# Real analysis and accumulation points of a series

1. Aug 19, 2012

### gottfried

I was reading a text book on real analysis and came across this definition:Given a real sequence we say x is an accumulation point if given any $\in$ greater than 0 we can find infinitely many natural numbers n such that |xn-x| is less than $\in$.

I also found a theorem that stated if a real sequence is bounded by a and b then it has an accumulation point c between a and b.

This confused me because if a sequence is bounded surely it is finite or at least could be finite in which case how does one find a value x such that there are INFINITELY many numbers in the sequence within $\in$ distance from it.

If somebody could explain conceptually how this is possible I would appreciate it.

2. Aug 19, 2012

### estro

This statement is not true:

a_n=1/n is clearly bounded however this sequence is obviously infinite...

Last edited: Aug 19, 2012
3. Aug 19, 2012

### voko

Consider the sequence 1, 2, 3, 1, 2, 3, 1, 2, 3, etc. It is obviously bounded and it is obviously infinite. Can you see the accumulation points?

4. Aug 19, 2012

### gottfried

I'm not actually completely sure would any number arbitrarily near (dependant on epsilon)to 1,2 or 3 be an accumulation point.

5. Aug 19, 2012

### voko

The sequence does not contain any numbers "arbitrarily near to 1, 2 or 3" - it contains only 1, 2, and 3, infinitely many of them, and they are accumulation points because they satisfy the condition (1 - 1 = 0, which is less than any epsilon).

Consider a more complex sequence:

1 - 1/2, 2 - 1/2, 3 - 1/2, 1 - 1/3, 2 - 1/3, 3 - 1/3, 1 - 1/4, 2 - 1/4, 3 - 1/4, etc.

What can you say about it? Is it bounded? Accumulation points?

6. Aug 19, 2012

### LCKurtz

Remember $|x_n-x|=0$ counts as being less than $\epsilon$.

7. Aug 19, 2012

### gottfried

This sequence looks to be bounded below by 1/2 and above by 3. Again the three accumulation points are 1,2,3 if I correct. Just as an aside does your accumulation point have to be an element of sequence?

Thanks for the help.

8. Aug 19, 2012

### voko

The first sequence has them as elements, the second does not. The definition does not require that.

9. Aug 19, 2012

### gottfried

Thanks for the help