# [SOLVED]open/closed

#### dwsmith

##### Well-known member
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.

#### Sudharaka

##### Well-known member
MHB Math Helper
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.
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.

#### Amer

##### Active member
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 true

#### dwsmith

##### Well-known member
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.
As with the complex example, why aren't the points of $1/n$ accumulation points as well?

#### Sudharaka

##### Well-known member
MHB Math Helper
As with the complex example, why aren't the points of $1/n$ accumulation points as well?
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?

Kind Regards,
Sudharaka.