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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.
$1/n = (0,1)$ 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.
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.
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?