Boundary of {1/n : n is a natural number}

In summary, the conversation discusses finding the boundary of the set of all 1/n, where n is a natural number, in the context of a subset of R with its usual metric. Two definitions of the boundary of a set are discussed and it is shown that using either definition results in different solutions. After realizing an error in the second definition, it is determined that all points in A are boundary points except for one point that is not in A. The conversation concludes with a question asking for the identity of this point.
  • #1
fluxions
51
0

Homework Statement


I'm trying to figure out the the boundary of the set of all 1/n, where n is a natural number.
Consider this as a subset of R with its usual metric, nothing fancy.

Homework Equations


There are at least two "equivalent" definitions of the boundary of a set:
1. the boundary of a set A is the intersection of the closure of A and the closure of the complement of A.
2. the boundary of a set A is the set of all elements x of R (in this case) such that every neighborhood of x contains at least one point in A and one point not in A.

The Attempt at a Solution

\
The problem is that I get a different solution depending on which definition I use. Please help me find my error!

Let A = {1/n : n is a natural number}.

If I use the definition 1 above:
- closure of A = {0} union A (since 0 is the only accumulation point of A)
- complement of A = (-infty,0] union [1,infty) union ( (0,1] - {1/n: n is a natural number} )
- closure of the complement of A = (infty, 0] union [1, infty)
union ( (0,1] - {1/n: n is a natural number} )
(there are no accumulation points for the complement of A in the interval (0,1] )
- hence the boundary of A is {0,1}

However! using definition 2 it is clear that:
- every neighborhood of 0 contains a point of A (Archimedean property) and a point not in A
- if x is an element of A then every neighborhood of x contains a point contains a point of A (namely x) and a point not in A.
- hence the boundary of A is {0} union A.
 
Last edited:
Physics news on Phys.org
  • #2
fluxions said:
- closure of the complement of A = (infty, 0] union [1, infty)
union ( (0,1] - {1/n: n is a natural number} )
(there are no accumulation points for the complement of A in the interval (0,1] )
- hence the boundary of A is {0,1}

This sure doesn't sound right. Why isn't for example 1/2 an accumulation point for A complement?
 
  • #3
clamtrox said:
This sure doesn't sound right. Why isn't for example 1/2 an accumulation point for A complement?

Well, it most certainly is an accumulation point for A complement. Indeed, every member of A is an accumulation point for A complement. I don't know what I was thinking. Thanks!
 
  • #4
So A= {1, 1/2, 1/3, 1/4, ...}. It should be obvious that, around each point in A is possible to construct a neighborhood with small enough radius (less than the distance to the next number in the sequence) that does not contain any other members of A. Thus, every point in A is a "boundary point". But there is one point [/b]not[/b] in A that is a boundary point of A. Can you think what that is?
 

Related to Boundary of {1/n : n is a natural number}

1. What is the meaning of "Boundary of {1/n : n is a natural number}"?

The boundary of {1/n : n is a natural number} refers to the set of numbers that are the closest to the set {1/n : n is a natural number} but are not included in the set itself.

2. How is the boundary of {1/n : n is a natural number} calculated?

The boundary of {1/n : n is a natural number} is calculated by finding the limit of the sequence {1/n} as n approaches infinity.

3. Does the boundary of {1/n : n is a natural number} include any numbers?

No, the boundary of {1/n : n is a natural number} does not include any numbers. It is a set of points that are infinitely close to the set {1/n : n is a natural number} but are not actually part of the set.

4. Can the boundary of {1/n : n is a natural number} be graphed?

Yes, the boundary of {1/n : n is a natural number} can be graphed on a number line. It would appear as a dotted line approaching but never touching the set {1/n : n is a natural number}.

5. How is the boundary of {1/n : n is a natural number} useful in mathematics?

The boundary of {1/n : n is a natural number} is useful in various mathematical concepts such as limits, continuity, and topology. It helps to define the "edge" or limit of a set and can aid in understanding the behavior of functions and sequences near that set.

Similar threads

  • Calculus and Beyond Homework Help
Replies
22
Views
582
  • Calculus and Beyond Homework Help
Replies
6
Views
497
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
2K
  • Differential Geometry
Replies
3
Views
318
  • Calculus and Beyond Homework Help
Replies
1
Views
640
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
650
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
597
Back
Top