Accumulation points must be interior or boundary points.

[stripes]

Homework Statement

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

Homework Equations

None

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.

Thank you in advance!

 

[mighty2000]

Your proof seems to make sense, but it could be simplified and made more clear. Here is a possible revised proof:

Suppose x is an accumulation point of a set S. Then for any neighborhood N of x, there exists a point y in S such that y is also in N (since x is an accumulation point). This means that N is not a subset of S, since it contains a point outside of S. Therefore, x cannot be an interior point of S.

Now suppose x is not a boundary point of S. This means that there exists a neighborhood N' of x such that either N' intersects S or N' intersects the complement of S (ℝ\S) in an empty set. But since x is an accumulation point, all deleted neighborhoods N'* around x must intersect S, meaning that N'* ∩ S is nonempty. Therefore, N' intersects S and not the complement of S, so x must be a boundary point of S.

Thus, we have shown that an accumulation point of a set S is either an interior point or a boundary point of S. ♦