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Compact Sets

Date: 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 

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 

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   
Associated Topics:
College Analysis

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