Originally Posted by billym
Boundary points:

If I have a subset: {1/2, 1/3, 1/4, ... 1/n}, then all of its points are boundary points because none of its elements are interior points of the subset(because they all have nbhd's that are not in the subset), nor are they interior points of the subset's complement (obviously).

Correct?
Yes, in fact, for [b]any[b] "discrete" set such as this (there are no intervals) every member of the set is a boundary point. It is also true, in this case, that 0 is a boundary point. Every neighborhood of 0 contains 0, of course, so every neighborhood contains a point not in the set. But given any [tex]\epsilon> 0[/itex], there exist n such that $\frac{1}{n}< \epsilon$ so there exist some points in the set in every neighborhood of 0.