proposition of derivate Let $(X,{\tau})$ a topological space and $A$ a subset of $X$, show that:
$\operatorname{der}A=\{x\in \operatorname{cl}A\mid \operatorname{cl}A=\operatorname{cl}(A-\{x\})\}$
So I try to prove this. I suppose $x\in \operatorname{der}A$ by definition, for all $V\in\mathcal{N}(x)$, $V\cap(A-\{x\})\neq \emptyset$, so $x\in \operatorname{cl}(A-\{x\})$ but I don't know how I can continue. If you give me some hint or suggest, I will be grateful. Thank you.
($\mathcal{N}(x)$ is the neighborhood system of $x$ and $\operatorname{der} A=\{x\in X\mid \forall V\in \mathcal{N}(x),V\cap (A-\{x\})\neq \emptyset\}$).

You also need that $x \in \operatorname{cl}(A)$ iff for all $V \in \mathcal{N}(x)$, $V \cap A \neq \emptyset$, the closure of $A$ is the set of all adherent points of $A$.
So $x \in \operatorname{der}(A)$ means that $x$ is an adherent point of $A - \{x\} $, or $x \in \operatorname{cl}(A-\{x\})\subseteq \operatorname{cl}(A)$ and this is equivalent to
$$\operatorname{cl}(A)= \operatorname{cl}(A-\{x\})$$
