# Homework Help: Any Cauchy sequence converges.

1. Nov 1, 2012

### peripatein

Hello,

My instructor, whilst trying to prove that liminf of sequence a_n = limsup of sequence a_n = A,
_
wrote that since we know that a_n0-ε<an<a_n0+ε → a_n0-ε ≤ A ≤ A ≤ a_n0+ε.
Why is that true? I mean, how do we know that if a sequence is bounded then its lim inf (i.e. the lowest amongst the limits of its subsequences) and lim sup (the highest amongst the limits of its subsequences; do forgive me for potential misnomers) are indeed within that area bounded by a_n0-ε and a_n0+ε?
I hope you may assit in clarifying this. Thank you!

2. Nov 1, 2012

### peripatein

The inequality did not come out right. The second A (following a_n0-epsilon <= A_) should be upper A (or, if I understand it aright, lim sup.

3. Nov 1, 2012

### haruspex

This is quite hard to read. You'll get much better response if you take the trouble to make it legible, e.g. by writing it out in latex.

4. Nov 1, 2012

### peripatein

Hello,

My instructor, whilst trying to prove that liminf of sequence an = limsup of sequence an = A, stated that since we know that an0-ε<an<an0+ε $\Rightarrow$ an0-ε ≤ A ≤ A-- ≤ an0+ε.

Why is that true? I mean, how do we know that if a sequence is bounded then its lim inf (i.e. the lowest amongst the limits of its subsequences) and lim sup (the highest amongst the limits of its subsequences; do forgive me for potential misnomers) are indeed within that neighbourhood bounded by an0-ε and an0+ε?
I hope you may assit in clarifying this. Thank you!

5. Nov 2, 2012

### peripatein

Following haruspex's advice, the content is now far more legible. May someone please help calrifying this issue? Why is that inequality valid given that the sequence is bounded?

6. Nov 2, 2012

### HallsofIvy

What are your definitions for "limsup" and "liminf" of a sequence?

7. Nov 2, 2012

### peripatein

I have explained that above, in parenthesis.

8. Nov 2, 2012

### HallsofIvy

Thanks. For some reason I didn't notice that before. So if limsup= liminf= A, you know that all subsequential limits are less than or equal to A and larger than or equal to A. Well, a subsequential limit cannot be both "less than A" and "larger than A" can it? So this means that all subsequential limits are equal to A which means that the sequence itself converges to A.