Intuition behind the difference between derived sets and closed sets? I missed the lecture from my Analysis class where my professor talked about derived sets. Furthermore, nothing about derived sets is in my textbook. Upon looking in many topology textbooks, few even have the term "derived set" in their index and many books only say "$A'$ is the set of limit points of $A$". But I am really confused on the difference between $A'$ and $\bar{A}$. For example,
Let $A=\{(x,y)∈ \mathbb{R}^{2} ∣x^2+y^2<1\}$, certainly $(1,0) \in A'$, but shouldn't $(0,0)∈A'$ too? In this way it would seem that $A \subseteq A'$.
The definition of a limit point is a point $x$ such that every neighborhood of $x$ contains a point of $A$ other than $x$ itself. Then wouldn't $(0,0)$ fit this criterion? If I am wrong, why? And if I am not, can someone please give me some more intuitive examples that clearly illustrate, the subtle difference between $A'$, and $\bar{A}$?
 A: The key to the difference is the notion of an isolated point. If $X$ is a space, $A\subseteq X$, and $x\in A$, $x$ is an isolated point of $A$ if there is an open set $U$ such that $U\cap A=\{x\}$. If $X$ is a metric space with metric $d$, this is equivalent to saying that there is an $\epsilon>0$ such that $B(x,\epsilon)\cap A=\{x\}$, where $B(x,\epsilon)$ is the open ball of radius $\epsilon$ centred at $x$. It’s not hard to see that $x$ is an isolated point of $A$ if and only if $x\in A$ and $x$ is not a limit point of $A$. This means that in principle $\operatorname{cl}A$ contains three kinds of points:


*

*isolated points of $A$;  

*points of $A$ that are limit points of $A$; and  

*points of $X\setminus A$ that are limit points of $A$.


If $A_I$ is the set of isolated points of $A$, $A_L$ is the set of limit points of $A$ that are in $A$, and $L$ is the set of limit points of $A$ that are not in $A$, then 


*

*$A=A_I\cup A_L$;  

*$A'=A_L\cup L$; and  

*$\operatorname{cl}A=A_I\cup A_L\cup L$.


In particular, if $A$ has no isolated points, so that $A_I=\varnothing$, then $A'=\operatorname{cl}A$. If, on the other hand, $A$ consists entirely of isolated points, like the sets $\Bbb Z$ and $\left\{\frac1n:n\in\Bbb Z^+\right\}$ in $\Bbb R$, then $A_L=\varnothing$, $A'=L$, and $\operatorname{cl}A=A\cup L$.
A: In the example you give, the derived set is indeed the same as the closure. However, that need not always be the case. 
Let's start with a silly example. Consider the set $A=\{17\}$. This is a closed set, so $\overline{A}=A$. But $A$ has no limit points, so $A'=\emptyset$.
Here is a slightly more complicated example. Let $A$ be the set of all numbers of the shape $\frac{1}{n}$, where $n$ ranges over the positive integers. Then the closure of $A$ is just $A$ together with the number $0$. But the derived set of $A$ is just $\{0\}$.
