Convergence of Finite Sets: A Limit on Repeated Elements?

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Homework Help Overview

The problem involves a finite subset A of real numbers and a sequence x_n drawn from A. The task is to demonstrate that if the sequence is convergent, it must eventually become constant.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of A being finite and the nature of convergent sequences. Some explore the relationship between the distances of elements in the sequence and the concept of limits. Others propose establishing a bijection to argue about the finiteness of A.

Discussion Status

The discussion is ongoing, with various ideas being presented. Some participants express uncertainty about their understanding, while others offer mathematical insights and approaches to the problem.

Contextual Notes

There is mention of the original poster's feelings of frustration regarding their understanding of the problem, indicating a potential lack of confidence in their mathematical skills. The finite nature of set A is a key constraint in the discussion.

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Homework Statement



Let A be a finite subset of R. For each n in N, let x_n be in A. Show that if the sequence x_n is convergent then it must become a constant sequence after a while.

Homework Equations



The definition of limit.

The Attempt at a Solution



As A is finite, at least one element of A will appear in the sequence more than once after some N. As this sequence is convergent there is an M for any ε such that |x_n - x| < ε for every n with n>M. Let M>N... Help please.

I can't decide whether this problem is too hard or I'm stupid or this is just because I'm a beginner?
 
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If A= {x_n} is finite then the set of distances {|x_i- x_j|} is finite and so has a smallest value. If \epsilon is smaller than that ...
 
Ah okay, thank you
 
Idea: Establish a bijection f: N -> A
n |-> f(n)=x_{n}
If there exists no N: \forall n &gt; N, x_{n} = const, then A must be infinite -> hence we obtain a contradiction. :smile:
 

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