Welcome to our community

Be a part of something great, join today!

open/closed (2)

dwsmith

Well-known member
Feb 1, 2012
1,673
All numbers of the form $(-1)^n + (1/m)$, $n,m\in\mathbb{Z}^+$.

Is this true $(-1)^n + (1/m) = (-1,1)$? If so, the accumulation points are $x\in [-1,1]$ and the set is open.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
All numbers of the form $(-1)^n + (1/m)$, $n,m\in\mathbb{Z}^+$.

Is this true $(-1)^n + (1/m) = (-1,1)$? If so, the accumulation points are $x\in [-1,1]$ and the set is open.
$3/4 \in (-1,1)$, but for what $n, m \in \mathbb{Z}^+$ is $3/4=(-1)^n+(1/m)$ ?

Moreover for $n$ even for what $m$ is $(-1)^n+(1/m) \in (-1,1)$ ?

CB
 

dwsmith

Well-known member
Feb 1, 2012
1,673
If I can't write this an interval, how can I determine if it is open or closed and the accumulation points?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
If I can't write this an interval, how can I determine if it is open or closed and the accumulation points?
There can only be two accumulation points. The $(-1)^{n}\in\{-1,1\}$, and the $(1/m)$ are small positive numbers getting ever smaller.

As for open or closed, you can consider either the set itself or its complement. I would advise looking at the complement, and thinking about a union you could use to write the entire complement. Recall that the union of an arbitrary number (including infinite) of open sets is open. Can you write the complement of your set as an open set? If so, then the set itself must be ...
 

dwsmith

Well-known member
Feb 1, 2012
1,673
There can only be two accumulation points. The $(-1)^{n}\in\{-1,1\}$, and the $(1/m)$ are small positive numbers getting ever smaller.

As for open or closed, you can consider either the set itself or its complement. I would advise looking at the complement, and thinking about a union you could use to write the entire complement. Recall that the union of an arbitrary number (including infinite) of open sets is open. Can you write the complement of your set as an open set? If so, then the set itself must be ...
It is open since it doesn't contain its limit points. Correct?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
It is open since it doesn't contain its limit points. Correct?
What is "it" in this sentence: the original set, or the complement?
 

dwsmith

Well-known member
Feb 1, 2012
1,673

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Hmm. Well, I'm not sure I agree that the original set is open. Here's an isolated point in the original set: $n=1, m=2$ yields the point $-1/2$. An isolated point is a closed set, because it is a boundary point. So the original set contains some of its boundary points. The question is: does it contain all of its boundary points? Or does it contain all of its limit points?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hmm. Well, I'm not sure I agree that the original set is open. Here's an isolated point in the original set: $n=1, m=2$ yields the point $-1/2$. An isolated point is a closed set, because it is a boundary point. So the original set contains some of its boundary points. The question is: does it contain all of its boundary points? Or does it contain all of its limit points?
It doesn't contain -1 or 1
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
So if it doesn't contain its limit points, but it does contain some of its boundary points, then is the set open, closed, or neither?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
So if it doesn't contain its limit points, but it does contain some of its boundary points, then is the set open, closed, or neither?
Neither. So sets that are of the form (1/n) will always contain some boundary points but wont contain their limit point so they will always be neither?