Exercise 15.  Regarding the relation between closed sets and accumulation points,  

15 (a).  Prove that if a set is closed, then it contains all its accumulations points.  

Solution 15 (a).

See text and/or instructor's solution manual.

We prove the contrapositive.  

Suppose    is accumulation point of  S,  but that    does not belong to  S.  

By definition of an accumulation point, every deleted neighborhood,  ,  contains at least one point of  S.  

Therefore, every (non-deleted) neighborhood    also contains at least one point of  S  and at least one point not in  S  (namely, ).  

This condition implies that , which does not belong to  S,  is a boundary point of  S.  (Show the details for this last assertion).  

Thus, the set  S  is not closed.

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell