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- Apr 14, 2013

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Let [tex] x_{n} [/tex] a sequence of positive numbers.How could I show that it has a decreasing subsequence that converges to 0,knowing that inf{[tex] x_{n} [/tex],n ε N} =0??

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- Thread starter
- #1

- Apr 14, 2013

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Let [tex] x_{n} [/tex] a sequence of positive numbers.How could I show that it has a decreasing subsequence that converges to 0,knowing that inf{[tex] x_{n} [/tex],n ε N} =0??

I really think this theorem will help you:Hi !

Let [tex] x_{n} [/tex] a sequence of positive numbers.How could I show that it has a decreasing sub-sequence that converges to 0,knowing that inf{[tex] x_{n} [/tex],n ε N} =0??

Theorem:

A bounded sequence of [tex]\mathbb{R}[/tex] has a convergent sub sequence.

If a sequence X is bounded,all its sub-sequences will be bounded. Now since every sequence has a monotone sub-sequence (i.e either decreasing or increasing), X will also have a monotone sub-sequence.

Therefore By Monotone Convergence Theorem the sub-sequence being bounded and Monotone will converge.

Your sequence is decreasing, its obvious it will tend to its infimum.

- Feb 13, 2012

- 1,704

If $\displaystyle \text{inf} [x_{n}] = 0$ and for all n is $\varepsilon > 0$ then by definition for a $\displaystyle \varepsilon > 0$ it exists at least an n for which is $\displaystyle x_{n} < \varepsilon$ and that means that for all n it exists at least one m for which is $\displaystyle x_{m} < x_{n}$...

Let [tex] x_{n} [/tex] a sequence of positive numbers.How could I show that it has a decreasing subsequence that converges to 0,knowing that inf{[tex] x_{n} [/tex],n ε N} =0??

Kind regards

$\chi$ $\sigma$

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- #4

- Apr 14, 2013

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Ok!Thank you for your help!