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

[fluxions]

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

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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.

 

[clamtrox]

- 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?

 

[fluxions]

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!

 

[HallsofIvy]

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?