What is the relationship between a closed set and a converging sequence?

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Homework Help Overview

The discussion revolves around the relationship between closed sets and converging sequences in the context of a proof. Participants are examining a specific proof related to a sequence and its convergence properties within a closed set.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are exploring the implications of a sequence converging to a point within a closed set and questioning the necessity of certain conditions stated in the proof. There is a focus on understanding the relationship between the radius of a ball and the behavior of the converging sequence.

Discussion Status

Some participants have provided insights into the proof's logic, particularly regarding the conditions under which the sequence remains within the closed set. There is an ongoing examination of the necessity of specific elements in the proof, indicating a productive exploration of the concepts involved.

Contextual Notes

Participants are grappling with the definitions of closed sets and the properties of converging sequences, particularly in relation to the proof's requirements and assumptions. The discussion reflects a need for clarity on these foundational concepts.

dirk_mec1
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Homework Statement


http://img527.imageshack.us/img527/6049/48193240ao5.png

I don't understand this proof. The first two lines are clear to me: the sequence xn is in F and F is closed so its complement is open so there is a ball with radius r around x in Fc.

But I don't understand the last two lines. Of course there y larger then the radius od the ball but what's the relation with the converging sequence?
 
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If x_n -> x then the sequence x_n is eventually in O.
 
But then you do not need the line that there are y which are greater than the radius and inside F, right?
 
Yes, you do. That's the whole point. Since {xn} converges to x, given any r> 0, there exist N such that if n> N, d(x, xn)< r. Taking r such that Br(x) is in O, it follows that xn, for n> N s in Br(x) so in O. IF, as in the hypothesis, all xn are in F, then some members of O (they "y" they mention) are in F, a contradiction.
 

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