# Real analysis and accumulation points of a series

• gottfried
In summary, the conversation discusses the definition of an accumulation point in real analysis and a theorem stating that a bounded real sequence has an accumulation point. The confusion arises when considering how an infinite sequence can have accumulation points within a certain distance from it. However, the example of a bounded sequence with accumulation points (1, 2, 3) is given to illustrate this concept. Another example is given of a more complex sequence with accumulation points not necessarily being elements of the sequence. The definition of an accumulation point does not require it to be an element of the sequence.

#### gottfried

I was reading a textbook 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.

gottfried said:
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This confused me because if a sequence is bounded surely it is finite...
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This statement is not true:

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

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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?

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

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

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?

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

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

voko said:
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?

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.

gottfried said:
Just as an aside does your accumulation point have to be an element of sequence?

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

Thanks for the help