# Understanding a Proof in Topology of the Reals

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?

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 $$x_n \to x$$.