Understanding a Proof in Topology of the Reals

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Good evening fellow math-friends (or morning, depending on where you are),



I am having trouble understanding a proof in the topology of the reals, i.e. a subset F of the reals is closed if and only if the limit of every convergent sequence in F belongs to F. In particular, I was trying to prove that "if the limit of every convergent sequence in F belongs to F, then F is closed. I was trying to do a proof by contradiction, and then for some help I looked at the proof here (under proposition 5.18):
https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch5.pdf



I don't really understand why they say to assume $x in F^c$, and $x$ has to have a neighbourhood belonging to $F^c$ otherwise $forall n in N, exists x_n in F$ such that $x_n in (x - frac1n, x + frac1n)$, so $x = lim x_n$ and $x$ is the limit of a sequence in $F$. I don't really follow through with the "otherwise" bit or see how it is a contradiction, may someone clarify this or further explain it to me?



Thank you in advance.






real-analysis sequences-and-series general-topology







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    up vote
    2
    down vote

    favorite












    Good evening fellow math-friends (or morning, depending on where you are),



    I am having trouble understanding a proof in the topology of the reals, i.e. a subset F of the reals is closed if and only if the limit of every convergent sequence in F belongs to F. In particular, I was trying to prove that "if the limit of every convergent sequence in F belongs to F, then F is closed. I was trying to do a proof by contradiction, and then for some help I looked at the proof here (under proposition 5.18):
    https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch5.pdf



    I don't really understand why they say to assume $x in F^c$, and $x$ has to have a neighbourhood belonging to $F^c$ otherwise $forall n in N, exists x_n in F$ such that $x_n in (x - frac1n, x + frac1n)$, so $x = lim x_n$ and $x$ is the limit of a sequence in $F$. I don't really follow through with the "otherwise" bit or see how it is a contradiction, may someone clarify this or further explain it to me?



    Thank you in advance.






    real-analysis sequences-and-series general-topology







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Good evening fellow math-friends (or morning, depending on where you are),



      I am having trouble understanding a proof in the topology of the reals, i.e. a subset F of the reals is closed if and only if the limit of every convergent sequence in F belongs to F. In particular, I was trying to prove that "if the limit of every convergent sequence in F belongs to F, then F is closed. I was trying to do a proof by contradiction, and then for some help I looked at the proof here (under proposition 5.18):
      https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch5.pdf



      I don't really understand why they say to assume $x in F^c$, and $x$ has to have a neighbourhood belonging to $F^c$ otherwise $forall n in N, exists x_n in F$ such that $x_n in (x - frac1n, x + frac1n)$, so $x = lim x_n$ and $x$ is the limit of a sequence in $F$. I don't really follow through with the "otherwise" bit or see how it is a contradiction, may someone clarify this or further explain it to me?



      Thank you in advance.






      real-analysis sequences-and-series general-topology







      share|cite|improve this question













      Good evening fellow math-friends (or morning, depending on where you are),



      I am having trouble understanding a proof in the topology of the reals, i.e. a subset F of the reals is closed if and only if the limit of every convergent sequence in F belongs to F. In particular, I was trying to prove that "if the limit of every convergent sequence in F belongs to F, then F is closed. I was trying to do a proof by contradiction, and then for some help I looked at the proof here (under proposition 5.18):
      https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch5.pdf



      I don't really understand why they say to assume $x in F^c$, and $x$ has to have a neighbourhood belonging to $F^c$ otherwise $forall n in N, exists x_n in F$ such that $x_n in (x - frac1n, x + frac1n)$, so $x = lim x_n$ and $x$ is the limit of a sequence in $F$. I don't really follow through with the "otherwise" bit or see how it is a contradiction, may someone clarify this or further explain it to me?



      Thank you in advance.






      real-analysis sequences-and-series general-topology




      real-analysis sequences-and-series general-topology






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      asked 3 hours ago









      GingerKittyLover

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          1 Answer
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          Suppose that $x$ does not have a neighbourhood contained in $F^c$. Then, for each interval $I_n = (x-frac1n,x+frac1n)$ which is a neighbourhood of $x$, there must exist a point $x_n in I_n cap F$. Otherwise, we would have $I_n subseteq F^c$. In particular, $|x_n - x| < frac1n$ for each $n$, and so $x_n to x$.






          share|cite|improve this answer




















          • Ohhhh perfect, that makes much more sense now; thank you so much!! :)
            – GingerKittyLover
            3 hours ago










          • No problem, glad I could help :)
            – Guido A.
            3 hours ago










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          Suppose that $x$ does not have a neighbourhood contained in $F^c$. Then, for each interval $I_n = (x-frac1n,x+frac1n)$ which is a neighbourhood of $x$, there must exist a point $x_n in I_n cap F$. Otherwise, we would have $I_n subseteq F^c$. In particular, $|x_n - x| < frac1n$ for each $n$, and so $x_n to x$.






          share|cite|improve this answer




















          • Ohhhh perfect, that makes much more sense now; thank you so much!! :)
            – GingerKittyLover
            3 hours ago










          • No problem, glad I could help :)
            – Guido A.
            3 hours ago














          up vote
          3
          down vote













          Suppose that $x$ does not have a neighbourhood contained in $F^c$. Then, for each interval $I_n = (x-frac1n,x+frac1n)$ which is a neighbourhood of $x$, there must exist a point $x_n in I_n cap F$. Otherwise, we would have $I_n subseteq F^c$. In particular, $|x_n - x| < frac1n$ for each $n$, and so $x_n to x$.






          share|cite|improve this answer




















          • Ohhhh perfect, that makes much more sense now; thank you so much!! :)
            – GingerKittyLover
            3 hours ago










          • No problem, glad I could help :)
            – Guido A.
            3 hours ago












          up vote
          3
          down vote










          up vote
          3
          down vote









          Suppose that $x$ does not have a neighbourhood contained in $F^c$. Then, for each interval $I_n = (x-frac1n,x+frac1n)$ which is a neighbourhood of $x$, there must exist a point $x_n in I_n cap F$. Otherwise, we would have $I_n subseteq F^c$. In particular, $|x_n - x| < frac1n$ for each $n$, and so $x_n to x$.






          share|cite|improve this answer












          Suppose that $x$ does not have a neighbourhood contained in $F^c$. Then, for each interval $I_n = (x-frac1n,x+frac1n)$ which is a neighbourhood of $x$, there must exist a point $x_n in I_n cap F$. Otherwise, we would have $I_n subseteq F^c$. In particular, $|x_n - x| < frac1n$ for each $n$, and so $x_n to x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Guido A.

          5,6181728




          5,6181728











          • Ohhhh perfect, that makes much more sense now; thank you so much!! :)
            – GingerKittyLover
            3 hours ago










          • No problem, glad I could help :)
            – Guido A.
            3 hours ago
















          • Ohhhh perfect, that makes much more sense now; thank you so much!! :)
            – GingerKittyLover
            3 hours ago










          • No problem, glad I could help :)
            – Guido A.
            3 hours ago















          Ohhhh perfect, that makes much more sense now; thank you so much!! :)
          – GingerKittyLover
          3 hours ago




          Ohhhh perfect, that makes much more sense now; thank you so much!! :)
          – GingerKittyLover
          3 hours ago












          No problem, glad I could help :)
          – Guido A.
          3 hours ago




          No problem, glad I could help :)
          – Guido A.
          3 hours ago

















           

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