Isolated points are equal to boundary points

[stripes]

Homework Statement

Prove: If x is an isolated point of a set S, then x is an element of bd S.

I believe this could be reworded as, if x is an isolated point of S, then x is also a boundary point of S.

Homework Equations

None

The Attempt at a Solution

My book says that an isolated point of S is a point that is NOT an accumulation point AND is itself a member of the set S. So x is an isolated point of S if x \in S AND x \notin S', where S' is the set of all accumulation points of S.

But I also know (or think) that a boundary point does not necessarily have to be in the set itself. I know what I've said so far might be confusing, so let me give you an example:

Let S = (1, 2), we have bd S = {1, 2}. Does this mean that the open set (1, 2) does not have any isolated points, 1 and 2 are NOT isolated points (due to the fact that they are NOT in S)?

An accumulation point of S is a point x where every deleted neighborhood N of that point x, that is, N\{x}, contains an element from S. That means that all interior points are accumulation points, and that also means all boundary points are accumulation points, since all boundary points' deleted neighborhoods will always contain an element in S. How can all boundary points of a set be accumulation points AND be isolation points, when a requirement of an isolation point is in fact NOT being an accumulation point?

I'm certain that this "conjecture" is in fact true for all nonempty subsets S of R, but from my understanding of each of these definitions, it cannot be true.

If someone could perhaps guide me in the right direction in a proof or show me where my definitions are wrong, I would really appreciate it. Thank you all so much in advance.

 

[Eynstone]

That means that all interior points are accumulation points, and that also means all boundary points are accumulation points, since all boundary points' deleted neighborhoods will always contain an element in S.

No, a boundary point may not be an accumulation point.Since an isolated point has a neighbourhood containing no other points of the set, it's not an interior point. The boundary is, by definition , A\intA & hence an isolated point is regarded as a boundary point.

 

[stripes]

Okay, so shall i start off by saying that a set S can only have isolated points iff the set S is closed? I can show that if a set is open, the bd points are not contained in the set, thus by definition cannot be isolated points. The open set has no isolated points since all other points (interior points) are accumulation points.

Then I need to show that given a closed interval/set, its boundary points cannot be accumulation points. THEN, it is obvious that boundary points of a closed set are included in the set itself. Thus the bd points are isolated points. Is that kind of correct?

 

[stripes]

But boundary points of a closed set CAN be accumulation points, since I CAN find a deleted neighborhood of a boundary point such that N*(x; epsilon) intersected with S is NOT nonempty...

 

[stripes]

This thread will get lost in the pile if I don't bumpity bump bump. I am so confused and my assignment is due soon.

 

[Trevor Vadas]

I believe you need to loof at the definitions.
A boundary point of S is a point such that every open set containing the point intersects S and the compliment of S.
An isolated point x in S is a point such that there exists an open set containing x that does not intersect another point in S.
Think then about any open set that contains an isolation point.