Proof verification that the sequence $x_n = frac{1}{n}$ converges to every point of $mathbb{R}$ on the...












5












$begingroup$


Let $a in mathbb{R}$. Then any open set $U$ in the cofinite topology containing $a$ is of the form $U = mathbb{R} - {alpha_1, alpha_2, cdots, alpha_p}$ for some $p in mathbb{N}$ (and of course $alpha_i neq a forall 1 leq i leq p$). Now, take $N = c left(displaystyle{frac{1}{min alpha_i}}right) + 1$, where $c(x)$ stands for the ceiling of $x$ (i.e, $c(2.1) = 3$). Then it's clear that $x_n in U forall n geq N$ and we're done.



Is this alright? I actually think I could do away with all of this and just make the argument that the tail of $x_n$ is eventually contained in $U$ since $mathbb{R} - U$ has only finitely many elements and ${x_n}_{n in mathbb{N}}$ is infinite, but I'm not sure if that's fine too.






real-analysis general-topology proof-verification







share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    Let $a in mathbb{R}$. Then any open set $U$ in the cofinite topology containing $a$ is of the form $U = mathbb{R} - {alpha_1, alpha_2, cdots, alpha_p}$ for some $p in mathbb{N}$ (and of course $alpha_i neq a forall 1 leq i leq p$). Now, take $N = c left(displaystyle{frac{1}{min alpha_i}}right) + 1$, where $c(x)$ stands for the ceiling of $x$ (i.e, $c(2.1) = 3$). Then it's clear that $x_n in U forall n geq N$ and we're done.



    Is this alright? I actually think I could do away with all of this and just make the argument that the tail of $x_n$ is eventually contained in $U$ since $mathbb{R} - U$ has only finitely many elements and ${x_n}_{n in mathbb{N}}$ is infinite, but I'm not sure if that's fine too.






    real-analysis general-topology proof-verification







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      Let $a in mathbb{R}$. Then any open set $U$ in the cofinite topology containing $a$ is of the form $U = mathbb{R} - {alpha_1, alpha_2, cdots, alpha_p}$ for some $p in mathbb{N}$ (and of course $alpha_i neq a forall 1 leq i leq p$). Now, take $N = c left(displaystyle{frac{1}{min alpha_i}}right) + 1$, where $c(x)$ stands for the ceiling of $x$ (i.e, $c(2.1) = 3$). Then it's clear that $x_n in U forall n geq N$ and we're done.



      Is this alright? I actually think I could do away with all of this and just make the argument that the tail of $x_n$ is eventually contained in $U$ since $mathbb{R} - U$ has only finitely many elements and ${x_n}_{n in mathbb{N}}$ is infinite, but I'm not sure if that's fine too.






      real-analysis general-topology proof-verification







      share|cite|improve this question









      $endgroup$




      Let $a in mathbb{R}$. Then any open set $U$ in the cofinite topology containing $a$ is of the form $U = mathbb{R} - {alpha_1, alpha_2, cdots, alpha_p}$ for some $p in mathbb{N}$ (and of course $alpha_i neq a forall 1 leq i leq p$). Now, take $N = c left(displaystyle{frac{1}{min alpha_i}}right) + 1$, where $c(x)$ stands for the ceiling of $x$ (i.e, $c(2.1) = 3$). Then it's clear that $x_n in U forall n geq N$ and we're done.



      Is this alright? I actually think I could do away with all of this and just make the argument that the tail of $x_n$ is eventually contained in $U$ since $mathbb{R} - U$ has only finitely many elements and ${x_n}_{n in mathbb{N}}$ is infinite, but I'm not sure if that's fine too.






      real-analysis general-topology proof-verification




      real-analysis general-topology proof-verification






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 13 at 19:37









      Matheus AndradeMatheus Andrade

      1,259417




      1,259417






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Your argument is fine, but you are right that such specifics are unnecessary. Any sequence that does not contain a particular point infinitely many times converges to every point in an infinite space with the cofinite topology.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            It's good to know my intuition was correct. Thanks!
            $endgroup$
            – Matheus Andrade
            Jan 13 at 19:57






          • 1




            $begingroup$
            @Matheus No problem.
            $endgroup$
            – Matt Samuel
            Jan 13 at 19:57






          • 1




            $begingroup$
            Your final claim is not completely correct. Let $X$ be a space with the cofinite topology. Then a sequence $x_1, x_2, ldots$ converges to every point in $X$ if and only if for every $x in X$ it holds true that there are only finitely many $i$ for which $x = x_i$. For example, the sequence $x_i = (-1)^i$ is not eventually constant, but doesn't converge at all, let alone to every point of $mathbb{R}$.
            $endgroup$
            – Woett
            Jan 13 at 21:13












          • $begingroup$
            @Woett Agree with the edit?
            $endgroup$
            – Matt Samuel
            Jan 13 at 21:15










          • $begingroup$
            yes! $text{ }$
            $endgroup$
            – Woett
            Jan 13 at 21:21



















          1












          $begingroup$

          The only thing we are using of the sequence here is that $x_n neq x_m$ whenever $n neq m$, so that ${x_n: n in mathbb{N}}$ is infinite.



          If then $U=mathbb{R}-F$ (so $F subseteq mathbb{R}$ finite) is any non-empty open set of the cofinite topology, then there is some $N$ such that ${x_n: n ge N}$ contains no points of $F$: we skip as many $n$ as we need to avoid any of the finitely many points in $F cap {x_n: n in mathbb{N}}$, if there are any. And then for all $n ge N$, $x_n in U$. This suffices to show that $x_n to x$ for any $x in mathbb{R}$.



          So also $1,2,3,4,ldots to pi$, e.g. in the cofinite topology.






          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









            2












            $begingroup$

            Your argument is fine, but you are right that such specifics are unnecessary. Any sequence that does not contain a particular point infinitely many times converges to every point in an infinite space with the cofinite topology.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It's good to know my intuition was correct. Thanks!
              $endgroup$
              – Matheus Andrade
              Jan 13 at 19:57






            • 1




              $begingroup$
              @Matheus No problem.
              $endgroup$
              – Matt Samuel
              Jan 13 at 19:57






            • 1




              $begingroup$
              Your final claim is not completely correct. Let $X$ be a space with the cofinite topology. Then a sequence $x_1, x_2, ldots$ converges to every point in $X$ if and only if for every $x in X$ it holds true that there are only finitely many $i$ for which $x = x_i$. For example, the sequence $x_i = (-1)^i$ is not eventually constant, but doesn't converge at all, let alone to every point of $mathbb{R}$.
              $endgroup$
              – Woett
              Jan 13 at 21:13












            • $begingroup$
              @Woett Agree with the edit?
              $endgroup$
              – Matt Samuel
              Jan 13 at 21:15










            • $begingroup$
              yes! $text{ }$
              $endgroup$
              – Woett
              Jan 13 at 21:21
















            2












            $begingroup$

            Your argument is fine, but you are right that such specifics are unnecessary. Any sequence that does not contain a particular point infinitely many times converges to every point in an infinite space with the cofinite topology.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It's good to know my intuition was correct. Thanks!
              $endgroup$
              – Matheus Andrade
              Jan 13 at 19:57






            • 1




              $begingroup$
              @Matheus No problem.
              $endgroup$
              – Matt Samuel
              Jan 13 at 19:57






            • 1




              $begingroup$
              Your final claim is not completely correct. Let $X$ be a space with the cofinite topology. Then a sequence $x_1, x_2, ldots$ converges to every point in $X$ if and only if for every $x in X$ it holds true that there are only finitely many $i$ for which $x = x_i$. For example, the sequence $x_i = (-1)^i$ is not eventually constant, but doesn't converge at all, let alone to every point of $mathbb{R}$.
              $endgroup$
              – Woett
              Jan 13 at 21:13












            • $begingroup$
              @Woett Agree with the edit?
              $endgroup$
              – Matt Samuel
              Jan 13 at 21:15










            • $begingroup$
              yes! $text{ }$
              $endgroup$
              – Woett
              Jan 13 at 21:21














            2












            2








            2





            $begingroup$

            Your argument is fine, but you are right that such specifics are unnecessary. Any sequence that does not contain a particular point infinitely many times converges to every point in an infinite space with the cofinite topology.






            share|cite|improve this answer











            $endgroup$



            Your argument is fine, but you are right that such specifics are unnecessary. Any sequence that does not contain a particular point infinitely many times converges to every point in an infinite space with the cofinite topology.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 13 at 21:15

























            answered Jan 13 at 19:55









            Matt SamuelMatt Samuel

            38k63666




            38k63666












            • $begingroup$
              It's good to know my intuition was correct. Thanks!
              $endgroup$
              – Matheus Andrade
              Jan 13 at 19:57






            • 1




              $begingroup$
              @Matheus No problem.
              $endgroup$
              – Matt Samuel
              Jan 13 at 19:57






            • 1




              $begingroup$
              Your final claim is not completely correct. Let $X$ be a space with the cofinite topology. Then a sequence $x_1, x_2, ldots$ converges to every point in $X$ if and only if for every $x in X$ it holds true that there are only finitely many $i$ for which $x = x_i$. For example, the sequence $x_i = (-1)^i$ is not eventually constant, but doesn't converge at all, let alone to every point of $mathbb{R}$.
              $endgroup$
              – Woett
              Jan 13 at 21:13












            • $begingroup$
              @Woett Agree with the edit?
              $endgroup$
              – Matt Samuel
              Jan 13 at 21:15










            • $begingroup$
              yes! $text{ }$
              $endgroup$
              – Woett
              Jan 13 at 21:21


















            • $begingroup$
              It's good to know my intuition was correct. Thanks!
              $endgroup$
              – Matheus Andrade
              Jan 13 at 19:57






            • 1




              $begingroup$
              @Matheus No problem.
              $endgroup$
              – Matt Samuel
              Jan 13 at 19:57






            • 1




              $begingroup$
              Your final claim is not completely correct. Let $X$ be a space with the cofinite topology. Then a sequence $x_1, x_2, ldots$ converges to every point in $X$ if and only if for every $x in X$ it holds true that there are only finitely many $i$ for which $x = x_i$. For example, the sequence $x_i = (-1)^i$ is not eventually constant, but doesn't converge at all, let alone to every point of $mathbb{R}$.
              $endgroup$
              – Woett
              Jan 13 at 21:13












            • $begingroup$
              @Woett Agree with the edit?
              $endgroup$
              – Matt Samuel
              Jan 13 at 21:15










            • $begingroup$
              yes! $text{ }$
              $endgroup$
              – Woett
              Jan 13 at 21:21
















            $begingroup$
            It's good to know my intuition was correct. Thanks!
            $endgroup$
            – Matheus Andrade
            Jan 13 at 19:57




            $begingroup$
            It's good to know my intuition was correct. Thanks!
            $endgroup$
            – Matheus Andrade
            Jan 13 at 19:57




            1




            1




            $begingroup$
            @Matheus No problem.
            $endgroup$
            – Matt Samuel
            Jan 13 at 19:57




            $begingroup$
            @Matheus No problem.
            $endgroup$
            – Matt Samuel
            Jan 13 at 19:57




            1




            1




            $begingroup$
            Your final claim is not completely correct. Let $X$ be a space with the cofinite topology. Then a sequence $x_1, x_2, ldots$ converges to every point in $X$ if and only if for every $x in X$ it holds true that there are only finitely many $i$ for which $x = x_i$. For example, the sequence $x_i = (-1)^i$ is not eventually constant, but doesn't converge at all, let alone to every point of $mathbb{R}$.
            $endgroup$
            – Woett
            Jan 13 at 21:13






            $begingroup$
            Your final claim is not completely correct. Let $X$ be a space with the cofinite topology. Then a sequence $x_1, x_2, ldots$ converges to every point in $X$ if and only if for every $x in X$ it holds true that there are only finitely many $i$ for which $x = x_i$. For example, the sequence $x_i = (-1)^i$ is not eventually constant, but doesn't converge at all, let alone to every point of $mathbb{R}$.
            $endgroup$
            – Woett
            Jan 13 at 21:13














            $begingroup$
            @Woett Agree with the edit?
            $endgroup$
            – Matt Samuel
            Jan 13 at 21:15




            $begingroup$
            @Woett Agree with the edit?
            $endgroup$
            – Matt Samuel
            Jan 13 at 21:15












            $begingroup$
            yes! $text{ }$
            $endgroup$
            – Woett
            Jan 13 at 21:21




            $begingroup$
            yes! $text{ }$
            $endgroup$
            – Woett
            Jan 13 at 21:21











            1












            $begingroup$

            The only thing we are using of the sequence here is that $x_n neq x_m$ whenever $n neq m$, so that ${x_n: n in mathbb{N}}$ is infinite.



            If then $U=mathbb{R}-F$ (so $F subseteq mathbb{R}$ finite) is any non-empty open set of the cofinite topology, then there is some $N$ such that ${x_n: n ge N}$ contains no points of $F$: we skip as many $n$ as we need to avoid any of the finitely many points in $F cap {x_n: n in mathbb{N}}$, if there are any. And then for all $n ge N$, $x_n in U$. This suffices to show that $x_n to x$ for any $x in mathbb{R}$.



            So also $1,2,3,4,ldots to pi$, e.g. in the cofinite topology.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The only thing we are using of the sequence here is that $x_n neq x_m$ whenever $n neq m$, so that ${x_n: n in mathbb{N}}$ is infinite.



              If then $U=mathbb{R}-F$ (so $F subseteq mathbb{R}$ finite) is any non-empty open set of the cofinite topology, then there is some $N$ such that ${x_n: n ge N}$ contains no points of $F$: we skip as many $n$ as we need to avoid any of the finitely many points in $F cap {x_n: n in mathbb{N}}$, if there are any. And then for all $n ge N$, $x_n in U$. This suffices to show that $x_n to x$ for any $x in mathbb{R}$.



              So also $1,2,3,4,ldots to pi$, e.g. in the cofinite topology.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The only thing we are using of the sequence here is that $x_n neq x_m$ whenever $n neq m$, so that ${x_n: n in mathbb{N}}$ is infinite.



                If then $U=mathbb{R}-F$ (so $F subseteq mathbb{R}$ finite) is any non-empty open set of the cofinite topology, then there is some $N$ such that ${x_n: n ge N}$ contains no points of $F$: we skip as many $n$ as we need to avoid any of the finitely many points in $F cap {x_n: n in mathbb{N}}$, if there are any. And then for all $n ge N$, $x_n in U$. This suffices to show that $x_n to x$ for any $x in mathbb{R}$.



                So also $1,2,3,4,ldots to pi$, e.g. in the cofinite topology.






                share|cite|improve this answer









                $endgroup$



                The only thing we are using of the sequence here is that $x_n neq x_m$ whenever $n neq m$, so that ${x_n: n in mathbb{N}}$ is infinite.



                If then $U=mathbb{R}-F$ (so $F subseteq mathbb{R}$ finite) is any non-empty open set of the cofinite topology, then there is some $N$ such that ${x_n: n ge N}$ contains no points of $F$: we skip as many $n$ as we need to avoid any of the finitely many points in $F cap {x_n: n in mathbb{N}}$, if there are any. And then for all $n ge N$, $x_n in U$. This suffices to show that $x_n to x$ for any $x in mathbb{R}$.



                So also $1,2,3,4,ldots to pi$, e.g. in the cofinite topology.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 7:07









                Henno BrandsmaHenno Brandsma

                107k347114




                107k347114






























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