I have a question about cluster points that would like to ask you, this question is one of the exercises in my textbook.
Question: If $A$ is any subset of $R^d$, then $der(der($A$))$ is the set of all cluster points of $der$ ($A$)
(a) Given an example of a subset $A$ of $R$ for which der(der($A$))= empty set and der(A) is not empty
(b) Given an example of a suset $A$ of $R$ for which der(der($A$)) = der($A$)
(c)Given an example of a subset $A$ of $R$ for which empty set does not equal to $der(der($A$))$ does not equal to der($A$)
(d) Show that, if $A$ is a nonempty subset of $R^d$, then der(der($A$)) is a subset of der($A$).
der($A$) is defined as the set of cluster points of $A$, and the definition for cluster point is that for all $r$>0, there exist a punctured ball of radius $r$ such that the intersection of this punctured ball and the set A is not empty.
My ideas:
(a) I am thinking about an interval, for example [-1,0] and we only consider about the rational number in betwee. I am stuck here because I am not sure whether the cluster point for tational number is empty or non-empty.
(b) I am thinking about an interval [-1,1 ]as well, but this time we consider all the real numbers in between. All the numbers within this interval are cluster points and the set of all cluster points of der ($A$) are also all the points of this interval.
(c) I haven't come up anything for this one yet, so any hints would be really appreciated.
(d) for the proof part, I am thinking that maybe i need to start by listing the definitions of cluster points.
Any suggestions and help would be really appreciated.
Thanks a lot
Let me know if the text is not readable.
Fishing
For (a), look for a set $A$ with just one limit point.
Your (b) is correct.
For (c), try to combine (a) and (b).
I can't make sense of (d).