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 $\displaystyle \frac{1}{n}< \epsilon$ so there exist some points in the set in every neighborhood of 0.