Prove that the complement of a closed set is open

[Mr Davis 97]

Homework Statement

Suppose that S is a closed set. We claim that S^c is open. Take any p ∈ S^c. If
there fails to exist an r > 0 such that
d(p, q) < r ⇒ q ∈ S^c
then for each r = 1/n with n = 1, 2, . . . there exists a point p_n ∈ S such that
d(p, pn) < 1/n. This sequence in S converges to p ∈ S^c, contrary to closedness of S.
Therefore there actually does exist an r > 0 such that
d(p, q) < r ⇒ q ∈ S^c
which proves that S^c is an open set.

Homework Equations

The Attempt at a Solution

So my question here is not how to do the prove, because I have that proof right here. I am just confused about the proof. I understand that the argument in general is a proof by contradiction, by assuming that S^c is not open and deriving a contradiction. However, I am confused about the reasoning here:

If there fails to exist an r > 0 such that
d(p, q) < r ⇒ q ∈ S^c
then for each r = 1/n with n = 1, 2, . . . there exists a point p_n ∈ S such that
d(p, p_n) < 1/n. This sequence in S converges to p ∈ S^c.

Can someone explain the logic here? Where does the 1/n comes from? Why does that sequence converge to p?

 

[Orodruin]

You take a point p in the complement where the condition for being an open set does not hold. You then show that there is a sequence in S that converges to that point in S^c, implying that S is not a closed set.

This question is significantly simpler using the general topological definition of a closed set: A closed set is the complement of an open set.