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 $x_{n} \in (x - \frac{1}{n}, x + \frac{1}{n})$, 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.


Suppose that $x$ does not have a neighbourhood contained in $F^c$. Then, for each interval $I_n = (x-\frac{1}{n},x+\frac{1}{n})$ 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, $|x_n - x| < \frac{1}{n}$ for each $n$, and so $x_n \to x$.

  • $\begingroup$ Ohhhh perfect, that makes much more sense now; thank you so much!! :) $\endgroup$ – GingerKittyLover Oct 20 '18 at 3:26
  • $\begingroup$ No problem, glad I could help :) $\endgroup$ – Guido A. Oct 20 '18 at 3:27

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