- #1

stripes

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## 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 [itex]\in[/itex] S AND x [itex]\notin[/itex] 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.