# Set Theory Proof

dgonnella89

## Homework Statement

I have been asked to prove the following statement:
An accumulation point of a set S is either an interior point or a boundary point of S.

Here is my attempt at a solution:
I started from the definition of accumulation points:
A point is an accumulation point ($$x_0$$) when $$\forall\epsilon > 0, N_{\epsilon}'(x_0)\capA\neq$$empty set

By using the definition of a deleted neighborhood:
$$(N_\epsilon(x_0)-\{x_0\})\capA\neq$$empty set

so $$\forall a \in S' a\inN_\epsilon(x_0), a \notin\{x_0\}, a \inA$$

and that's as far as I got. I'm not sure where to go from here.

## Answers and Replies

aPhilosopher
I'm not sure of all your notation but what would happen if $$x_{0}$$ were not contained in $$S$$ or its boundary? Specifically, what would the intersections of sufficiently small neighborhoods of $$x_{0}$$ with $$S$$ be?

Elucidus
It appears you are speaking of relative deleted neighborhoods. I.e. $$N'_\epsilon(x_0) = \{x \in S-\{x_0\}: |x-x_0| < \epsilon \}$$.

The last line in the original post unfortunately is not syntactically well formed. What was the intended meaning? Was it close to:

$$\forall a \in S' a \in N_\epsilon(x_0), a \notin \{x_0\}, a \in A$$? (Modified from OP)

Recommendation: Attack this by contradiction. You are asked to show that an accumulation point of S must either lie in the interior of S or on its boundary. The only other possible place it could be is in its exterior (the complement of the closure of S). I'd start with:

"Assume otherwise. Let x be an accumulation point of S that lies in the exterior of S..." and show this leads to contradiction. (Hint: think closed vs. open sets).

--Elucidus