Show $mathcal{B}( mathbb R)$ contains all countable subsets of $mathbb R$












0














We defined $mathcal{B}(mathbb R):=sigma({[a,b[}),$ where $a < b in mathbb R.$



I have been able to prove:



i) for any $b in mathbb R$, ${b} inmathcal{B}(mathbb R)$



and subsequently ii) $[a,b] in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$



iii) $]a,b[ in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$



Now for my problem, how can I prove that $mathcal{B}(mathbb R)$ contains all countable subsets of $mathbb R$?



My idea was to take any countable subset $C$ of $mathbb R$,



it can be written as $C = bigcup_{n in mathbb N}{a_{n}}$. Since we were able to prove (i), $C$ is immediately as the countable union $in mathcal{B}(mathbb R)$. Is this ok? Does not seem clean enough to me?






real-analysis measure-theory borel-sets borel-measures







share|cite|improve this question





























    0














    We defined $mathcal{B}(mathbb R):=sigma({[a,b[}),$ where $a < b in mathbb R.$



    I have been able to prove:



    i) for any $b in mathbb R$, ${b} inmathcal{B}(mathbb R)$



    and subsequently ii) $[a,b] in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$



    iii) $]a,b[ in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$



    Now for my problem, how can I prove that $mathcal{B}(mathbb R)$ contains all countable subsets of $mathbb R$?



    My idea was to take any countable subset $C$ of $mathbb R$,



    it can be written as $C = bigcup_{n in mathbb N}{a_{n}}$. Since we were able to prove (i), $C$ is immediately as the countable union $in mathcal{B}(mathbb R)$. Is this ok? Does not seem clean enough to me?






    real-analysis measure-theory borel-sets borel-measures







    share|cite|improve this question



























      0












      0








      0







      We defined $mathcal{B}(mathbb R):=sigma({[a,b[}),$ where $a < b in mathbb R.$



      I have been able to prove:



      i) for any $b in mathbb R$, ${b} inmathcal{B}(mathbb R)$



      and subsequently ii) $[a,b] in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$



      iii) $]a,b[ in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$



      Now for my problem, how can I prove that $mathcal{B}(mathbb R)$ contains all countable subsets of $mathbb R$?



      My idea was to take any countable subset $C$ of $mathbb R$,



      it can be written as $C = bigcup_{n in mathbb N}{a_{n}}$. Since we were able to prove (i), $C$ is immediately as the countable union $in mathcal{B}(mathbb R)$. Is this ok? Does not seem clean enough to me?






      real-analysis measure-theory borel-sets borel-measures







      share|cite|improve this question















      We defined $mathcal{B}(mathbb R):=sigma({[a,b[}),$ where $a < b in mathbb R.$



      I have been able to prove:



      i) for any $b in mathbb R$, ${b} inmathcal{B}(mathbb R)$



      and subsequently ii) $[a,b] in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$



      iii) $]a,b[ in mathcal{B}(mathbb R)$ for any $a, b in mathbb R$



      Now for my problem, how can I prove that $mathcal{B}(mathbb R)$ contains all countable subsets of $mathbb R$?



      My idea was to take any countable subset $C$ of $mathbb R$,



      it can be written as $C = bigcup_{n in mathbb N}{a_{n}}$. Since we were able to prove (i), $C$ is immediately as the countable union $in mathcal{B}(mathbb R)$. Is this ok? Does not seem clean enough to me?






      real-analysis measure-theory borel-sets borel-measures




      real-analysis measure-theory borel-sets borel-measures






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 at 20:32









      Lord Shark the Unknown

      100k958131




      100k958131










      asked Nov 21 at 20:29









      SABOY

      522311




      522311






















          2 Answers
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          3














          The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.






          share|cite|improve this answer





















          • So my version is correct?
            – SABOY
            Nov 21 at 20:37










          • Yes, it is fine.
            – Ross Millikan
            Nov 21 at 20:50



















          1














          Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:



          Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.






            share|cite|improve this answer





















            • So my version is correct?
              – SABOY
              Nov 21 at 20:37










            • Yes, it is fine.
              – Ross Millikan
              Nov 21 at 20:50
















            3














            The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.






            share|cite|improve this answer





















            • So my version is correct?
              – SABOY
              Nov 21 at 20:37










            • Yes, it is fine.
              – Ross Millikan
              Nov 21 at 20:50














            3












            3








            3






            The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.






            share|cite|improve this answer












            The usual definition of a sigma-algebra requires that it be closed under countable unions. Once that you have that every singleton is in $mathcal{B}( mathbb R)$, every countable subset of $Bbb R$ is a countable union of singletons.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 21 at 20:35









            Ross Millikan

            290k23196370




            290k23196370












            • So my version is correct?
              – SABOY
              Nov 21 at 20:37










            • Yes, it is fine.
              – Ross Millikan
              Nov 21 at 20:50


















            • So my version is correct?
              – SABOY
              Nov 21 at 20:37










            • Yes, it is fine.
              – Ross Millikan
              Nov 21 at 20:50
















            So my version is correct?
            – SABOY
            Nov 21 at 20:37




            So my version is correct?
            – SABOY
            Nov 21 at 20:37












            Yes, it is fine.
            – Ross Millikan
            Nov 21 at 20:50




            Yes, it is fine.
            – Ross Millikan
            Nov 21 at 20:50











            1














            Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:



            Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$






            share|cite|improve this answer


























              1














              Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:



              Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$






              share|cite|improve this answer
























                1












                1








                1






                Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:



                Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$






                share|cite|improve this answer












                Your answer is entirely correct, since $sigma$-algebras are closed under countable unions and singeltons are in this $sigma$-algebra:



                Indeed, if $ain mathbb{R}$, then $${a} = bigcap_{n=1}^infty{[a,a + 1/n[} = bigcap_{n=1}^infty{[a,a + 1/n]}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 at 20:43









                Math_QED

                6,96731449




                6,96731449






























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