Topology: Finding the set of limit points

[Damascus Road]

Determine the set of limit points of:

$$A = { \frac{1}{m} + \frac{1}{n} \in R | m,n \in Z_{+} }$$


I can see that everything less than one can't be reached by this set.
Is my set of limit points (0,1) ?

 

[radou]

If x is in <0, 1>, can x be a limit point of this set? What's the definition of a limit point?

 

[SammyS]

Determine the set of limit points of:

$$A = { \frac{1}{m} + \frac{1}{n} \in R | m,n \in Z_{+} }$$

I can see that everything less than one can't be reached by this set.
Is my set of limit points (0,1) ?

Perhaps you meant to say: Not everything less than one can be reached by this set.

Not sure what you mean by "reach".

From Wolfram Math World:

Limit Point: A number x is a limit point of a set if for all ε>0, there exists a member y of the set different from x such that |y-x|<ε.​

The rational number 3/4 ∈ A, (n=2, m=4), but is 3/4 a limit point of A? Why not?

on the other hand 1/2 is a limit point of A.

 

[Robert1986]

Determine the set of limit points of:

$$A = { \frac{1}{m} + \frac{1}{n} \in R | m,n \in Z_{+} }$$


I can see that everything less than one can't be reached by this set.
Is my set of limit points (0,1) ?

Hmm, I think you mean that everything more than one can't be reached by this set. For example, 3/2 is not a limit point of of this set, though it is in the set (take m = 1, n = 2). If that is what you meant, then yes, no points above 1 are limit points of this set.


However, the interval (0,1) is not the answer, either. For example, take any x in (0,1) that is not in A. Then there will always be a neighborhood of x that contains no points of A. Now, let x be a fixed 1/n, in any neighborhood of 1/n, there is always a point of A (different from 1/n); think about why this is. So, right now, I am saying the the points 1/n are all limit points of A. But there is one more than that. Do you know what it is?

 

[Damascus Road]

Hm, ok. I can see how I was thinking about this the wrong way.

It seems as though the limit points may be of the form 1/v, where v is a positive integer..

 

[Robert1986]

Yes, points of the form 1/v are limit points, but that is no all of them. There is exactly one more.

 

[Damascus Road]

and the empty set!

 

[Robert1986]

Nope. It is a number.

 

[Damascus Road]

Wait, what?
We can't approximate the empty set with that set... So it's a limit point, isn't it?

 

[Robert1986]

Wait, what?
We can't approximate the empty set with that set... So it's a limit point, isn't it?

A limit point, p, of a set, S, is a point such that every neighborhood of that p contains a point of S different from p. The empty set is not a point. To get the missing limit point, go along with your 1/v idea and think about a REALLY big v.

 

[Damascus Road]

1/ really big v = zero.. ?

 

[Robert1986]

yep, 0 is a limit point of A.

 

[Damascus Road]

lol, doh. My brain was thinking zero, and my fingers typed empty set.

I'm going to go teach them a lesson, thanks very much!
I'll probably end up posting with other questions from this assignment :P

 

[SammyS]

I'm pretty sure that 0 is NOT a limit point of A because 0 ∉ A.

 

[Robert1986]

I'm pretty sure that 0 is NOT a limit point of A because 0 ∉ A.

A limit point does not have to be in the set of which it is a limit point.

For example, consider the set A={1/n:n\in N} then 0 is a limit point because every neighborhood of 0 has a point from A. Thus, 0 is a limit point.

 

[SammyS]

I stand corrected. DUH!