Why is $E$ discrete?











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I am reading through the following proof in Aupetit's book A Primer on Spectral Theory.



enter image description here



I understand the proof, except for the very last sentence:




Consequently its set of zeros, precisely $E$, is discrete.




I understand that $Omega = { lambda in D : phi(
lambda)=0 }=E$ and that, since $phi$ is continuous on $D$ and holomorphic on $Omega=E$ we have, by Rado's Theorem, that $phi$ is holomorphic on $D$. I do not see how this shows that $E$ is discrete though?






complex-analysis functional-analysis spectral-theory banach-algebras







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    up vote
    1
    down vote

    favorite












    I am reading through the following proof in Aupetit's book A Primer on Spectral Theory.



    enter image description here



    I understand the proof, except for the very last sentence:




    Consequently its set of zeros, precisely $E$, is discrete.




    I understand that $Omega = { lambda in D : phi(
    lambda)=0 }=E$     and that, since $phi$ is continuous on $D$ and holomorphic on $Omega=E$ we have, by Rado's Theorem, that $phi$ is holomorphic on $D$. I do not see how this shows that $E$ is discrete though?






    complex-analysis functional-analysis spectral-theory banach-algebras







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am reading through the following proof in Aupetit's book A Primer on Spectral Theory.



      enter image description here



      I understand the proof, except for the very last sentence:




      Consequently its set of zeros, precisely $E$, is discrete.




      I understand that $Omega = { lambda in D : phi(
      lambda)=0 }=E$       and that, since $phi$ is continuous on $D$ and holomorphic on $Omega=E$ we have, by Rado's Theorem, that $phi$ is holomorphic on $D$. I do not see how this shows that $E$ is discrete though?






      complex-analysis functional-analysis spectral-theory banach-algebras







      share|cite|improve this question













      I am reading through the following proof in Aupetit's book A Primer on Spectral Theory.



      enter image description here



      I understand the proof, except for the very last sentence:




      Consequently its set of zeros, precisely $E$, is discrete.




      I understand that $Omega = { lambda in D : phi(
      lambda)=0 }=E$       and that, since $phi$ is continuous on $D$ and holomorphic on $Omega=E$ we have, by Rado's Theorem, that $phi$ is holomorphic on $D$. I do not see how this shows that $E$ is discrete though?






      complex-analysis functional-analysis spectral-theory banach-algebras




      complex-analysis functional-analysis spectral-theory banach-algebras






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