Let $a \in A$. Prove that $a$ is an isolated point of $A$ if and only if there exists an $\epsilon$-neighborhood $V_\epsilon(a)$ such that $V_\epsilon(x) \cap A = \{a\}$.
I am not understanding if $V_\epsilon(x)$ is different from $V_\epsilon(a)$. Because $V_\epsilon(a)$ will always contain $a$. I am totally confused by this.
So if it is an typo, the definition of an isolated point is that isn't a limit point. So that means that we can find a neighborhood that doesn't contain any other elements from $A$. Hence the intersection $V_\epsilon(a) \cap A = {a}$