Equivalence of Baire Category Theorem for Complete Metric Spaces.











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In my studies of functional analysis I have come across a statement of the Baire Category Theorem which states:




"The Baire Category Theorem says that if a complete
metric space $X$ is a countable union
of closed sets $S_n$ then at least one of the $S_n$ contains
an open ball".




And I am struggling with how to relate this to the more common form of the theorem for example from wikipedia:




A Baire space is a topological space with the following property: for each countable collection of open dense sets ${displaystyle {U_{n}}_{n=1}^{infty }}$, their intersection ${displaystyle textstyle bigcap _{n=1}^{infty }U_{n}}$ is dense. And Every complete metric space is Baire space.




Any insight to how these two are equivalent would be greatly appreciated.










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    In my studies of functional analysis I have come across a statement of the Baire Category Theorem which states:




    "The Baire Category Theorem says that if a complete
    metric space $X$ is a countable union
    of closed sets $S_n$ then at least one of the $S_n$ contains
    an open ball".




    And I am struggling with how to relate this to the more common form of the theorem for example from wikipedia:




    A Baire space is a topological space with the following property: for each countable collection of open dense sets ${displaystyle {U_{n}}_{n=1}^{infty }}$, their intersection ${displaystyle textstyle bigcap _{n=1}^{infty }U_{n}}$ is dense. And Every complete metric space is Baire space.




    Any insight to how these two are equivalent would be greatly appreciated.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      In my studies of functional analysis I have come across a statement of the Baire Category Theorem which states:




      "The Baire Category Theorem says that if a complete
      metric space $X$ is a countable union
      of closed sets $S_n$ then at least one of the $S_n$ contains
      an open ball".




      And I am struggling with how to relate this to the more common form of the theorem for example from wikipedia:




      A Baire space is a topological space with the following property: for each countable collection of open dense sets ${displaystyle {U_{n}}_{n=1}^{infty }}$, their intersection ${displaystyle textstyle bigcap _{n=1}^{infty }U_{n}}$ is dense. And Every complete metric space is Baire space.




      Any insight to how these two are equivalent would be greatly appreciated.










      share|cite|improve this question















      In my studies of functional analysis I have come across a statement of the Baire Category Theorem which states:




      "The Baire Category Theorem says that if a complete
      metric space $X$ is a countable union
      of closed sets $S_n$ then at least one of the $S_n$ contains
      an open ball".




      And I am struggling with how to relate this to the more common form of the theorem for example from wikipedia:




      A Baire space is a topological space with the following property: for each countable collection of open dense sets ${displaystyle {U_{n}}_{n=1}^{infty }}$, their intersection ${displaystyle textstyle bigcap _{n=1}^{infty }U_{n}}$ is dense. And Every complete metric space is Baire space.




      Any insight to how these two are equivalent would be greatly appreciated.







      general-topology functional-analysis metric-spaces complete-spaces baire-category






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      edited Nov 21 at 14:42









      Davide Giraudo

      124k16150258




      124k16150258










      asked Nov 21 at 12:32









      Matthew

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      19912






















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          Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.






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            Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.






            share|cite|improve this answer



























              up vote
              4
              down vote



              accepted










              Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.






              share|cite|improve this answer

























                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.






                share|cite|improve this answer














                Consider the assertion “every countable intersection of open dense sets is dense”. Passing to complements, it is seen to be equivalent to “every countable union of closed sets with empty interiors has empty interior”. But the interior of whole space $X$ is certainly not empty! Therefore, if it can be written has the union of a countable family of closed subsets, then at least one of these sets must have non-empty interior, which is equivalent to the assertion that it contains an open ball.







                share|cite|improve this answer














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                share|cite|improve this answer








                edited Nov 21 at 14:44

























                answered Nov 21 at 12:39









                José Carlos Santos

                144k20113214




                144k20113214






























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