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 $\bar{A}$?

  • $\begingroup$ Sometimes the definition of limit point requires that it's not in $A$ - I think somebody once gave me a sensible reason to make this part of the definition, but I've forgotten it now. That would make $A'=\bar{A}\setminus A$, but you should check the definition from your class. $\endgroup$ – mdp Apr 22 '13 at 16:27

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:

  1. isolated points of $A$;
  2. points of $A$ that are limit points of $A$; and
  3. 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 $\operatorname{cl}A=A\cup L$.

| cite | improve this answer | |
  • $\begingroup$ Perfect answer :) Just the explanation I was looking for. Thank you so much! $\endgroup$ – Eric Apr 23 '13 at 12:31
  • $\begingroup$ @Eric: You’re very welcome. $\endgroup$ – Brian M. Scott Apr 23 '13 at 12:32

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 $\{0\}$.

| cite | improve this answer | |
  • $\begingroup$ Thanks. This does help. I will wait a while to seem what other answers are given. :) $\endgroup$ – Eric Apr 22 '13 at 18:11
  • $\begingroup$ You can produce fancier examples. For example, in addition to the $\frac{1}{n}$ of the second example, add for each $n$ a sequence that approaches $\frac{1}{n}$, in the same way that the $\frac{1}{n}$ approach $1$. Call the resulting set $B$. Then $B'=A$ and B''=\{0\}$. Cantor dealt with much fancier versions. He worked with them because he needed them in his investigations of trigonometric series. This was before his Set Theory, and led directly to ordinals. $\endgroup$ – André Nicolas Apr 22 '13 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.