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/
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