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- Thread starter dwsmith
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- Feb 5, 2012

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Hi dwsmith,All numbers of the form $1/n$, $(n = 1,2,3,\ldots)$.

$1/n = (0,1)$

The accumulation points are $x\in [0,1]$.

This set is open.

So the set under consideration is, \(S=\{\frac{1}{n}:n\in\mathbb{Z}^+\}\). The definition of a limit point (accumulation point) taking the set of real numbers as the reference space is given >>here<<. If you go by that definition you can prove that the only accumulation point of \(S\) is zero.

Kind Regards,

Sudharaka.

$1/n = (0,1)$ this is not trueAll numbers of the form $1/n$, $(n = 1,2,3,\ldots)$.

$1/n = (0,1)$

The accumulation points are $x\in [0,1]$.

This set is open.

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As with the complex example, why aren't the points of $1/n$ accumulation points as well?Hi dwsmith,

So the set under consideration is, \(S=\{\frac{1}{n}:n\in\mathbb{Z}^+\}\). The definition of a limit point (accumulation point) taking the set of real numbers as the reference space is given >>here<<. If you go by that definition you can prove that the only accumulation point of \(S\) is zero.

Kind Regards,

Sudharaka.

- Feb 5, 2012

- 1,621

Let me try to visualize things for you. The neighboring points of \(\frac{1}{n}\) are \(\frac{1}{n+1}\) and \(\frac{1}{n-1}\). You can take a neibourhood around \(\frac{1}{n}\) such that both of these points are not included. But to be a limit point every neighborhood of \(\frac{1}{n}\) should contain at least one point (in \(S\)) distinct from \(\frac{1}{n}\), which is not the case here. Does this clarify things for you?As with the complex example, why aren't the points of $1/n$ accumulation points as well?

Kind Regards,

Sudharaka.