MHB Clovis's question at Yahoo Answers (bounded sequence)

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A bounded sequence can exist without a maximum or minimum element, exemplified by the sequence a_n = (-1)^n n/(n+1), which remains within the range of -1 to 1. Such a sequence will have an infimum and supremum but will not converge due to the presence of infinitely many elements near both bounds. Since the infimum and supremum must differ for the sequence to lack maximum and minimum elements, it indicates that the sequence cannot stabilize at a single limit. Therefore, a bounded sequence without a maximum or minimum does not converge. This highlights the relationship between boundedness and convergence in real analysis.
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Here is the question:

Construct a bounded sequence that has no maximum and no minimum element. Could a sequence like this converge? Why or why not?

Here is a link to the question:

Real Analysis Sequences 10pts? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Clovis,

Consider the sequence $a_n=\dfrac{(-1)^nn}{n+1}$. Easily proved, $-1\leq a_n\leq 1$ for all $n$ and the sequence has no maximum and no minimum element.

Now, suppose $x_n$ is bounded sequence, then it has a infimum $m$ and a supremum $M$. If $x_n$ has no maximum and minimum, it has infinitely many elements close both infimum and supremum, so we can construct two subsequences $x_{n_k}\to m$ and $x_{n_r}\to M$. But $m\neq M$ (otherwise, $x_n$ would be a constant sequence, i.e. with maximum and minimum). This means that $x_n$ does not converge.
 
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