If D = {1/n : n belongs to Natural Numbers} then Why D' = R - D is not also the neighborhood of 0?

Question

If D = {1/n : n belongs to Natural Numbers} Then while finding the exterior of D why D' = R - D is a neighborhood of each of it's points except 0. Why D' = R - D is not also the neighborhood of 0.

Answer 1

In order for D' = R - D to contain a neighborhood of 0, it would by definition have to contain an entire interval

{ x | epsilon1 < x and x < epsilon2 }

where epsilon1 < 0 and 0 < epsilon2.

But every such interval intersects with D and so cannot be completely within D'. For "M" sufficiently large all of the 1/n for n > M will be in (0, epsilon2).

Dan

Answer 2

Because any open interval containing 0 will also contain an element of D.

Let A be an open interval containing 0 and let r > 0 ∈ A (we know such an r must exist). Then let N be any natural number greater than 1/r; it follows that 1/N is an element of D between 0 and r, so it is in A also.