Welcome to our community

Be a part of something great, join today!

Show that the sequence has a decreasing subsequence

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,134
Hi ! :)

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??
 

shen07

Member
Aug 14, 2013
54
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??
I really think this theorem will help you:

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.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hi ! :)

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??
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

Well-known member
MHB Site Helper
Apr 14, 2013
4,134
Ok!Thank you for your help! :)