Is F a Complete Ordered Field?

[╔(σ_σ)╝]

Homework Statement

Let F be an ordered field in which every strictly monotone increasing sequence bounded above converges. Prove that F is complete

Homework Equations

Definitions:

Monotone Sequence property:
Let F be an ordered field. We say that F has the monotone sequence property if every monotone increasing sequence bounded above converges.


Completeness Property:
An ordered field is said to be complete if it obeys the monotone sequence property

The Attempt at a Solution

Approach 1
I'm not sure what exactly to prove. The question says that "strictly monotone increasing sequence bounded above converges" which is pretty much the monotone sequence property. And by the completeness property, F is complete . So what exactly am I supposed to do ? It seems trivial.

Approach 2
I could also get any strictly increasing sequence and extract an increasing subsequence which is bounded above and thus converges by monotone sequence property

 

[Dick]

The 'strictly monotone' sequences are a subset of the 'monotone' sequences. I think you have to take a monotone sequence and show it converges because it either contains a strictly monotone subsequence, or it's eventually constant.

 

[╔(σ_σ)╝]

The 'strictly monotone' sequences are a subset of the 'monotone' sequences. I think you have to take a monotone sequence and show it converges because it either contains a strictly monotone subsequence, or it's eventually constant.

I see what you are trying to say but the problem with this approach is that I cannot readily find a way of doing this.

Can I perhaps repeat some terms in the strictly monotone sequence to obtain a "regular" monotone sequence.

Eg
Let $$< a_{n}>$$ be a strictly monotone sequence then we define
$$<a*_{n}>$$
where $$<a*_{n}> = <a_{0},a_{0},a_{1},a_{1}...>$$
where $$a_{0}, ... ,a_{n}$$ are terms of the strictly monotone sequence $$<a_{n}>$$.
$$<a*_{n}>$$ is a "regular" monotone sequence and is increase and bounded above thus, it converges.

Does this work ?
PS I took a look at the back of the book I'm using and it says
"From a( nontrivial) monotone sequence $$<x_{n}>$$, extract a subsequence that is strictly monotone ". WHAT ? . I understand the writers hint.

 

[Dick]

I see what you are trying to do but it's easier if you follow the books hint. Start with a monotone sequence and delete terms to get a strictly monotone sequence. Then show since the strictly monotone sequence converges, so does the original.

 

[╔(σ_σ)╝]

I see what you are trying to do but it's easier if you follow the books hint. Start with a monotone sequence and delete terms to get a strictly monotone sequence. Then show since the strictly monotone sequence converges, so does the original.

That seems good but considering that the chapter on cauchy sequences is further along I don't know any theorem to support my claim. I would have to show that the convergence, of a subsequence of a monotone increasing sequence, is enough to establish convergence of the sequence itself. I'm I over complicating things ?

 

[Dick]

That seems good but considering that the chapter on cauchy sequences is further along I don't know any theorem to support my claim. I would have to show that the convergence, of a subsequence of a monotone increasing sequence, is enough to establish convergence of the sequence itself. I'm I over complicating things ?

Overcomplicating. Just think epsilons and deltas. Use the definition of limit. You hardly need a theorem here. The terms you deleted are between the terms in the strictly monotone sequence.

 

[╔(σ_σ)╝]

Overcomplicating. Just think epsilons and deltas. Use the definition of limit. You hardly need a theorem here. The terms you deleted are between the terms in the strictly monotone sequence.

Perfect, I see it.

Thanks a lot for the help.

 

[Eynstone]

This can also seen to hold true as every Cauchy sequence is bounded & has a monotonic subsequence .