I need help with this question cause I don't know where to start off:
Prove that if a set contains each of its accumulation points, then it must be a closed set.
Maybe I am overlooking something here, but I thought that in a topological space $\displaystyle X$ A set $\displaystyle E$ is closed iff $\displaystyle E=\bar{E}$ and $\displaystyle \bar{E}=E\cup D\left(E\right)$, where $\displaystyle D\left(E\right)$ is the derived set (set of all limit points of $\displaystyle E$, and so $\displaystyle D\left(E\right)\subseteq E\implies E=E\cup D\left(E\right)=\bar{E}$ and so $\displaystyle E$ is closed?
I am relatively new to general topology.
All of that is very true. But that is not the question.
The question is: Prove that the derived set of a set is closed. And that is not true.
Think of the space of real numbers, $\displaystyle \mathbb{R}$ with a topology generated by the collection $\displaystyle \left\{ {( - \infty ,a):a \in \mathbb{R}} \right\}$.
The derived set of $\displaystyle \{0\}$ is the set $\displaystyle (0,\infty)$ which is not closed because $\displaystyle 0$ is limit point of that set.