Welcome to our community

Be a part of something great, join today!

[SOLVED] open/closed

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Feb 5, 2012
1,621
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
Mar 1, 2012
275
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
Feb 1, 2012
1,673
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
Feb 5, 2012
1,621
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.