# Prove: If x is an accumulation point of the set S, then every neighborhood of x contains infinitely many points of S.

### 1 Answer | Add Yours

This demonstration consists of two parts, both being based on contradiction principle such that:

1) If x is an accumulation point of the set A,B is the neighborhood of x that consists of a finite number of points and x results from intersection of both sets A and B such that:

A `nn` B = x

If you remove x, then you will remove the neighborhood, hence, there are no common points in A and B.

But the absence of common points leads to a contradiction of definition of accumulation point x.

2) You need to consider that in B are a finite number of points from A. Considering the closest point to x at the distance `epsilon > ` 0, then in the neighborhood B `(0, epsilon/2)` there are no points from A, contradicting the definition of accumulation point.

**Hence, since 1) and 2) were based on contradiction principle, thus, the neighborhood of x consists of an infinit number of points.**