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## 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.

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