Help one question about analysis

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Discussion Overview

The discussion revolves around the compactness of the set A={1/n : n in N} within the standard topology of the real numbers. Participants are tasked with providing proofs for the non-compactness of the set and exploring different families of open sets that cover A.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose a family of open sets {U_n = (1/n, infinity) | n in N} to show that A is not compact, noting that any finite subcollection will miss points in A.
  • Another participant suggests using the open sets defined by neighborhoods around each point 1/n with radius 1/(n(n+1)), arguing that no finite subset will cover all points in A.
  • For part B, one participant mentions that the set {(-infinity, infinity)} covers A and has a finite subcover, while another suggests using the set {(0, 2)} as a single set that covers A.
  • There is a correction regarding the interpretation of compactness, with one participant emphasizing that A is still not compact despite some arguments suggesting otherwise.
  • A later reply introduces the idea that adding the point 0 to the set A would make it compact, indicating a shift in the discussion towards conditions for compactness.

Areas of Agreement / Disagreement

Participants generally agree that the set A is not compact, but there are differing views on the implications of certain open covers and the conditions under which compactness can be achieved.

Contextual Notes

Some arguments depend on the definitions of open covers and compactness, and there are unresolved aspects regarding the implications of different families of open sets.

ShengyaoLiang
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The set A={1/n : n in N} is not compact.
A) prove this by explicitly finding a family of open sets which covers A but has no finite subfamily whcih also covers A.
B) Find another family of open sets which covers A and does have a finite subfamily which cobers A.
 
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i have no idea on how it works...
i am learning by myself right now...so could some one help me out?...thank you very much...
 
I assume you are working in the standard topology for R. Then just take the family of open sets defined by

{U_n = (1/n,infinity) | n in N}.

This does not have a finite subcover. To see this notice that U_n+1 contains U_n for all n. Then any finite subcollection of these sets will have a maximal lement U_M where M is in N. Clear the point 1/(M+1) is not in any of the sets in the finite subcollection.

B. Take the collection {(-infinty, infinity)}.

Try to extend this argument for the set {(x,y) | |x| + |y| < 1} and to be able to say any open set in R or R^2 is not compact.
 
Here's another example:
The distance from 1/n to 1/(n+1) is 1/(n(n+1)) (the distance from 1/n to 1/(n-1) is larger). Take as your open sets each Un the neighborhood of 1/n with radius 1/(n(n+1)). Each of those contains exactly one point in {1/n} and so no finite subset will cover all of {1/n}.

For B, just take { (0, 2)}. That is a single set that covers all of {1/n} by itself. Of course, that tells us nothing about the compactness. {1/n} is "still" compact because not every open cover has a finite subcover.
 
is still NOT compact is what halls of ivy means.
 
Oops, thanks! My fingers don't always connect to my brain.
 
here is a clue. if you throw in 0, it becomes compact.
 

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