Is the linear operator $(Tx)_n = fracx_nn$ a homeomorphism?

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Let $X$ be an Hilbert subspace of $ell^2$ over $mathbbC$



$forall v in ell^2$ let indicate with $v_n$ the nth element of the sequence $v$



Let $Y = y in ell^2 backslash exists x in X: y_n = fracx_nn $



Let $T: X to Y$ be the linear operator such that $(Tx)_n = fracx_nn$



My question is: Is $T$ a homeomorphism?




functional-analysis hilbert-spaces




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    up vote
    2
    down vote

    favorite












    Let $X$ be an Hilbert subspace of $ell^2$ over $mathbbC$



    $forall v in ell^2$ let indicate with $v_n$ the nth element of the sequence $v$



    Let $Y = y in ell^2 backslash exists x in X: y_n = fracx_nn $



    Let $T: X to Y$ be the linear operator such that $(Tx)_n = fracx_nn$



    My question is: Is $T$ a homeomorphism?




    functional-analysis hilbert-spaces




    share|cite|improve this question





















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $X$ be an Hilbert subspace of $ell^2$ over $mathbbC$



      $forall v in ell^2$ let indicate with $v_n$ the nth element of the sequence $v$



      Let $Y = y in ell^2 backslash exists x in X: y_n = fracx_nn $



      Let $T: X to Y$ be the linear operator such that $(Tx)_n = fracx_nn$



      My question is: Is $T$ a homeomorphism?




      functional-analysis hilbert-spaces




      share|cite|improve this question











      Let $X$ be an Hilbert subspace of $ell^2$ over $mathbbC$



      $forall v in ell^2$ let indicate with $v_n$ the nth element of the sequence $v$



      Let $Y = y in ell^2 backslash exists x in X: y_n = fracx_nn $



      Let $T: X to Y$ be the linear operator such that $(Tx)_n = fracx_nn$



      My question is: Is $T$ a homeomorphism?





      functional-analysis hilbert-spaces





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 26 at 16:30









      Matey Math

      827413




      827413




















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          It need not be for an infinite-dimensional subspace. Suppose the zero sequence and at least a countable subset of the countable family of sequences $mathcalF = s_k_k=1^infty, textwhere (s_k)_n = delta_kn$ is in X. ($s_k$ is the sequence that has zeros as entries except for the kth entry, where it is 1.)



          Note that $Ts_k rightarrow 0 in ell^2$, but $s_k nrightarrow 0$. This shows that $T^-1$ is not continuous. Hence T is not a homeomorphism.






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          • thanks so much for your answer
            – Matey Math
            Jul 26 at 23:47










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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          It need not be for an infinite-dimensional subspace. Suppose the zero sequence and at least a countable subset of the countable family of sequences $mathcalF = s_k_k=1^infty, textwhere (s_k)_n = delta_kn$ is in X. ($s_k$ is the sequence that has zeros as entries except for the kth entry, where it is 1.)



          Note that $Ts_k rightarrow 0 in ell^2$, but $s_k nrightarrow 0$. This shows that $T^-1$ is not continuous. Hence T is not a homeomorphism.






          share|cite|improve this answer





















          • thanks so much for your answer
            – Matey Math
            Jul 26 at 23:47














          up vote
          4
          down vote



          accepted










          It need not be for an infinite-dimensional subspace. Suppose the zero sequence and at least a countable subset of the countable family of sequences $mathcalF = s_k_k=1^infty, textwhere (s_k)_n = delta_kn$ is in X. ($s_k$ is the sequence that has zeros as entries except for the kth entry, where it is 1.)



          Note that $Ts_k rightarrow 0 in ell^2$, but $s_k nrightarrow 0$. This shows that $T^-1$ is not continuous. Hence T is not a homeomorphism.






          share|cite|improve this answer





















          • thanks so much for your answer
            – Matey Math
            Jul 26 at 23:47












          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          It need not be for an infinite-dimensional subspace. Suppose the zero sequence and at least a countable subset of the countable family of sequences $mathcalF = s_k_k=1^infty, textwhere (s_k)_n = delta_kn$ is in X. ($s_k$ is the sequence that has zeros as entries except for the kth entry, where it is 1.)



          Note that $Ts_k rightarrow 0 in ell^2$, but $s_k nrightarrow 0$. This shows that $T^-1$ is not continuous. Hence T is not a homeomorphism.






          share|cite|improve this answer













          It need not be for an infinite-dimensional subspace. Suppose the zero sequence and at least a countable subset of the countable family of sequences $mathcalF = s_k_k=1^infty, textwhere (s_k)_n = delta_kn$ is in X. ($s_k$ is the sequence that has zeros as entries except for the kth entry, where it is 1.)



          Note that $Ts_k rightarrow 0 in ell^2$, but $s_k nrightarrow 0$. This shows that $T^-1$ is not continuous. Hence T is not a homeomorphism.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 26 at 17:40









          ertl

          445110




          445110











          • thanks so much for your answer
            – Matey Math
            Jul 26 at 23:47
















          • thanks so much for your answer
            – Matey Math
            Jul 26 at 23:47















          thanks so much for your answer
          – Matey Math
          Jul 26 at 23:47




          thanks so much for your answer
          – Matey Math
          Jul 26 at 23:47












           

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