Show that the sequence has a decreasing subsequence

[mathmari]

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??

 

[shen07]

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.

 

[chisigma]

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

$\chi$ $\sigma$

 

[mathmari]

Ok!Thank you for your help! :)

 

[blue_raver22]

To show that the sequence has a decreasing subsequence, we can use the fact that the infimum of a set is the greatest lower bound. This means that for any positive number, there exists a term in the sequence that is smaller than that number.

Since the infimum of the sequence is 0, we know that for any positive number, there exists a term in the sequence that is smaller than that number. This means that we can choose a term x_{n} such that x_{n} < 1, and then choose another term x_{m} such that x_{m} < x_{n}. We can continue this process, choosing terms x_{k} such that x_{k} < x_{k-1} until we have a subsequence that is strictly decreasing.

Furthermore, since the infimum of the sequence is 0, we know that this subsequence will converge to 0 as well. This is because for any positive number \epsilon, we can choose a term in the subsequence that is smaller than \epsilon. As we continue to choose smaller and smaller terms, the subsequence will approach 0 as its limit.

Therefore, we have shown that the sequence has a decreasing subsequence that converges to 0, as desired. This demonstrates the decreasing nature of the subsequence and its convergence to 0.