Compact SetsDate: 02/04/2001 at 12:39:35 From: Paul Subject: Introduction to Topology - compact sets I've been asked to answer the following: Let A be an infinite set in R^1 (the real numbers) with a single accumulation point in A. Must A be compact? Here's what I have come up with: The first example of such a set that I thought of was as follows: A = {0} U {1/n | n is a natural number} Here, the single accumulation point is {0} and this set is indeed compact (the proof is given as an example of a compact set in the textbook). To show A is compact, I need to show that A is closed and bounded. Since A has only one accumulation point and this accumulation point is contained in A, it follows trivially that A contains all of its accumulation points. This means that A must be closed (we proved in class that a set A is closed iff A contains all of its accumulation points). The answer for this problem in the back of the book says "No." Since I've shown that A must be closed (and I don't think there's anything wrong with my argument there), the problem must be that A doesn't have to be bounded. But I'm having a hard time envisioning a set which is closed but not bounded. It seems to me that closed would imply bounded. Here's my argument: A set A in R^1 is bounded if there exists r > 0 such that A is a subset of [-r,r]. Let [a,b] be a closed set in R^1. Let x = (a+b)/2 and let y = d(a,b) = the distance from a to b. Then [x-y, x+y] contains [a,b] and it follows that [a,b] is bounded. Maybe the problem with this argument is that [a,b] contains an infinite number of accumulation points and so this argument doesn't hold for the type of set I've been asked to consider. But I don't know how to modify my argument for the appropriate kind of set. Date: 02/04/2001 at 20:30:00 From: Doctor Schwa Subject: Re: Introduction to Topology - compact sets Hi Paul, I agree with your argument that it's closed. As for the boundedness, your assumption of "let [a,b] be a closed set" is assuming way too much! If it's not bounded, then certainly the set can't be written in interval notation. Here's what occurs to me: take the set that you had (of 1/n, together with 0) and add to it some ISOLATED points, infinitely many of them, that are not bounded but are still closed. A good example of that would be the integers. Since they are isolated, the integers have no extra accumulation points. Also since they are isolated, they are a closed set. And of course they are not bounded. In fact I think I've solved the problem! - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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