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Which one of these points is accumulation point, which not and why? I read the definition x-times but I'm quite confused :-/ I also found this post which is relevant to my question but it seems to me strange.

The problem is with natural numbers and what is epsilon? Real or natural?

Lets see the problem:

$A=\{\frac{1}{n}; n\in\mathbb{N}\}$

  1. $0.999999...$
  2. $0$
  3. $1$
  4. $10$
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It's 0 because we have $$ \forall \epsilon > 0 \exists N \in \mathbb N \forall n \geq N: \left |0-\frac 1n \right | < \epsilon$$ where $\epsilon \in \mathbb R$. In general $x_0$ is an accumulation point of the set $A$ if there is a sequence $(x_n)_{n=0}^{\infty}$ in $A$ which converges to $x_0$. – Epsilon Oct 21 '12 at 14:35
up vote 0 down vote accepted

No, you are wrong with your guess. Remember that an accumulation point may not be in the set. If $A=\{1,1/2,1/3,1/4,...\}$ then $0$ (though $0\notin A)$ is an accumulation of $A$. In fact we cannot find any other accumulation point of $A$ other than $0$.

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Here is what I would recommend.

Draw a number line. Mark the points $10, 1, .99999, 0$ on your number line. Then mark down the first, say, $20$ elements of the sequence $1/n$.

You'll even get to see visually why it's called an "accumulation point."

As for your other questions:

$\epsilon$ is real, not necessarily a natural number. And it can be arbitrarily small, so long as it is positive.

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I have already did it, but I´m still lost. It's why I asked here. – user1097772 Oct 21 '12 at 13:56
I guess none of them are accummulation points. – user1097772 Oct 21 '12 at 13:58

There are different definitions of what acumulation point for a sequence is. Maybe the most clear for this case is: It's such a point that for any neighborhood that you take with center in this point there are infinitly many points of the sequence inside this neighborhood. Let's denote circle neighborhood like this: $$N(p,\epsilon,R)$$ where $p$ is a point in question, $\epsilon$ is the radius of the circle, and $R$ is the field to which the points belong. Now just test each of you options a),b),c),d), and try to figure out which contain infinitly many points of the sequence, no matter how smole $\epsilon$ you take.

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0 is the accumulation point. By Arhimedean property, for any $\varepsilon>0$, choose $N\in \mathbb{N}$ such that $N>\frac{1}{\varepsilon}$, hence $A$-{0} intersect infinitely many points in $(-\varepsilon,\varepsilon)$

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