Answer to Question #190649 in Real Analysis for Nikhil

Question

Prove that the complement of every closed set is open.

Answer 1

Suppose that S is a closed set. We claim that $S^c$ is a open set. Take any $p \in S^c$ . If there fails to exist an $r> 0$ such that

$d(p,q) <r \implies q \in S^c$

then for each $r = \dfrac{1}{n}$ with $n = 1,2,3....$ there exist a point $p_n \in S$ such that

$d(p,pn) < \dfrac{1}{n}$ . This sequence in S converges to $p \in S^c$ , contrary to closeness of S.

Therefore there actually does not exist an $r>0$ such that

$d(p,q)<r \implies q \in S^c$

which proves that $S^c$ is an open set.