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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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