# I Completeness Property (and Monotone Sequence)

1. Sep 17, 2016

### KT KIM

I am studying Classical Analysis with Marsden book.
At very first chapter it covers sequence, field, etc...

The book has theorems
1."Let F be an ordered field. We say that the monotone sequence property if every monotone increasing sequence bounded above converges."

2."An ordered field is said to be complete if it obeys the monotone sequence property"

3."There is a unique complete ordered field called the real number system."

so I wonder

for 1. as I know monotone increasing sequence has a definition of
A sequence (an) is monotonic increasing if an+1≥ an for all nN

then what about e.g (an) = n , a1=1 a2=2 a3=3 .... ? this sequence is monotonic increasing but not bounded.

In the same reason, according to 2 and 3. Any sequences in real number system has to obey monotone sequence property and it means it has to be bounded. But I can come up with lots of sequence which has real numbers as it's element that explodes.
like (an) = 2n , or Harmonic sequence.

I am pretty sure I have a critical misconception. but don't know what it is. help me please.

2. Sep 17, 2016

### Staff: Mentor

"every monotone increasing sequence bounded above converges."
Your sequence is not bounded, so this sequence is irrelevant for that statement.

"If it rains, the street gets wet" doesn't make any statement about days where it does not rain.
No. It means "if it is bounded, then it converges".

3. Sep 17, 2016

### KT KIM

I finally found out that actually I misread it!, Haha sorry, English is not my primary language. I did read like "sequence 'bounded' above converges." , have thought converges is just a noun.
Anyway I really appreciate your help.
Thank you

4. Sep 17, 2016

### Staff: Mentor

There is another possible trap around.
"... then it converges" should better be read as
"... then it converges and this implies that the limit exists within F".

Otherwise you could find an example of a sequence which is bounded, but does not converge.
E.g. if F = $\mathbb{Q}$ then $a_n = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+ \dots +\frac{1}{n^2}$ is bounded, e.g. by $2$, but it does not converge, because its limit $\frac{\pi^2}{6}$ does not exist in $\mathbb{Q}$. However, it exists in $\mathbb{R}$. Therefore $\mathbb{R}$ is complete, whereas $\mathbb{Q}$ is not.

5. Sep 18, 2016

### KT KIM

Really good one it is, Thank you.