Can A be a limit point if there exists an epsilon neighborhood

[kathrynag]

Homework Statement

Let a be an element of A. Prove that A is an isolated point of A iff there exists an epsilon neighborhood V(a) such that V(a)$$\cap$$A={a}

Homework Equations

The Attempt at a Solution

A point is an isolated point if it is not a limit point.
Let a be an element of A.
Let be an isolated point. We want to show V(a)$$\cap$$A={a}.
Since a is not a limit point, we say x=lim$$a_{n}$$ satisfying $$a_{n}$$=x

 

[jbunniii]

Try proving this equivalent statement:

A is a limit point iff there does not exist an epsilon neighborhood $V(a)$ such that $V(a) \cap A = \{a\}$.

Notice that the following mean the same thing:

"there does not exist an epsilon neighborhood $V(a)$ such that $V(a) \cap A = \{a\}$"

"every epsilon neighborhood $V(a)$ contains at least one point of $A$ that is distinct from $a$."