0

Please may you help with the following question:

let E be a non-empty subset of R. Let E' be its derived set(the set of all the limit points of E). How to prove that E' is a closed set.

closed set is the set containing all its limit points. limit points p of a set A are points p in which all deleted neigborhoods intersect A

Thank you

Charlie
  • 71
  • 1
    What definition of closed set are you using? – Dan Rust Sep 14 '13 at 20:17
  • 1
    @StefanH.: For someone working in $\mathbb{R}$, it may be a bit of a leap to understand that the duplicate target is indeed a duplicate? – copper.hat Sep 14 '13 at 20:39
  • Isn't the question more of a duplicate of this one: http://math.stackexchange.com/questions/15766/proving-that-the-set-of-limit-points-of-a-set-is-closed – user10444 Sep 15 '13 at 23:01

1 Answers1

1

Suppose $x\notin E'$. Then $x$ is not a limit point of $E$. This means there exists a neighborhood $U$ of $x$ such that $(E \cap U) \setminus \{x\} = \emptyset$. Let $y \in U$. Then $U\setminus \{x\}$ is an open neighborhood of $y$ that does not intersect $E.$ Hence $y$ is not a limit point. In particular, every point of $U$ is not a limit point. Hence $(E')^c$ is open.

Stefan Hamcke
  • 27,733
copper.hat
  • 172,524