MHB Show that the sequence has a decreasing subsequence

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A sequence of positive numbers with an infimum of zero will have a decreasing subsequence that converges to zero. The Monotone Convergence Theorem states that a bounded sequence has a convergent subsequence, which implies that any monotone subsequence must also converge. Given that the infimum is zero, for any positive epsilon, there exists an index where the sequence values fall below epsilon. This ensures that a decreasing subsequence can be constructed that approaches the infimum. Thus, the sequence indeed contains a decreasing subsequence converging to zero.
mathmari
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Hi ! :)

Let x_{n} a sequence of positive numbers.How could I show that it has a decreasing subsequence that converges to 0,knowing that inf{x_{n},n ε N} =0??
 
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mathmari said:
Hi ! :)

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

I really think this theorem will help you:

Theorem:
A bounded sequence of \mathbb{R} 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.
 
mathmari said:
Hi ! :)

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

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 $\displaystyle x_{m} < x_{n}$...

Kind regards

$\chi$ $\sigma$
 
Ok!Thank you for your help! :)
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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