# Accumulation points must be interior or boundary points.

1. Dec 4, 2011

### stripes

1. The problem statement, all variables and given/known data

Prove the following: an accumulation point of a set S is either an interior point of S or a boundary point of S.

2. Relevant equations

None

3. The attempt at a solution

Suppose x is not an interior point. Then you cannot find a neighborhood around x such that N is a subset of S. Suppose further that x is not a boundary point, then there exists a neighborhood N' around x such that N' intersecting S is empty OR N' intersecting ℝ\S is empty. But recall that if x is an accumulation point, all deleted neighborhoods N'* around x contain a point in S, which means N'* intersecting S is nonempty. So indeed, N' intersecting S is nonempty. So we must have that N' intersecting ℝ\S is empty. But this means that N' must be a subset of S, which contradicts our statement earlier, saying we cannot find a neighborhood N around x such that N is a subset of S. Thus, if x is not an interior point and x is an accumulation point, x is a boundary point.

If x is not a boundary point, then there exists a neighborhood N around x such that N intersecting S is empty OR N intersecting ℝ\S is empty. But since x is an accumulation point, for all deleted neighborhoods N* around x, there exists a point in both S and N*. So N* intersecting S is nonempty, thus N intersecting S must be nonempty. So we must have that N intersecting ℝ\S is empty. Suppose further that x is not an interior point. Then there does not exist any neighborhood N' around x such that N is a subset of S.If N is never a subset of S, then N intersecting with ℝ\S is nonempty, a contradiction. Thus, if x is not a boundary point and x is an accumulation point, x is an interior point. ♦

Not sure if my proof makes sense or not...it's so long that even i've gotten lost along the way. Any help would be greatly appreciated.