Prove that a set $\displaystyle S \subset \bold{R} $ is closed if and only if it contains all of its limit points.

Suppose that $\displaystyle S $ is closed, and $\displaystyle x_0 $ is a limit point of $\displaystyle S $. Then $\displaystyle S^{C} $ is open so that there is an $\displaystyle \varepsilon > 0 $ such that $\displaystyle (x_{0} - \varepsilon, x_{0} + \varepsilon) \cap S = \emptyset $ which is a contradiction.

If $\displaystyle S $ contains all its limit points, then there exists an $\displaystyle \varepsilon > 0 $ such that $\displaystyle (x_{0}- \varepsilon, x_{0} + \varepsilon) \cap S \neq \emptyset $. So $\displaystyle S^{C}$ is open $\displaystyle \implies S $ is closed.

Is this correct?