(Prove or Disprove) If $X$ is a metric space, then the Borel $sigma$-algebra ${cal B}_X$ contains every...












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I am tasked with determining whether the following statement is true or false:



If $X$ is a metric space, then the Borel $sigma$-algebra ${cal B}_X$ contains every countable subset of $X$.



I understand how the Borel $sigma$-algebra is constructed, but I have very little intuition into what sets it contains.






measure-theory







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    1












    $begingroup$


    I am tasked with determining whether the following statement is true or false:



    If $X$ is a metric space, then the Borel $sigma$-algebra ${cal B}_X$ contains every countable subset of $X$.



    I understand how the Borel $sigma$-algebra is constructed, but I have very little intuition into what sets it contains.






    measure-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am tasked with determining whether the following statement is true or false:



      If $X$ is a metric space, then the Borel $sigma$-algebra ${cal B}_X$ contains every countable subset of $X$.



      I understand how the Borel $sigma$-algebra is constructed, but I have very little intuition into what sets it contains.






      measure-theory







      share|cite|improve this question









      $endgroup$




      I am tasked with determining whether the following statement is true or false:



      If $X$ is a metric space, then the Borel $sigma$-algebra ${cal B}_X$ contains every countable subset of $X$.



      I understand how the Borel $sigma$-algebra is constructed, but I have very little intuition into what sets it contains.






      measure-theory




      measure-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 21 at 6:57









      johnny133253johnny133253

      321110




      321110






















          2 Answers
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          $begingroup$

          First, note that it contains every singleton subset of $X$. For take any point $x in X$; then the intersection of all open balls around $x$ with radius $1/n$ lies in the Borel sigma-algebra, but this only contains $x$ since $X$ is a metric space.



          Now take any countable subset $A$ of $X$. Since the Borel sigma-algebra contains all singletons ${x}$ for $xin A$ and it’s closed under countable unions, it contains $A$.






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            $begingroup$

            Show that ${x}$ is closed for every $x$ in a metric space. It's also a countable intersection of open sets $B(x,frac1n)$ so in either case, singletons are in the Borel $sigma$-algebra. Countable sets are countable unions of those.






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              2 Answers
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              2 Answers
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              2












              $begingroup$

              First, note that it contains every singleton subset of $X$. For take any point $x in X$; then the intersection of all open balls around $x$ with radius $1/n$ lies in the Borel sigma-algebra, but this only contains $x$ since $X$ is a metric space.



              Now take any countable subset $A$ of $X$. Since the Borel sigma-algebra contains all singletons ${x}$ for $xin A$ and it’s closed under countable unions, it contains $A$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                First, note that it contains every singleton subset of $X$. For take any point $x in X$; then the intersection of all open balls around $x$ with radius $1/n$ lies in the Borel sigma-algebra, but this only contains $x$ since $X$ is a metric space.



                Now take any countable subset $A$ of $X$. Since the Borel sigma-algebra contains all singletons ${x}$ for $xin A$ and it’s closed under countable unions, it contains $A$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  First, note that it contains every singleton subset of $X$. For take any point $x in X$; then the intersection of all open balls around $x$ with radius $1/n$ lies in the Borel sigma-algebra, but this only contains $x$ since $X$ is a metric space.



                  Now take any countable subset $A$ of $X$. Since the Borel sigma-algebra contains all singletons ${x}$ for $xin A$ and it’s closed under countable unions, it contains $A$.






                  share|cite|improve this answer









                  $endgroup$



                  First, note that it contains every singleton subset of $X$. For take any point $x in X$; then the intersection of all open balls around $x$ with radius $1/n$ lies in the Borel sigma-algebra, but this only contains $x$ since $X$ is a metric space.



                  Now take any countable subset $A$ of $X$. Since the Borel sigma-algebra contains all singletons ${x}$ for $xin A$ and it’s closed under countable unions, it contains $A$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 21 at 7:02









                  SvanNSvanN

                  2,0661422




                  2,0661422























                      3












                      $begingroup$

                      Show that ${x}$ is closed for every $x$ in a metric space. It's also a countable intersection of open sets $B(x,frac1n)$ so in either case, singletons are in the Borel $sigma$-algebra. Countable sets are countable unions of those.






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        Show that ${x}$ is closed for every $x$ in a metric space. It's also a countable intersection of open sets $B(x,frac1n)$ so in either case, singletons are in the Borel $sigma$-algebra. Countable sets are countable unions of those.






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          Show that ${x}$ is closed for every $x$ in a metric space. It's also a countable intersection of open sets $B(x,frac1n)$ so in either case, singletons are in the Borel $sigma$-algebra. Countable sets are countable unions of those.






                          share|cite|improve this answer









                          $endgroup$



                          Show that ${x}$ is closed for every $x$ in a metric space. It's also a countable intersection of open sets $B(x,frac1n)$ so in either case, singletons are in the Borel $sigma$-algebra. Countable sets are countable unions of those.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 21 at 7:00









                          Henno BrandsmaHenno Brandsma

                          112k348120




                          112k348120






























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