingroup$
    Linear span of a ball = ?
    $endgroup$
    – metamorphy
    Jan 2 at 5:08






  • 1




    $begingroup$
    @metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
    $endgroup$
    – epsilon_delta
    Jan 2 at 5:18








  • 1




    $begingroup$
    I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
    $endgroup$
    – metamorphy
    Jan 2 at 5:21






  • 2




    $begingroup$
    Sorry, please disregard my close vote; I misread the question.
    $endgroup$
    – angryavian
    Jan 2 at 5:27






  • 1




    $begingroup$
    Nice question. Glad you didn't give up on it. :) +1.
    $endgroup$
    – DisintegratingByParts
    Jan 2 at 6:02
















  • 3




    $begingroup$
    Linear span of a ball = ?
    $endgroup$
    – metamorphy
    Jan 2 at 5:08






  • 1




    $begingroup$
    @metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
    $endgroup$
    – epsilon_delta
    Jan 2 at 5:18








  • 1




    $begingroup$
    I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
    $endgroup$
    – metamorphy
    Jan 2 at 5:21






  • 2




    $begingroup$
    Sorry, please disregard my close vote; I misread the question.
    $endgroup$
    – angryavian
    Jan 2 at 5:27






  • 1




    $begingroup$
    Nice question. Glad you didn't give up on it. :) +1.
    $endgroup$
    – DisintegratingByParts
    Jan 2 at 6:02










3




3




$begingroup$
Linear span of a ball = ?
$endgroup$
– metamorphy
Jan 2 at 5:08




$begingroup$
Linear span of a ball = ?
$endgroup$
– metamorphy
Jan 2 at 5:08




1




1




$begingroup$
@metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
$endgroup$
– epsilon_delta
Jan 2 at 5:18






$begingroup$
@metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
$endgroup$
– epsilon_delta
Jan 2 at 5:18






1




1




$begingroup$
I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
$endgroup$
– metamorphy
Jan 2 at 5:21




$begingroup$
I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
$endgroup$
– metamorphy
Jan 2 at 5:21




2




2




$begingroup$
Sorry, please disregard my close vote; I misread the question.
$endgroup$
– angryavian
Jan 2 at 5:27




$begingroup$
Sorry, please disregard my close vote; I misread the question.
$endgroup$
– angryavian
Jan 2 at 5:27




1




1




$begingroup$
Nice question. Glad you didn't give up on it. :) +1.
$endgroup$
– DisintegratingByParts
Jan 2 at 6:02






$begingroup$
Nice question. Glad you didn't give up on it. :) +1.
$endgroup$
– DisintegratingByParts
Jan 2 at 6:02












2 Answers
2






active

oldest

votes


















2












$begingroup$

No.



Let $N$ be a normed space and



$M subsetneq N tag 1$



a proper subspace. Then $M$ contains no nonempty open set. For if



$emptyset ne U subset M tag 2$



were open, with



$M ni m in U, tag 3$



we could find $rho > 0$ such that the open ball



$B(m, rho) subset U; tag 4$



then picking any



$0 ne v in N setminus M tag 5$



the vector



$m + alpha (v - m) in B(m, rho) tag 6$



if $0 ne alpha in Bbb R$ is sufficiently small, since



$Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



for



$vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



but then



$m + alpha(v - m) in M, tag 9$



whence



$alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



whence



$v - m in M, tag{11}$



whence



$v = v - m + m in M, tag{12}$



in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
    $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
    But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.






    share|cite|improve this answer











    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      No.



      Let $N$ be a normed space and



      $M subsetneq N tag 1$



      a proper subspace. Then $M$ contains no nonempty open set. For if



      $emptyset ne U subset M tag 2$



      were open, with



      $M ni m in U, tag 3$



      we could find $rho > 0$ such that the open ball



      $B(m, rho) subset U; tag 4$



      then picking any



      $0 ne v in N setminus M tag 5$



      the vector



      $m + alpha (v - m) in B(m, rho) tag 6$



      if $0 ne alpha in Bbb R$ is sufficiently small, since



      $Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



      for



      $vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



      but then



      $m + alpha(v - m) in M, tag 9$



      whence



      $alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



      whence



      $v - m in M, tag{11}$



      whence



      $v = v - m + m in M, tag{12}$



      in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        No.



        Let $N$ be a normed space and



        $M subsetneq N tag 1$



        a proper subspace. Then $M$ contains no nonempty open set. For if



        $emptyset ne U subset M tag 2$



        were open, with



        $M ni m in U, tag 3$



        we could find $rho > 0$ such that the open ball



        $B(m, rho) subset U; tag 4$



        then picking any



        $0 ne v in N setminus M tag 5$



        the vector



        $m + alpha (v - m) in B(m, rho) tag 6$



        if $0 ne alpha in Bbb R$ is sufficiently small, since



        $Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



        for



        $vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



        but then



        $m + alpha(v - m) in M, tag 9$



        whence



        $alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



        whence



        $v - m in M, tag{11}$



        whence



        $v = v - m + m in M, tag{12}$



        in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          No.



          Let $N$ be a normed space and



          $M subsetneq N tag 1$



          a proper subspace. Then $M$ contains no nonempty open set. For if



          $emptyset ne U subset M tag 2$



          were open, with



          $M ni m in U, tag 3$



          we could find $rho > 0$ such that the open ball



          $B(m, rho) subset U; tag 4$



          then picking any



          $0 ne v in N setminus M tag 5$



          the vector



          $m + alpha (v - m) in B(m, rho) tag 6$



          if $0 ne alpha in Bbb R$ is sufficiently small, since



          $Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



          for



          $vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



          but then



          $m + alpha(v - m) in M, tag 9$



          whence



          $alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



          whence



          $v - m in M, tag{11}$



          whence



          $v = v - m + m in M, tag{12}$



          in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.






          share|cite|improve this answer









          $endgroup$



          No.



          Let $N$ be a normed space and



          $M subsetneq N tag 1$



          a proper subspace. Then $M$ contains no nonempty open set. For if



          $emptyset ne U subset M tag 2$



          were open, with



          $M ni m in U, tag 3$



          we could find $rho > 0$ such that the open ball



          $B(m, rho) subset U; tag 4$



          then picking any



          $0 ne v in N setminus M tag 5$



          the vector



          $m + alpha (v - m) in B(m, rho) tag 6$



          if $0 ne alpha in Bbb R$ is sufficiently small, since



          $Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



          for



          $vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



          but then



          $m + alpha(v - m) in M, tag 9$



          whence



          $alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



          whence



          $v - m in M, tag{11}$



          whence



          $v = v - m + m in M, tag{12}$



          in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 6:15









          Robert LewisRobert Lewis

          49.1k23168




          49.1k23168























              3












              $begingroup$

              If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
              $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
              But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
                $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
                But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
                  $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
                  But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.






                  share|cite|improve this answer











                  $endgroup$



                  If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
                  $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
                  But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 2 at 7:02









                  Henno Brandsma

                  117k350128




                  117k350128










                  answered Jan 2 at 6:16









                  pitariverpitariver

                  469213




                  469213






























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