Accumulation Points of $(-1)^n + (1/m)$: Is it True?

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Discussion Overview

The discussion revolves around the nature of the set defined by the expression $(-1)^n + (1/m)$ for positive integers $n$ and $m$. Participants explore whether this set can be characterized as open or closed, and they examine its accumulation points.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the accumulation points of the set are $x \in [-1, 1]$, suggesting that the set is open.
  • Others question how to express the set as an interval and whether it can be classified as open or closed.
  • One participant notes that there are only two accumulation points, $-1$ and $1$, and that the terms $(1/m)$ are small positive numbers approaching zero.
  • Another participant argues that the original set contains isolated points, such as $-1/2$, which may imply that the set is not open.
  • A later reply suggests that if the set does not contain its limit points but contains some boundary points, it may be classified as neither open nor closed.

Areas of Agreement / Disagreement

Participants express differing views on whether the original set is open, closed, or neither, with no consensus reached on the classification of the set or its accumulation points.

Contextual Notes

Participants highlight the complexity of determining the nature of the set due to the presence of isolated points and the relationship between limit points and boundary points.

Dustinsfl
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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.
 
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dwsmith said:
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
 
If I can't write this an interval, how can I determine if it is open or closed and the accumulation points?
 
dwsmith said:
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 ...
 
Ackbach said:
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?
 
dwsmith said:
It is open since it doesn't contain its limit points. Correct?

What is "it" in this sentence: the original set, or the complement?
 
Ackbach said:
What is "it" in this sentence: the original set, or the complement?

Original set.
 
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?
 
Ackbach said:
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
 
  • #10
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?
 
  • #11
Ackbach said:
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 won't contain their limit point so they will always be neither?
 

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