Prove S' Closed in R^n

Date: 10/07/2002 at 20:36:16
From: Sandy
Subject: Prove S' closed in R^n

Prove that S' (the derived set or set of all accumulation points of S)
is closed in R^n.

I think I need to show (S')' is a subset of S'. I started by letting
x be any point in (S')' and then used the definition of accumulation
point to say that any n-ball around x contains points in S' distinct
from x. I can't think of any way to get from there to x having to be
in S'.

Thank you,
Sandy

Date: 11/19/2002 at 18:25:11
From: Doctor Nitrogen
Subject: Re: Prove S' closed in R^n

Hello Sandy:

Let R^n\S' denote the complement of S' in R^n. Let r be any element of
R^n\S'. Then you must show that every n-ball around r is contained 
entirely in R^n\S' 

(Why? Hint: look at S'. What does S' have that R^n\S' doesn't? Put 
another way: "R^n\S' is a set in R^n which does not have all its...?")

Next you can apply the following theorem, which you might have to 
prove, but it is not too hard and can even be found in books on 
general topology:

Theorem: A set is closed if and only if ("iff") its complement is 
open.

From this theorem, IF S' is closed in R^n, then its complement R^n\S' 
in R^n is open. CONVERSELY, IF R^n\S' is open in R^n, then S' is 
closed in R^n.

The converse part is what you must look at. If every n-ball around 
any r in R^n\S' is entirely in R^n\S', it means R^n\S' must be .....?

Next, applying the theorem, since R^n\S' is.....(?) in R^n, 

then S' must be.....(?)in R^n.

- Doctor Nitrogen, The Math Forum
  http://mathforum.org/dr.math/