Hi,

I had a problem proving this theorem:

Let {sn} be a bounded sequence of real numbers. Assume S = {sn: n= 1, 2, ...} is infinite. Then {sn} is convergent if and only if S has exactly one accumulation point.

But in another page of my notebook, it says:
Accumulation points can be more than one or even infinitely many while the limit is unique.

Are they opposite? Thanks for any help.

2. Originally Posted by truevein
I had a problem proving this theorem:
Let {sn} be a bounded sequence of real numbers. Assume S = {sn: n= 1, 2, ...} is infinite. Then {sn} is convergent if and only if S has exactly one accumulation point. But in another page of my notebook, it says: Accumulation points can be more than one or even infinitely many while the limit is unique. Are they opposite?
No, they are not opposite. A convergent sequence has exactly one accumulation point. A set can have many accumulation points.
The set $[0,1]$ has infinitely many accumulation points.
The sequence $\left(\frac{1}{n}\right),~n\in \mathbb{Z}^+$ has exactly one.

3. Thanks! Then the problem is one being a sequence and other a set. I'll try it again.

4. That doesn't quite resolve the issue. For example, the sequence $a_n=(-1)^n(1-1/n)$ has two accumulation points: 1 and -1.

5. Originally Posted by roninpro
That doesn't quite resolve the issue. For example, the sequence $a_n=(-1)^n(1-1/n)$ has two accumulation points: 1 and -1.
You missed the qualifying adjective, convergent .

6. I was actually responding the post above mine. I thought that the poster did not understand that a sequence can also have multiple accumulation points. Maybe I misunderstood the spirit of the question. Either way, sorry for the confusion.

7. Originally Posted by roninpro
I was actually responding the post above mine.
This why we hope that you will use the "reply with quote" option.