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.


  • $\begingroup$ What does der($A$) mean? $\endgroup$ – TonyK Feb 6 '16 at 20:42
  • $\begingroup$ it means the cluster points of subset A. $\endgroup$ – user311290 Feb 6 '16 at 20:43
  • $\begingroup$ Your "refers to" is wrong here; I have edited it to make sense. (Also, your "cluster point" is usually called a "limit point".) $\endgroup$ – TonyK Feb 6 '16 at 20:49
  • $\begingroup$ Now that's sorted out, I looked at the question, and it's obvious that (a) and (d) can't both be right! $\endgroup$ – TonyK Feb 6 '16 at 20:53
  • $\begingroup$ Is cluster point the same as limit point? They refer to two different things according to my textbook $\endgroup$ – user311290 Feb 6 '16 at 20:57

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).

  • $\begingroup$ (d) makes sense, it just happens to be false. $\endgroup$ – zhw. Feb 6 '16 at 21:06
  • $\begingroup$ I had a typo for (d), not der(der(A)) = der (A), it is der(der(A)) is a subset of der (A) $\endgroup$ – user311290 Feb 6 '16 at 21:09

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