Topology of the Set {1/n} for Positive Integers Z

In summary, the set {1/n : n is in the positive integers Z} has an empty interior because every ball drawn around an interior point would contain points that are not from the integers. The boundary would be all of the integers Z of the form 1/n, and the closure is the union of the set and the boundary. This means that the closure is the set itself, except for the point 0, which is a limit point of the set. Therefore, the closure of the set is not the set itself and the set is not a closed set.
  • #1
trap101
342
0
Determine the interior, boujdary, and closure for the set:

{ 1/n : n is in the positive integers Z}.

Attempt: two things bothering me.

1) if i am in the set of positive integers, how does 1/n even exist?

2) now let's say it does exist, then the inteior would be empty because every ball drawn around an interior point would contain points that are not from the integers i.e. rationals and reals.

Moving to the boundary, the boundary would contain all of the points from the positive integers Z but again how would this be the case given that the function is 1/n?

The closure is defined as the union of the set and the boundary, the set is empty, but the boundary is Z positive, so it would be Z positive?
 
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  • #2
trap101 said:
Determine the interior, boujdary, and closure for the set:

{ 1/n : n is in the positive integers Z}.

Attempt: two things bothering me.

1) if i am in the set of positive integers, how does 1/n even exist?

2) now let's say it does exist, then the inteior would be empty because every ball drawn around an interior point would contain points that are not from the integers i.e. rationals and reals.

Moving to the boundary, the boundary would contain all of the points from the positive integers Z but again how would this be the case given that the function is 1/n?

The closure is defined as the union of the set and the boundary, the set is empty, but the boundary is Z positive, so it would be Z positive?

Your set doesn't consist of integers, n is an integer. The set is {1/n}={1,1/2,1/3,1/4,...}. It only contains one integer. Now try and think about those questions again.
 
  • #3
I had a feeling that's what it was meaning. If that's tge case,

Interior is (0,1) , boundary is the points 0 and 1, and tge closure is the union of those 2.

Then why would my instructor say that the set itself is the set of integers?
 
  • #4
trap101 said:
I had a feeling that's what it was meaning. If that's tge case,

Interior is (0,1) , boundary is the points 0 and 1, and tge closure is the union of those 2.

Then why would my instructor say that the set itself is the set of integers?

You probably understood the instructor, the given problem doesn't say that at all. And your answers are way off. Why would you think the interior is (0,1)?? The interior has to be a subset of {1/n}. Is (0,1)?
 
  • #5
Oh, well now i have a feeling i get what the instructor was getting at. The interior would be empty because between every integer n, there will be a rational number or an irrational number in which case any ball around a
1 /n would have a rational or irrational in it.

The boundry would be all of the integers Z of the form 1/n.

Closure: is the union of the two above, so all the integers Z of the form 1/n
 
  • #6
trap101 said:
Oh, well now i have a feeling i get what the instructor was getting at. The interior would be empty because between every integer n, there will be a rational number or an irrational number in which case any ball around a
1 /n would have a rational or irrational in it.

The boundry would be all of the integers Z of the form 1/n.

Closure: is the union of the two above, so all the integers Z of the form 1/n

Yes, the interior is empty. If you take any point 1/n then a ball around it contains other points that don't have the form 1/(some integer). The rest of what you are saying is nonsense. Did you draw a sketch of the set of points {1/n}? Does it have any limit points? What's your definition of 'boundary'?
 
  • #7
Hey Dick, I have a question about this as well.

Would it be true that the set is its own boundary? That is, that each element of the set {1/n : n is a positive integer} would be a boundary point? If that is true is the set not already a closed set, and therefore the closure of the set is the set itself?

I'm not sure what I'm talking about as I have very little background (ie. no background) in topology.
 
  • #8
Tsunoyukami said:
Hey Dick, I have a question about this as well.

Would it be true that the set is its own boundary? That is, that each element of the set {1/n : n is a positive integer} would be a boundary point? If that is true is the set not already a closed set, and therefore the closure of the set is the set itself?

I'm not sure what I'm talking about as I have very little background (ie. no background) in topology.

The set is not a closed set. There's point in the closure of the set that's not in the set. Just one. Where is it?
 
  • #9
Would it be 0?

EDIT: My reasoning for this is that as you increase n you get increasingly close to 0 but never actually reach 0. I feel like this is "intuitively" correct, but am unsure how I would justify this "rigorously".
 
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  • #10
Tsunoyukami said:
Would it be 0?

It would. Explain why? To make it rigorous use the definition of closure.
 
  • #11
I believe the definition of closure is as follows.

The closure of a set S is the union of S with its own boundary, or:

[itex]\overline{S} = S \cup \delta S[/itex]


So we know that S contains all elements of [itex]\delta[/itex] S with the exception of the point 0, which is contained within the boundary because the value 1/n converges to this point as n approaches infinity? So the difference between 1/∞ and 0 is simply that 1/∞ is contained within the set but is not actually a boundary point.

I feel as if I'm just repeating myself in a convoluted manner now...
 
  • #12
Tsunoyukami said:
I believe the definition of closure is as follows.

The closure of a set S is the union of S with its own boundary, or:

[itex]\overline{S} = S \cup \delta S[/itex]


So we know that S contains all elements of [itex]\delta[/itex] S with the exception of the point 0, which is contained within the boundary because the value 1/n converges to this point as n approaches infinity? So the difference between 1/∞ and 0 is simply that 1/∞ is contained within the set but is not actually a boundary point.

I feel as if I'm just repeating myself in a convoluted manner now...

That is getting convoluted. My definition says that a point is a boundary point if every neighborhood of that point contains points in S and outside of S.
 
  • #13
At the moment we haven't discussed limit points, but my definitiin of a boundary point is any ball drawn around a boundary point contains elements from the set and elements from the complement of the set, i.e. the ball around our boundary point that intersects the set is non empty and the same ball intersects with the complement is nonempty.

But with that being the case, every ball around the point of the form 1/n from Z would obviously include points not in our set. Particularly all points not of 1/n not in Z, but as well wouldn't the point 1/n be in the set?
Hence why I say that the boundary points are the set of 1/n where n = Z.
 
  • #14
trap101 said:
At the moment we haven't discussed limit points, but my definitiin of a boundary point is any ball drawn around a boundary point contains elements from the set and elements from the complement of the set, i.e. the ball around our boundary point that intersects the set is non empty and the same ball intersects with the complement is nonempty.

But with that being the case, every ball around the point of the form 1/n from Z would obviously include points not in our set. Particularly all points not of 1/n not in Z, but as well wouldn't the point 1/n be in the set?
Hence why I say that the boundary points are the set of 1/n where n = Z.

Yes, all of the points 1/n are boundary points. But there's also a boundary point that is not in your set. BTW don't say 'integers of the form 1/n'. They aren't integers. n is an integer, 1/n isn't.
 
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  • #15
I just realized you had semt a reply to @Tsuno, so if my conclusion is correct for the boundary and moving to the closure, are we able to have the closure as being only one point, in this case 0? I always thought that the closure had to be a the collection of both the points in the set and the boundary. Hence why i said all of the boundary is the closure.
 
  • #16
trap101 said:
I just realized you had semt a reply to @Tsuno, so if my conclusion is correct for the boundary and moving to the closure, are we able to have the closure as being only one point, in this case 0? I always thought that the closure had to be a the collection of both the points in the set and the boundary. Hence why i said all of the boundary is the closure.

You are not being very clear here. Is 0 a boundary point or not?
 
  • #17
Im tempted to say 0 is a boundary point because a boundary point doesn't have to be in the set. As such any ball drawn around the point 0 would have a point in the set amd a point in the complement of the set.
 
  • #18
trap101 said:
Im tempted to say 0 is a boundary point because a boundary point doesn't have to be in the set. As such any ball drawn around the point 0 would have a point in the set amd a point in the complement of the set.

Yes, 0 is a boundary point. Now can you describe the boundary and the closure?
 
  • #19
Then the closure would be the union of the set and the boundary.

The boundary is the set of 1/n with n from Z and 0

The set is well like above the set of 1/n where n is fro Z

That means the closure is essentially 1/n from Z and 0...i.e. just the boundary points.
 
  • #20
trap101 said:
Then the closure would be the union of the set and the boundary.

The boundary is the set of 1/n with n from Z and 0

The set is well like above the set of 1/n where n is fro Z

That means the closure is essentially 1/n from Z and 0...i.e. just the boundary points.

Yes. The closure and the boundary are both the same. {1/n: n a positive integer} U {0}. Another BTW, Z is the symbol usually used to denote ALL integers, not just the positive ones. N is usual for the positive ones.
 
  • #21
Yes. I apologize for that. I've had to reply to this thread over my phone so i haven't been granted the luxury of all the notation a d since my latex sucks...ive been left to writing thr long winded paragraphs.

So i have a more generL question related to this.
So i have a set A, then there exists an interior of A, a boundary of A, and a closure of A.

Here is my confusion: how can an element be in the set A but not in either the interior or the boindary of A? What part of the set is left for an element to be a part of?
 
  • #22
trap101 said:
Yes. I apologize for that. I've had to reply to this thread over my phone so i haven't been granted the luxury of all the notation a d since my latex sucks...ive been left to writing thr long winded paragraphs.

So i have a more generL question related to this.
So i have a set A, then there exists an interior of A, a boundary of A, and a closure of A.

Here is my confusion: how can an element be in the set A but not in either the interior or the boindary of A? What part of the set is left for an element to be a part of?

Look at the definitions. A point in A has to be either in the interior of A or the boundary of A. But the boundary can also contain points that aren't in A.
 
  • #23
I forgot to say thanks for the help Dick.
 
  • #24
trap101 said:
I forgot to say thanks for the help Dick.

Thanks! You're welcome!
 

Related to Topology of the Set {1/n} for Positive Integers Z

1. What is a simple topological argument?

A simple topological argument is a logical reasoning method that uses concepts from topology, a branch of mathematics that studies the properties of geometric shapes and spaces. It involves breaking down a complex problem into simpler components and then using topological principles to prove a statement or solve a problem.

2. How does a simple topological argument work?

A simple topological argument works by using basic topological properties such as continuity, compactness, and connectedness to prove a statement. It involves breaking down a problem into smaller components, identifying the relevant topological properties, and then using them to logically deduce the solution.

3. What are some examples of simple topological arguments?

Some examples of simple topological arguments include the Jordan Curve Theorem, which states that a continuous closed loop in a plane divides the plane into an inside and an outside region, and the Brouwer Fixed Point Theorem, which states that any continuous function from a closed disk to itself must have at least one fixed point.

4. What are the limitations of a simple topological argument?

Like any mathematical proof, a simple topological argument is subject to certain assumptions and limitations. For example, it may only be applicable to certain types of spaces or may require certain conditions to be met. Additionally, the application of topological principles may not always lead to a definitive solution.

5. How is a simple topological argument useful in scientific research?

Simple topological arguments are useful in scientific research because they provide a logical and rigorous framework for solving problems and proving statements. They can be applied to a wide range of scientific fields, such as physics, biology, and computer science, to analyze complex systems and phenomena. They also offer a way to visualize and understand abstract concepts in a concrete and intuitive manner.

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