Closed Sets A & B: Does A+B Follow Suit?

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Homework Help Overview

The discussion revolves around the properties of closed sets in the context of real numbers, specifically examining whether the sum of two closed sets, A and B, is also closed. The original poster presents a scenario involving closed sets and their limits.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of closed sets and their limits, questioning whether the sum A+B retains the property of being closed. Various examples and counterexamples are discussed, including specific sequences and their closures.

Discussion Status

The discussion is active, with participants sharing insights and examples that challenge the initial assumption. Some guidance has been offered regarding bounded sets and compactness, but no consensus has been reached on the general case.

Contextual Notes

Participants are considering specific examples of closed sets, including bounded and unbounded cases, and the implications of convergence in sequences related to these sets. There is an ongoing exploration of definitions and properties without a definitive resolution.

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Homework Statement


Given A and B are closed sets in R does it follow that A+B is closed? (A+B={a+b|a in A and b in B})


Homework Equations


A set X is closed iff all of its limiting points are in X.


The Attempt at a Solution


I don't think this is true. I've tried constructing convergent sequences A and B and having the limit of the sum not being contained in A+B. But then A and B can't be closed.
 
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Ok, let A=N where N={1,2,3...}. Let B={-n+1/(n+1)} for n in N. I claim 0 is in the closure of A+B. But is not in A+B. Can you prove me wrong? Whew, that took a while.
 
Gah. That's nice and clean. Thanks.
 
No problem. Though it did hurt. I figure if you are having problems, it's not going to be easy for me, either.
 
Incidentally, if A and B are both bounded, A+B is closed. I'm not sure about the case when only one of them is bounded though.
 
Well, if the C sequence is convergent and A is bounded then the A sequence has a convergent subsequence (compact). Doesn't that imply the corresponding B subsequence is convergent and seal everyone's fate?
 
Not bad. My idea for the case where both were compact was to send AxB under +:RxR->R, the product and images of compact sets being compact.
 

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