Can we find $V_n$ in such a way that $bigcaplimits_{n=1}^{infty} V_n$ is countable?












0












$begingroup$



If $V_n$ is open, dense subset of $mathbb R$ for each $nin mathbb N$ then by Baire's Category theorem we know that $displaystyle cap_{n=1}^{infty} V_n$ is dense in $mathbb R$.



My question is that : Can we find $V_n$ in such a way that $displaystyle bigcap_{n=1}^{infty} V_n$ is countable set?




Any help is appreciated. Thank you.






real-analysis baire-category







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    If $V_n$ is open, dense subset of $mathbb R$ for each $nin mathbb N$ then by Baire's Category theorem we know that $displaystyle cap_{n=1}^{infty} V_n$ is dense in $mathbb R$.



    My question is that : Can we find $V_n$ in such a way that $displaystyle bigcap_{n=1}^{infty} V_n$ is countable set?




    Any help is appreciated. Thank you.






    real-analysis baire-category







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$



      If $V_n$ is open, dense subset of $mathbb R$ for each $nin mathbb N$ then by Baire's Category theorem we know that $displaystyle cap_{n=1}^{infty} V_n$ is dense in $mathbb R$.



      My question is that : Can we find $V_n$ in such a way that $displaystyle bigcap_{n=1}^{infty} V_n$ is countable set?




      Any help is appreciated. Thank you.






      real-analysis baire-category







      share|cite|improve this question











      $endgroup$





      If $V_n$ is open, dense subset of $mathbb R$ for each $nin mathbb N$ then by Baire's Category theorem we know that $displaystyle cap_{n=1}^{infty} V_n$ is dense in $mathbb R$.



      My question is that : Can we find $V_n$ in such a way that $displaystyle bigcap_{n=1}^{infty} V_n$ is countable set?




      Any help is appreciated. Thank you.






      real-analysis baire-category




      real-analysis baire-category






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 29 '18 at 9:59









      Martin Sleziak

      44.7k10118272




      44.7k10118272










      asked Dec 20 '18 at 13:00









      nurun neshanurun nesha

      1,0242623




      1,0242623






















          2 Answers
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          active

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          4












          $begingroup$

          No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            We could also take $W_n=V_n setminus {r_n}$.
            $endgroup$
            – nurun nesha
            Dec 20 '18 at 13:12












          • $begingroup$
            @nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
            $endgroup$
            – José Carlos Santos
            Dec 20 '18 at 13:13










          • $begingroup$
            Yeah, sure edit your answer.
            $endgroup$
            – nurun nesha
            Dec 20 '18 at 13:16



















          3












          $begingroup$

          No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              We could also take $W_n=V_n setminus {r_n}$.
              $endgroup$
              – nurun nesha
              Dec 20 '18 at 13:12












            • $begingroup$
              @nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
              $endgroup$
              – José Carlos Santos
              Dec 20 '18 at 13:13










            • $begingroup$
              Yeah, sure edit your answer.
              $endgroup$
              – nurun nesha
              Dec 20 '18 at 13:16
















            4












            $begingroup$

            No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              We could also take $W_n=V_n setminus {r_n}$.
              $endgroup$
              – nurun nesha
              Dec 20 '18 at 13:12












            • $begingroup$
              @nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
              $endgroup$
              – José Carlos Santos
              Dec 20 '18 at 13:13










            • $begingroup$
              Yeah, sure edit your answer.
              $endgroup$
              – nurun nesha
              Dec 20 '18 at 13:16














            4












            4








            4





            $begingroup$

            No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.






            share|cite|improve this answer











            $endgroup$



            No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 20 '18 at 13:21

























            answered Dec 20 '18 at 13:07









            José Carlos SantosJosé Carlos Santos

            162k22128232




            162k22128232












            • $begingroup$
              We could also take $W_n=V_n setminus {r_n}$.
              $endgroup$
              – nurun nesha
              Dec 20 '18 at 13:12












            • $begingroup$
              @nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
              $endgroup$
              – José Carlos Santos
              Dec 20 '18 at 13:13










            • $begingroup$
              Yeah, sure edit your answer.
              $endgroup$
              – nurun nesha
              Dec 20 '18 at 13:16


















            • $begingroup$
              We could also take $W_n=V_n setminus {r_n}$.
              $endgroup$
              – nurun nesha
              Dec 20 '18 at 13:12












            • $begingroup$
              @nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
              $endgroup$
              – José Carlos Santos
              Dec 20 '18 at 13:13










            • $begingroup$
              Yeah, sure edit your answer.
              $endgroup$
              – nurun nesha
              Dec 20 '18 at 13:16
















            $begingroup$
            We could also take $W_n=V_n setminus {r_n}$.
            $endgroup$
            – nurun nesha
            Dec 20 '18 at 13:12






            $begingroup$
            We could also take $W_n=V_n setminus {r_n}$.
            $endgroup$
            – nurun nesha
            Dec 20 '18 at 13:12














            $begingroup$
            @nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
            $endgroup$
            – José Carlos Santos
            Dec 20 '18 at 13:13




            $begingroup$
            @nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
            $endgroup$
            – José Carlos Santos
            Dec 20 '18 at 13:13












            $begingroup$
            Yeah, sure edit your answer.
            $endgroup$
            – nurun nesha
            Dec 20 '18 at 13:16




            $begingroup$
            Yeah, sure edit your answer.
            $endgroup$
            – nurun nesha
            Dec 20 '18 at 13:16











            3












            $begingroup$

            No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.






                share|cite|improve this answer









                $endgroup$



                No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 20 '18 at 13:06









                Robert IsraelRobert Israel

                323k23213467




                323k23213467






























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